A second-order differential equation has $y''$ as its highest derivative; it may also contain $y'$ and $y$. For a linear equation, the normalized form is

$$y''+p(t)y'+q(t)y=g(t).$$

Here $p$, $q$, and $g$ are known functions of the independent variable $t$. The equation is homogeneous when $g(t)=0$ throughout the interval under consideration:

$$y''+p(t)y'+q(t)y=0.$$

If $g$ is not identically zero, the equation is nonhomogeneous. The function $g$ is the input or forcing term.

Because the equation is second order, an initial-value problem normally needs two pieces of information at the same time $t_0$:

$$y(t_0)=y_0, \qquad y'(t_0)=y'_0.$$

The first condition fixes the starting value. The second fixes the starting slope. We need two independent solutions to accommodate both conditions; later, the Wronskian will tell us whether a chosen pair does that.

Homogeneous equations with constant coefficients

Start with

$$ay''+by'+cy=0,$$

where $a$, $b$, and $c$ are constants. We assume $a\ne0$ so that this is genuinely a second-order equation. If $a=0$ but $b\ne0$, it reduces to the first-order equation

$$y'=-\frac cb y,$$

whose solutions are $y(t)=Ce^{-(c/b)t}$.

Why exponential trial functions work

The equation balances a function's value, slope, and second derivative. For $a\ne0$, we want a function whose derivatives keep the same shape, so all three terms can share a common factor. Exponentials do exactly that. Try

$$y(t)=e^{rt},$$

where $r$ is a constant that we still need to find. Exponentials are useful here because differentiating them only introduces powers of $r$:

$$y'(t)=re^{rt}, \qquad y''(t)=r^2e^{rt}.$$

Put these expressions into the differential equation:

$$ar^2e^{rt}+bre^{rt}+ce^{rt}=0.$$

Factor out the common exponential:

$$e^{rt}\bigl(ar^2+br+c\bigr)=0.$$

The exponential $e^{rt}$ is never zero, so the polynomial factor must vanish:

$$\boxed{ar^2+br+c=0}.$$

The polynomial

$$P(r)=ar^2+br+c$$

is the characteristic polynomial, and $P(r)=0$ is the characteristic equation. Each real root gives a real exponential solution. Repeated or non-real roots require an extra step, so the first example uses two distinct real roots.

Example: two distinct real roots

Consider

$$y''+5y'+6y=0.$$

Its characteristic equation is

$$r^2+5r+6=0.$$

Factor it:

$$(r+2)(r+3)=0.$$

The roots are

$$r_1=-2, \qquad r_2=-3,$$

so we get two solutions:

$$y_1(t)=e^{-2t}, \qquad y_2(t)=e^{-3t}.$$

The principle of superposition

For the homogeneous equation

$$y''+p(t)y'+q(t)y=0,$$

any linear combination of solutions is another solution. This is the principle of superposition: if $y_1$ and $y_2$ solve the equation, then

$$y(t)=C_1y_1(t)+C_2y_2(t)$$

also solves it for every $C_1,C_2\in\mathbb R$.

The reason is the linearity of differentiation. Define the differential operator

$$L[y]=y''+p(t)y'+q(t)y.$$

Then

$$\begin{aligned} L[C_1y_1+C_2y_2] &=C_1L[y_1]+C_2L[y_2]\\ &=C_1(0)+C_2(0)\\ &=0. \end{aligned}$$

This argument depends on the right side being zero. If $L[y_1]=g$ and $L[y_2]=g$, then

$$L[C_1y_1+C_2y_2]=(C_1+C_2)g,$$

which is not generally equal to $g$. Arbitrary superposition is therefore a property of the associated homogeneous equation, not of a nonhomogeneous equation with fixed forcing.

A complete initial-value problem

Return to

$$y''+5y'+6y=0.$$

Superposition gives the family

$$y(t)=C_1e^{-2t}+C_2e^{-3t}.$$

Now impose

$$y(0)=2, \qquad y'(0)=3.$$

Evaluating the solution and its derivative at $t=0$ gives

$$C_1+C_2=2,$$

and

$$-2C_1-3C_2=3.$$

The same system can be written as

$$\begin{pmatrix} 1&1\\ -2&-3 \end{pmatrix} \begin{pmatrix} C_1\\ C_2 \end{pmatrix} = \begin{pmatrix} 2\\ 3 \end{pmatrix}.$$

Its determinant is $-1$, so the system has one solution:

$$C_1=9, \qquad C_2=-7.$$

Therefore the unique solution of the IVP is

$$\boxed{y(t)=9e^{-2t}-7e^{-3t}}.$$

A quick check confirms both initial conditions:

$$\begin{aligned} y(0)&=9-7=2,\\ y'(t)&=-18e^{-2t}+21e^{-3t},\\ y'(0)&=-18+21=3. \end{aligned}$$

The initial slope lifts the solution before both modes decay

The positive initial slope explains why the curve rises at first. Both exponential terms decay as $t$ increases, so the solution eventually approaches zero.

Existence and uniqueness on an interval

Consider the linear IVP

$$\begin{aligned} y''+p(t)y'+q(t)y&=g(t),\\ y(t_0)&=y_0,\\ y'(t_0)&=y'_0. \end{aligned}$$

Let $I$ be an open interval with $t_0\in I$. If $p$, $q$, and $g$ are continuous on the interval $I$, then the IVP has exactly one solution on the entire interval $I$:

$$y:I\to\mathbb R.$$

This is stronger than a merely local statement. The solution does not stop at an interior point where the normalized coefficients remain continuous. The interval ends only where the chosen coefficient interval ends; it need not be all of $\mathbb R$.

If an equation is initially written as

$$a_2(t)y''+a_1(t)y'+a_0(t)y=h(t),$$

first divide by $a_2(t)$. The normalized coefficients are

$$p(t)=\frac{a_1(t)}{a_2(t)}, \qquad q(t)=\frac{a_0(t)}{a_2(t)}, \qquad g(t)=\frac{h(t)}{a_2(t)}.$$

The theorem applies on an interval where $a_2(t)\ne0$ and these three quotients are continuous. A zero of $a_2$ can split the real line into separate coefficient intervals.

The Wronskian and fundamental solutions

The Wronskian answers a concrete question: do two solutions provide enough freedom to match any starting value and slope? Given two differentiable functions $f$ and $g$, their Wronskian is

$$\begin{aligned} W[f,g](t) &=\det \begin{pmatrix} f(t)&g(t)\\ f'(t)&g'(t) \end{pmatrix}\\ &=f(t)g'(t)-f'(t)g(t). \end{aligned}$$

The Wronskian tells us whether two solution functions provide independent initial data. Suppose $y_1$ and $y_2$ solve the same homogeneous second-order linear equation on $I$, with $p$ and $q$ continuous there. To make

$$y=C_1y_1+C_2y_2$$

satisfy prescribed values at $t_0$, the constants must solve

$$\begin{pmatrix} y_1(t_0)&y_2(t_0)\\ y_1'(t_0)&y_2'(t_0) \end{pmatrix} \begin{pmatrix} C_1\\ C_2 \end{pmatrix} = \begin{pmatrix} y_0\\ y'_0 \end{pmatrix}.$$

The determinant of the coefficient matrix is exactly

$$W[y_1,y_2](t_0).$$

The matrix can match every pair of initial values if and only if

$$\boxed{W[y_1,y_2](t_0)\ne0}.$$

When this condition holds, the matrix is invertible, so every pair of initial values determines one pair of constants $C_1,C_2$. The existence-and-uniqueness theorem then shows that this linear combination is the only solution with those initial values.

In that case, $\{y_1,y_2\}$ is called a fundamental set of solutions on $I$, and

$$\boxed{y(t)=C_1y_1(t)+C_2y_2(t)}$$

is the general solution of the homogeneous equation: every solution on $I$ appears exactly once for a choice of $C_1$ and $C_2$.

For the example $y_1=e^{-2t}$ and $y_2=e^{-3t}$,

$$\begin{aligned} W[y_1,y_2](t) &=e^{-2t}\bigl(-3e^{-3t}\bigr) -\bigl(-2e^{-2t}\bigr)e^{-3t}\\ &=-e^{-5t}. \end{aligned}$$

This never vanishes, so the two exponentials form a fundamental set of solutions on $\mathbb R$.

Distinct real roots in general

Assume $a,b,c\in\mathbb R$ and $a\ne0$. The quadratic formula gives

$$r=\frac{-b\pm\sqrt{\Delta}}{2a}, \qquad \Delta=b^2-4ac,$$

where $\Delta$, read “delta,” is the discriminant. Its sign separates the three possible root patterns:

• $\Delta>0$: two distinct real roots;
• $\Delta<0$: two complex-conjugate roots;
• $\Delta=0$: one repeated real root.

For now, consider the first case. Let $r_1\ne r_2$ be the two real roots. Then

$$y_1(t)=e^{r_1t}, \qquad y_2(t)=e^{r_2t}$$

are solutions. Their Wronskian is

$$\begin{aligned} W[y_1,y_2](t) &= \begin{vmatrix} e^{r_1t}&e^{r_2t}\\ r_1e^{r_1t}&r_2e^{r_2t} \end{vmatrix}\\ &=(r_2-r_1)e^{(r_1+r_2)t}. \end{aligned}$$

The exponential is never zero, and $r_2-r_1\ne0$. Therefore the Wronskian never vanishes, so the two exponentials form a fundamental set. Every real solution is

$$\boxed{y(t)=C_1e^{r_1t}+C_2e^{r_2t}}.$$

Applying the distinct-root formula

Example: two opposite real roots

Consider the initial-value problem

$$y''-y=0, \qquad y(0)=1, \qquad y'(0)=-1.$$

The characteristic equation factors as

$$r^2-1=(r-1)(r+1)=0,$$

so the roots are $-1$ and $1$. Choose

$$y_1(t)=e^{-t}, \qquad y_2(t)=e^t.$$

Their Wronskian is

$$W[y_1,y_2](t) = \begin{vmatrix} e^{-t}&e^t\\ -e^{-t}&e^t \end{vmatrix} =2,$$

so the general solution is

$$y(t)=C_1e^{-t}+C_2e^t.$$

The initial conditions give

$$C_1+C_2=1, \qquad -C_1+C_2=-1.$$

Adding the equations gives $C_2=0$, and then $C_1=1$. Thus the unique solution is

$$\boxed{y(t)=e^{-t}}.$$

Complex characteristic roots

Now suppose $\Delta<0$. The quadratic formula can be written as

$$r= \frac{-b}{2a} \pm \frac{\sqrt{-\Delta}}{2a}i =\lambda\pm i\mu,$$

where $i$ is the imaginary unit, $i^2=-1$, and

$$\lambda=-\frac{b}{2a}, \qquad \mu=\frac{\sqrt{-\Delta}}{2a}\ne0.$$

Both $\lambda$ and $\mu$ are real. The two roots are complex conjugates: they have the same real part and opposite imaginary parts.

Euler's formula

To turn the complex exponentials into real functions, start with the Taylor series

$$e^z=\sum_{k=0}^{\infty}\frac{z^k}{k!}, \qquad z\in\mathbb C.$$

Here $\mathbb C$ is the set of complex numbers and $k!$ is the factorial of the nonnegative integer $k$. Put $z=ix$ for a real number $x$. Using $i^2=-1$, $i^3=-i$, and $i^4=1$ gives

$$\begin{aligned} e^{ix} &=1+ix-\frac{x^2}{2!}-i\frac{x^3}{3!} +\frac{x^4}{4!}+i\frac{x^5}{5!}-\cdots\\ &=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\right) +i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right). \end{aligned}$$

The first parenthesis is the Taylor series for cosine, and the second is the series for sine. Therefore

$$\boxed{e^{ix}=\cos x+i\sin x}.$$

This is Euler's formula.

Turning one complex solution into two real solutions

Use Euler's formula with $x=\mu t$:

$$\begin{aligned} e^{(\lambda+i\mu)t} &=e^{\lambda t}e^{i\mu t}\\ &=e^{\lambda t}\bigl(\cos(\mu t)+i\sin(\mu t)\bigr)\\ &=e^{\lambda t}\cos(\mu t) +i\,e^{\lambda t}\sin(\mu t). \end{aligned}$$

The real and imaginary parts are the real-valued functions:

$$y_1(t)=e^{\lambda t}\cos(\mu t), \qquad y_2(t)=e^{\lambda t}\sin(\mu t).$$

To see why each part solves the ODE, define

$$L[y]=ay''+by'+cy,$$

where $a$, $b$, and $c$ are real. If $y=y_1+iy_2$ solves the equation, linearity gives

$$0=L[y_1+iy_2]=L[y_1]+iL[y_2].$$

A complex number is zero only when both its real and imaginary parts are zero. Hence

$$L[y_1]=0, \qquad L[y_2]=0.$$

So the real and imaginary parts are each real-valued solutions of the original equation.

Their Wronskian is

$$\begin{aligned} W[y_1,y_2](t) &= \begin{vmatrix} e^{\lambda t}\cos(\mu t)&e^{\lambda t}\sin(\mu t)\\ e^{\lambda t}\bigl(\lambda\cos(\mu t)-\mu\sin(\mu t)\bigr)& e^{\lambda t}\bigl(\lambda\sin(\mu t)+\mu\cos(\mu t)\bigr) \end{vmatrix}\\ &=\mu e^{2\lambda t} \bigl(\cos^2(\mu t)+\sin^2(\mu t)\bigr)\\ &=\mu e^{2\lambda t}\ne0. \end{aligned}$$

Thus the two real functions form a fundamental set, and the real general solution is

$$\boxed{ y(t)=C_1e^{\lambda t}\cos(\mu t) +C_2e^{\lambda t}\sin(\mu t) }.$$

Applying the complex-root formula

Example: a complex-root initial-value problem

Solve

$$y''-2y'+5y=0, \qquad y(0)=1, \qquad y'(0)=0.$$

The characteristic equation is

$$r^2-2r+5=0,$$

and the quadratic formula gives

$$r=\frac{2\pm\sqrt{4-20}}{2}=1\pm2i.$$

Therefore $\lambda=1$, $\mu=2$, and

$$y(t)=C_1e^t\cos(2t)+C_2e^t\sin(2t).$$

Differentiate before applying the slope condition:

$$\begin{aligned} y'(t) &=C_1e^t\bigl(\cos(2t)-2\sin(2t)\bigr)\\ &\quad+C_2e^t\bigl(\sin(2t)+2\cos(2t)\bigr). \end{aligned}$$

At $t=0$,

$$y(0)=C_1=1, \qquad y'(0)=C_1+2C_2=0.$$

Hence $C_2=-1/2$, and the unique solution is

$$\boxed{ y(t)=e^t\cos(2t)-\frac12e^t\sin(2t) }.$$

Repeated roots

If the discriminant is zero, the characteristic equation has only one root, counted twice:

$$r=-\frac{b}{2a}, \qquad b^2-4ac=0.$$

The exponential $y_1(t)=e^{rt}$ is still a solution. But writing $C_1e^{rt}+C_2e^{rt}$ does not give two independent solutions: it is just $(C_1+C_2)e^{rt}$. We need a second function that is not a constant multiple of the first.

For example,

$$y''-2y'+y=0$$

has characteristic equation $(r-1)^2=0$, so $y_1(t)=e^t$. To find the missing solution, we keep the exponential and let its multiplier vary with time.

Constructing the second solution

For the general repeated root $r=-b/(2a)$, look for a solution of the form

$$y_2(t)=v(t)y_1(t),$$

where $y_1(t)=e^{rt}$ is known and $v(t)$ is unknown. This is the idea of reduction of order: use one known solution to turn the search for another into a first-order problem.

Apply the product rule:

$$\begin{aligned} y_2'&=v'y_1+vy_1',\\ y_2''&=v''y_1+2v'y_1'+vy_1''. \end{aligned}$$

Putting these into $ay_2''+by_2'+cy_2=0$ and grouping by derivatives of $v$ gives

$$ay_1v''+(2ay_1'+by_1)v' +(ay_1''+by_1'+cy_1)v=0.$$

The coefficient of $v$ is zero because $y_1$ already solves the equation. What remains is

$$ay_1v''+(2ay_1'+by_1)v'=0.$$

There is no undifferentiated $v$ in this equation. Setting $w=v'$ makes it first order:

$$ay_1w'+(2ay_1'+by_1)w=0.$$

For a repeated root, it simplifies even further. Since $y_1'=ry_1$ and $2ar+b=0$,

$$2ay_1'+by_1=(2ar+b)y_1=0.$$

Thus $ay_1w'=0$. Both $a$ and the exponential $y_1$ are nonzero, so $w'=0$. Integrating twice gives

$$w(t)=K, \qquad v(t)=Kt+D,$$

where $K$ and $D$ are constants. Therefore

$$v(t)y_1(t)=Kte^{rt}+De^{rt}.$$

The term $De^{rt}$ is a multiple of the solution we already have. Choosing $D=0$ and $K=1$ leaves the new solution

$$\boxed{y_2(t)=te^{rt}}.$$

Its Wronskian with $y_1$ is

$$\begin{aligned} W[y_1,y_2](t) &=\begin{vmatrix} e^{rt}&te^{rt}\\ re^{rt}&e^{rt}+rte^{rt} \end{vmatrix}\\ &=e^{rt}(e^{rt}+rte^{rt})-re^{rt}te^{rt}\\ &=e^{2rt}\ne0. \end{aligned}$$

So $\{e^{rt},te^{rt}\}$ is a fundamental set on $\mathbb R$. The general solution is

$$\boxed{y(t)=C_1e^{rt}+C_2te^{rt}=(C_1+C_2t)e^{rt}}.$$

In particular, the equation $y''-2y'+y=0$ has general solution $y(t)=(C_1+C_2t)e^t$.

Example: a repeated-root initial-value problem

Solve

$$y''-y'+\frac14y=0, \qquad y(0)=2, \qquad y'(0)=\frac13.$$

The characteristic polynomial factors as

$$r^2-r+\frac14=\left(r-\frac12\right)^2,$$

so the repeated root is $r=1/2$. The general solution and its derivative are

$$\begin{aligned} y(t)&=C_1e^{t/2}+C_2te^{t/2},\\ y'(t)&=\frac{C_1}{2}e^{t/2} +C_2e^{t/2}+\frac{C_2}{2}te^{t/2}. \end{aligned}$$

The starting value determines $C_1$; then the slope determines $C_2$:

$$C_1=2, \qquad \frac{C_1}{2}+C_2=\frac13, \qquad C_2=-\frac23.$$

Hence the unique solution is

$$\boxed{y(t)=\left(2-\frac23t\right)e^{t/2}}.$$

Although $e^{t/2}$ grows, the factor in front changes sign. The derivative makes the behavior clear:

$$y'(t)=\frac{1-t}{3}e^{t/2}.$$

For $t\ge0$, the solution rises until $t=1$, then falls. It crosses zero at $t=3$.

If we keep $y(0)=2$ but change the initial slope to $y'(0)=2$, the constants instead satisfy $C_1=2$ and $1+C_2=2$. This gives

$$y(t)=(2+t)e^{t/2},$$

which keeps increasing for $t\ge0$. Both curves solve the same ODE and start at the same point; the second initial condition determines which one we get.

$y'(0)=\frac13$$y=\left(2-\frac23t\right)e^{t/2}$

$y'(0)=2$$y=(2+t)e^{t/2}$

Choosing the right pair of solutions

For $ay''+by'+cy=0$ with real constant coefficients and $a\ne0$, the three characteristic-root cases give these fundamental sets:

In each case, combine the two functions as $y=C_1y_1+C_2y_2$, then use the initial value and initial slope to find the constants.

Reduction of order with variable coefficients

The same substitution works even when the coefficients are not constant. Start with

$$y''+p(t)y'+q(t)y=0,$$

where $p$ and $q$ are continuous on an open interval $I$. Suppose a solution $y_1$ is known. Work on an interval where $y_1(t)\ne0$, since we will divide by it. If the known solution has a zero, use a smaller interval that avoids that zero rather than dividing through it.

Again, set $y_2=vy_1$. The product-rule derivatives above give

$$y_1v''+\bigl(2y_1'+p(t)y_1\bigr)v' +\bigl(y_1''+p(t)y_1'+q(t)y_1\bigr)v=0.$$

The last bracket vanishes because $y_1$ solves the original equation. This leaves

$$y_1v''+\bigl(2y_1'+p(t)y_1\bigr)v'=0.$$

With $w=v'$, divide by $y_1$ to obtain

$$\boxed{w'+\left(2\frac{y_1'}{y_1}+p(t)\right)w=0}.$$

This is a first-order linear equation in $w$, and it is also separable. Unlike the repeated-root case, the coefficient of $w$ need not vanish.

Solve this equation for $w$, integrate $v'=w$, and multiply the result by $y_1$. A constant added to $v$ only adds a multiple of $y_1$, so it can be set to zero when choosing one second solution. The choice $w=0$ also gives only a multiple of $y_1$; to find an independent solution, choose a nonzero solution for $w$.

Example: finding a second power-function solution

Consider

$$y''+\frac{3}{2t}y'-\frac{1}{2t^2}y=0, \qquad t>0.$$

We are given $y_1(t)=1/t$. First check that it really is a solution:

$$y_1'=-\frac{1}{t^2}, \qquad y_1''=\frac{2}{t^3},$$

so

$$y_1''+\frac{3}{2t}y_1'-\frac{1}{2t^2}y_1 =\frac{2}{t^3}-\frac{3}{2t^3}-\frac{1}{2t^3}=0.$$

Here $p(t)=3/(2t)$ and $y_1$ is nonzero on $(0,\infty)$. Substituting into the equation for $w=v'$ gives

$$\frac1t w' +\left(-\frac{2}{t^2}+\frac{3}{2t^2}\right)w=0.$$

Multiply by $t$ and simplify:

$$w'-\frac{1}{2t}w=0.$$

For the nonzero solution we want, separate variables and integrate:

$$\frac{dw}{w}=\frac{dt}{2t}, \qquad \ln|w|=\frac12\ln t+C.$$

The absolute value allows either sign of $w$. Since $t>0$, no absolute value is needed around $t$. Exponentiating and absorbing the sign into a constant $K$ gives

$$w(t)=Kt^{1/2}.$$

There is one constant at this stage. The second constant appears when we integrate $v'=w$:

$$v(t)=\frac{2K}{3}t^{3/2}+B.$$

Consequently,

$$v(t)y_1(t) =\left(\frac{2K}{3}t^{3/2}+B\right)\frac1t =\frac{2K}{3}t^{1/2}+\frac Bt.$$

The term $B/t$ is already a multiple of $y_1$. Choose $B=0$ and $K=\frac32$ to obtain the convenient second solution

$$y_2(t)=t^{1/2}.$$

Choosing any other nonzero value of $K$ would only rescale this second solution. The Wronskian confirms independence:

$$\begin{aligned} W[y_1,y_2](t) &=\begin{vmatrix} t^{-1}&t^{1/2}\\ -t^{-2}&\frac12t^{-1/2} \end{vmatrix}\\ &=\frac12t^{-3/2}+t^{-3/2}\\ &=\frac32t^{-3/2}\ne0, \qquad t>0. \end{aligned}$$

Thus $\{1/t,t^{1/2}\}$ is a fundamental set on $(0,\infty)$, and

$$\boxed{y(t)=\frac{C_1}{t}+C_2t^{1/2},\qquad t>0}.$$

The zero solution of the first-order equation for $w$ gives the $C_2=0$ part of this family, so it has not been lost. The domain restriction remains essential: the original coefficients are undefined at $t=0$.

Nonhomogeneous equations

What changes when the right-hand side is not zero? Consider

$$y''+p(t)y'+q(t)y=g(t),$$

where $p$, $q$, and $g$ are continuous on an open interval $I$. Its associated homogeneous equation keeps the same left-hand side but replaces the forcing $g(t)$ with zero:

$$y''+p(t)y'+q(t)y=0.$$

Suppose $\{y_1,y_2\}$ is a fundamental set of solutions of the homogeneous equation on $I$. Its general solution, called the complementary solution, is

$$y_c(t)=C_1y_1(t)+C_2y_2(t).$$

A particular solution, written $y_p$, is any one function that solves the nonhomogeneous equation. It need not satisfy any prescribed initial conditions; its job is to produce the forcing $g(t)$ when put into the left-hand side.

Once we find one such $y_p$, the general nonhomogeneous solution is

$$\boxed{y(t)=C_1y_1(t)+C_2y_2(t)+y_p(t)=y_c(t)+y_p(t)}, \qquad C_1,C_2\in\mathbb R.$$

The constants in $y_c$ still give the freedom to match an initial value and slope. The particular solution does not add another arbitrary constant.

Why the solutions add this way

Use the linear differential operator

$$L[u]=u''+p(t)u'+q(t)u.$$

Both $y$ and $y_p$ solve $L[u]=g$. Their difference therefore solves the homogeneous equation:

$$L[y-y_p]=L[y]-L[y_p]=g-g=0.$$

Since $\{y_1,y_2\}$ is a fundamental set, this difference must be

$$y-y_p=C_1y_1+C_2y_2.$$

Rearranging gives the claimed formula. Conversely, adding any complementary solution to $y_p$ works because

$$L[y_c+y_p]=L[y_c]+L[y_p]=0+g=g.$$

This proves both that the formula gives solutions and that it includes every solution on $I$. There may be many choices for $y_p$, but any two differ by a homogeneous solution, so they give the same general family.

To solve a nonhomogeneous equation:

1. Find the complementary solution $y_c$ from the associated homogeneous equation.
2. Find one particular solution $y_p$ of the original equation.
3. Add them: $y=y_c+y_p$. If initial conditions are given, apply them to this full sum, not to $y_c$ alone.

The new work is the second step: finding $y_p$.

The method of undetermined coefficients

For an equation with constant coefficients,

$$ay''+by'+cy=g(t), \qquad a\ne0,$$

we can often choose a trial form for $y_p$ by looking at $g(t)$. The method of undetermined coefficients puts unknown constants into that trial form, then finds their values by substitution.

The standard method applies when $g(t)$ is built from polynomials, exponentials, sine, and cosine through finite sums and products. These functions are useful because their derivatives stay within a finite collection of the same kinds of terms. With constant coefficients, substituting the trial then gives algebraic equations for its unknown constants.

For the starting guesses below, $\lambda$ and $\mu$ are known real constants from the forcing, while $A$ and $B$ are the coefficients to determine:

An exponential keeps its form under differentiation. Sine and cosine turn into each other, so the trial includes both even when the forcing contains only one.

Before using a trial, check the complementary solution. If the trial already solves the homogeneous equation, substitution produces zero rather than the required nonzero forcing, so the trial must be adjusted. Neither example below has that overlap.

Unlike $C_1$ and $C_2$, the coefficients in a particular-solution trial are not left arbitrary: the equation determines them. Variation of parameters is a more general method for finding particular solutions, including in variable-coefficient problems, but it is a separate method from the coefficient matching used here.

Example: exponential forcing

Solve

$$y''-3y'-4y=3e^{2t}.$$

First find the complementary solution. The characteristic equation is

$$r^2-3r-4=(r-4)(r+1)=0,$$

so the two distinct roots are $4$ and $-1$. Thus

$$y_c(t)=C_1e^{4t}+C_2e^{-t}.$$

The forcing is a multiple of $e^{2t}$, which is not one of those homogeneous solutions. Try

$$y_p(t)=Ae^{2t}, \qquad y_p'(t)=2Ae^{2t}, \qquad y_p''(t)=4Ae^{2t}.$$

Substitution gives

$$\begin{aligned} y_p''-3y_p'-4y_p &=4Ae^{2t}-6Ae^{2t}-4Ae^{2t}\\ &=-6Ae^{2t}. \end{aligned}$$

This must equal $3e^{2t}$. Since the exponential is never zero,

$$-6A=3, \qquad A=-\frac12.$$

Therefore

$$y_p(t)=-\frac12e^{2t},$$

and the general solution is

$$\boxed{y(t)=C_1e^{4t}+C_2e^{-t}-\frac12e^{2t}}.$$

No initial conditions are given, so $C_1$ and $C_2$ remain arbitrary. All the coefficients and forcing are continuous for real $t$, and the solution family is valid on $\mathbb R$.

Example: trigonometric forcing

Now change only the forcing:

$$y''-3y'-4y=2\sin t.$$

The associated homogeneous equation has not changed, so neither has the complementary solution:

$$y_c(t)=C_1e^{4t}+C_2e^{-t}.$$

For the particular solution, include both sine and cosine:

$$\begin{aligned} y_p(t)&=A\sin t+B\cos t,\\ y_p'(t)&=A\cos t-B\sin t,\\ y_p''(t)&=-A\sin t-B\cos t. \end{aligned}$$

Put these into the ODE and group like terms:

$$\begin{aligned} y_p''-3y_p'-4y_p &=(-A\sin t-B\cos t)\\ &\quad-3(A\cos t-B\sin t)\\ &\quad-4(A\sin t+B\cos t)\\ &=(-5A+3B)\sin t+(-3A-5B)\cos t. \end{aligned}$$

We need this identity to hold for every $t$:

$$(-5A+3B)\sin t+(-3A-5B)\cos t=2\sin t.$$

Sine and cosine are linearly independent, so match their coefficients. The right-hand side has sine coefficient $2$ and cosine coefficient $0$:

$$\begin{cases} -5A+3B=2,\\ -3A-5B=0. \end{cases}$$

The second equation gives $B=-3A/5$. Put this into the first:

$$-5A-\frac95A=2 \quad\Longrightarrow\quad -\frac{34}{5}A=2.$$

Hence

$$A=-\frac5{17}, \qquad B=\frac3{17},$$

so a particular solution is

$$y_p(t)=-\frac5{17}\sin t+\frac3{17}\cos t.$$

Adding the complementary solution gives the general solution, again valid for all real $t$:

$$\boxed{ y(t)=C_1e^{4t}+C_2e^{-t} -\frac5{17}\sin t+\frac3{17}\cos t }.$$

As a check, the coefficients produce $-5A+3B=2$ and $-3A-5B=0$, so the left-hand side is exactly $2\sin t$, with no cosine term.

Choosing a complete trial

Matching the forcing means including every kind of term that differentiation can produce, not just copying the terms that happen to appear on the right-hand side.

• If $g(t)=t^2+7$, start with $y_p=At^2+Bt+C$. The forcing has no linear term, but derivatives of the quadratic can produce one, so the trial still needs $Bt$.
• If $g(t)=-8e^t\cos(2t)$, start with $y_p=e^t\bigl(A\cos(2t)+B\sin(2t)\bigr)$. Differentiation mixes the two trigonometric functions while retaining the exponential.
• If $g(t)=7te^t\sin(8t)$, start with

$$y_p=e^t\bigl[(A+Ct)\sin(8t)+(B+Dt)\cos(8t)\bigr].$$

In the last trial, each trigonometric function needs both a constant coefficient and a coefficient multiplying $t$. Differentiating $t$ creates a constant term. In all three cases, the capital letters are unknown constants, and the trial may still need an extra factor if it overlaps the complementary solution.

Splitting a sum of forcing terms

Linearity lets us find particular solutions in pieces. For the same operator $L[y]=ay''+by'+cy$, suppose

$$L[y_{p,1}]=g_1, \qquad L[y_{p,2}]=g_2.$$

The subscripts label two particular solutions, one for each forcing term. Their sum satisfies

$$L[y_{p,1}+y_{p,2}]=g_1+g_2.$$

This does not say that an arbitrary sum of solutions keeps the same forcing: the forcing terms add too.

For example, consider

$$y''-3y'-4y=3e^{2t}+2\sin t.$$

The previous two examples already did the coefficient matching. Add their particular solutions:

$$y_p(t)=-\frac12e^{2t}-\frac5{17}\sin t+\frac3{17}\cos t.$$

The complementary solution is unchanged, so the general solution is

$$\boxed{ y(t)=C_1e^{4t}+C_2e^{-t} -\frac12e^{2t}-\frac5{17}\sin t+\frac3{17}\cos t }.$$

Resonance in undetermined coefficients

What if the trial is already a homogeneous solution? The differential operator sends it to zero for every choice of coefficient. No coefficient matching can make that trial produce a nonzero forcing. This overlap is called resonance in the method of undetermined coefficients.

Example: a simple characteristic root

Solve

$$y''-y=7e^t.$$

The starting trial $y_p=Ae^t$ fails:

$$y_p''-y_p=Ae^t-Ae^t=0\ne7e^t.$$

The failure is in the trial, not in the existence of a solution. The associated homogeneous equation is $y''-y=0$, whose characteristic equation is

$$r^2-1=(r-1)(r+1)=0.$$

Thus

$$y_c(t)=C_1e^{-t}+C_2e^t.$$

The function $e^t$ is already in this family. Multiply the trial by $t$ to get a different function:

$$\begin{aligned} y_p(t)&=Ate^t,\\ y_p'(t)&=A(1+t)e^t,\\ y_p''(t)&=A(2+t)e^t. \end{aligned}$$

Now the $te^t$ terms cancel, but a nonzero exponential remains:

$$y_p''-y_p=2Ae^t=7e^t.$$

Therefore $A=7/2$, so

$$y_p(t)=\frac72te^t.$$

Adding the complementary solution gives

$$\boxed{y(t)=C_1e^{-t}+C_2e^t+\frac72te^t}.$$

Example: a repeated characteristic root

One factor of $t$ is not always enough. Consider

$$y''-2y'+y=7e^t.$$

The characteristic equation is

$$r^2-2r+1=(r-1)^2=0,$$

so $1$ is a double root and

$$y_c(t)=(C_1+C_2t)e^t.$$

Both $Ae^t$ and $Ate^t$ are homogeneous solutions. Each makes the left-hand side zero. The first trial that avoids those two functions is

$$y_p(t)=At^2e^t.$$

Differentiate and substitute:

$$\begin{aligned} y_p'&=A(t^2+2t)e^t,\\ y_p''&=A(t^2+4t+2)e^t,\\ y_p''-2y_p'+y_p &=Ae^t\bigl[(t^2+4t+2)-2(t^2+2t)+t^2\bigr]\\ &=2Ae^t. \end{aligned}$$

Again $2A=7$, giving

$$y_p(t)=\frac72t^2e^t,$$

and hence

$$\boxed{y(t)=\left(C_1+C_2t+\frac72t^2\right)e^t}.$$

Both resonance examples are valid for all real $t$. Neither has initial conditions, so the two complementary constants remain arbitrary.

The trial-form rule

The power of $t$ is determined by the multiplicity of a characteristic root: the number of times that root occurs in the characteristic polynomial. A value that is not a root has multiplicity zero.

Let $P_n(t)$ be a known polynomial of degree $n$. Let $Q_n(t)$ and $R_n(t)$ be polynomials with separate unknown coefficients for every power from $t^n$ down to the constant term. For example,

$$Q_n(t)=A_nt^n+A_{n-1}t^{n-1}+\cdots+A_0.$$

The symbols $\alpha$ and $\beta$ below are known real constants from the forcing. Use $\beta\ne0$ in the trigonometric row. The exponent $s$ is the multiplicity of the root listed in the last column. For $ay''+by'+cy=g(t)$ with constant coefficients and $a\ne0$:

The exponent $s$ is the smallest nonnegative integer that removes overlap with the homogeneous solution space. In a second-order equation it is $0$, $1$, or $2$; a nonreal root of a real quadratic can only be simple. Multiply the whole trial by $t^s$, not just one of its terms.

This recovers both examples: the root $1$ is simple for $r^2-1$, so use $te^t$; it is double for $(r-1)^2$, so use $t^2e^t$. For a sum of forcing components, choose the appropriate trial for each component and add the resulting particular solutions.

Variation of parameters

Undetermined coefficients works by recognizing a finite family of functions that stays closed under differentiation. Variation of parameters takes a different approach: start with two known homogeneous solutions and replace their constant multipliers with unknown functions. This works for more general forcing terms and for variable coefficients, provided we know a fundamental set for the homogeneous equation.

Consider the normalized equation

$$y''+p(t)y'+q(t)y=g(t),$$

where $p$, $q$, and $g$ are continuous on an open interval $I$. Let $\{y_1,y_2\}$ be a fundamental set for the associated homogeneous equation on $I$. Write its Wronskian as

$$W(t)=y_1(t)y_2'(t)-y_1'(t)y_2(t)\ne0, \qquad t\in I.$$

Normalization matters here. If the original equation is $a_2(t)y''+a_1(t)y'+a_0(t)y=h(t)$, divide by $a_2(t)$ on an interval where it is nonzero. The forcing used in this method is $g=h/a_2$, not the original $h$.

Choosing the two unknown functions

Keep the known functions $y_1$ and $y_2$, but replace their constant multipliers with unknown functions $u_1$ and $u_2$. Look for a particular solution of the form

$$y_p(t)=u_1(t)y_1(t)+u_2(t)y_2(t).$$

The product rule gives

$$y_p'=u_1'y_1+u_1y_1'+u_2'y_2+u_2y_2'.$$

We have introduced two unknown functions to construct one solution. Use that freedom to impose an auxiliary condition that cancels the first and third terms:

$$\boxed{y_1u_1'+y_2u_2'=0}.$$

This is a condition we choose for the construction, not an identity that holds for arbitrary $u_1,u_2$. It leaves

$$y_p'=u_1y_1'+u_2y_2',$$

so differentiating again gives

$$y_p''=u_1'y_1'+u_1y_1''+u_2'y_2'+u_2y_2''.$$

The benefit is that no second derivatives of $u_1$ or $u_2$ appear. Put these expressions into the nonhomogeneous equation and group the terms multiplying each unknown function:

$$\begin{aligned} y_p''+py_p'+qy_p &=u_1\bigl(y_1''+py_1'+qy_1\bigr)\\ &\quad+u_2\bigl(y_2''+py_2'+qy_2\bigr)\\ &\quad+u_1'y_1'+u_2'y_2'. \end{aligned}$$

Both parentheses vanish because $y_1$ and $y_2$ solve the homogeneous equation. To produce the required forcing, the remaining terms must satisfy

$$\boxed{y_1'u_1'+y_2'u_2'=g(t)}.$$

Solving the two conditions

Together, the auxiliary condition and the ODE give a linear system for the derivatives $u_1'$ and $u_2'$:

$$\begin{pmatrix} y_1&y_2\\ y_1'&y_2' \end{pmatrix} \begin{pmatrix} u_1'\\ u_2' \end{pmatrix} = \begin{pmatrix} 0\\ g \end{pmatrix}.$$

All functions in this system are evaluated at the same $t$. The determinant of its coefficient matrix is $W(t)$, which is nonzero on $I$. Its inverse is therefore

$$\begin{pmatrix} y_1&y_2\\ y_1'&y_2' \end{pmatrix}^{-1} = \frac1{W(t)} \begin{pmatrix} y_2'&-y_2\\ -y_1'&y_1 \end{pmatrix}.$$

Multiplying by the column with entries $0$ and $g$ gives

$$\boxed{ u_1'=-\frac{y_2g}{W}, \qquad u_2'=\frac{y_1g}{W} }.$$

The minus sign in the first formula comes from the upper-right entry of the inverse matrix. Keep the Wronskian in the same order, $W[y_1,y_2]$, in both formulas.

Integrate to find one choice of the two functions:

$$u_1(t)=-\int\frac{y_2(t)g(t)}{W(t)}\,dt, \qquad u_2(t)=\int\frac{y_1(t)g(t)}{W(t)}\,dt.$$

Why can we omit the integration constants? Adding constants $k_1,k_2$ to $u_1,u_2$ adds only $k_1y_1+k_2y_2$ to $y_p$. Those terms are already included in the complementary solution. Choose any antiderivatives on $I$ to obtain one particular solution.

The general solution

Putting $u_1$ and $u_2$ back into $y_p=u_1y_1+u_2y_2$ gives

$$\boxed{ \begin{aligned} y_p(t) &=-y_1(t)\int\frac{y_2(t)g(t)}{W(t)}\,dt\\ &\quad+y_2(t)\int\frac{y_1(t)g(t)}{W(t)}\,dt. \end{aligned} }$$

The two conditions ensure that this function satisfies the nonhomogeneous equation. The general solution on $I$ is

$$\boxed{ \begin{aligned} y(t) &=C_1y_1(t)+C_2y_2(t)\\ &\quad-y_1(t)\int\frac{y_2(t)g(t)}{W(t)}\,dt\\ &\quad+y_2(t)\int\frac{y_1(t)g(t)}{W(t)}\,dt, \end{aligned} }$$

where $C_1,C_2\in\mathbb R$. If initial conditions are given, apply them to this full solution. Variation of parameters supplies the integrals without guessing a trial family, though those integrals need not have elementary antiderivatives.

We can also choose a fixed $t_0\in I$ as the lower limit of both integrals and combine them:

$$y_p(t)=\int_{t_0}^t \frac{y_1(s)y_2(t)-y_1(t)y_2(s)}{W(s)}g(s)\,ds.$$

Here $s$ is just the integration variable; $t$ is the time at which we want the answer. This form gives $y_p(t_0)=y_p'(t_0)=0$. It is still a valid particular solution when the integral cannot be simplified further.

Example: recovering the resonant exponential

Return to $y''-y=7e^t$, now using variation of parameters rather than a modified trial. Choose the ordered pair

$$y_1=e^t,\qquad y_2=e^{-t},\qquad W=-2.$$

With $g(t)=7e^t$, the parameter integrals are

$$\begin{aligned} u_1&=-\int\frac{e^{-t}(7e^t)}{-2}\,dt=\frac72t,\\ u_2&=\int\frac{e^t(7e^t)}{-2}\,dt=-\frac74e^{2t}. \end{aligned}$$

Multiplying by the corresponding homogeneous solutions gives

$$y_p=u_1y_1+u_2y_2=\frac72te^t-\frac74e^t.$$

The last term is itself homogeneous. Absorb it into the arbitrary coefficient of $e^t$ to obtain

$$\boxed{y(t)=C_1e^t+C_2e^{-t}+\frac72te^t},\qquad t\in\mathbb R.$$

This is the same family found by undetermined coefficients. Variation of parameters produces the needed factor of $t$ through integration, without requiring us to guess it.

Example: tangent forcing

Now consider an input outside the usual undetermined-coefficients families:

$$y''+y=\tan t,\qquad -\frac\pi2<t<\frac\pi2.$$

The endpoints are excluded because $\tan t$ is undefined there. On this open interval, the coefficients and forcing are continuous. The homogeneous equation has roots $\pm i$, so choose

$$y_1=\sin t,\qquad y_2=\cos t,\qquad W=\sin t(-\sin t)-\cos^2t=-1.$$

Keep this order when applying the formulas. The first integral simplifies immediately:

$$u_1=-\int\frac{\cos t\tan t}{-1}\,dt =\int\sin t\,dt=-\cos t.$$

For the second, rewrite $\sin^2t=1-\cos^2t$:

$$\begin{aligned} u_2&=\int\frac{\sin t\tan t}{-1}\,dt =-\int\frac{\sin^2t}{\cos t}\,dt\\ &=\int(\cos t-\sec t)\,dt\\ &=\sin t-\ln|\sec t+\tan t|. \end{aligned}$$

To check the logarithm, differentiate it: $(\sec t+\tan t)'=\sec t(\sec t+\tan t)$, so its logarithmic derivative is $\sec t$.

The mixed sine–cosine terms cancel when we form the particular solution:

$$\begin{aligned} y_p&=(-\cos t)\sin t +\bigl(\sin t-\ln|\sec t+\tan t|\bigr)\cos t\\ &=-\cos t\ln|\sec t+\tan t|. \end{aligned}$$

Thus the general solution on the stated interval is

$$\boxed{y(t)=C_1\sin t+C_2\cos t-\cos t\ln|\sec t+\tan t|}.$$

For a direct check, let $H(t)=\ln|\sec t+\tan t|$. Since $H'=\sec t$,

$$y_p'=\sin t\,H-1,\qquad y_p''=\cos t\,H+\tan t,$$

and $y_p''+y_p=\tan t$, as required.

Choosing a particular-solution method

Use undetermined coefficients when the coefficients are constant and the forcing is built from polynomials, exponentials, sine, and cosine. Check for overlap with the homogeneous solutions and multiply the whole trial by the required power of $t$.

Use variation of parameters for more general forcing or variable coefficients when a fundamental homogeneous pair is known. It does not find that pair for us, and its integrals may need to remain unevaluated.

These methods now let us study mechanical vibrations and RLC circuits: the same second-order equation describes both systems.