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4.6 Variation of Parameters: Class Notes

4.6a

Consider the homogeneous second-order differential equation $a_2(x)y″+a_1(x)y\prime +a_0(x)y=0$. The general solution to this DE is $y=c_1{y}_1+c_2{y}_2$. When the input function of an associated nonhomogeneous DE is not restricted to the type we saw in 4.4, we consider particular solutions with coefficients that are variable functions. That is, we will find $y_p(x)=u_1(x)y_1(x)+u_2(x)y_2(x)$.

Starting with $a_2(x)y″+a_1(x)y\prime +a_0(x)y=g(x)$, divide by $a_2(x)$ to put the DE in standard form:

$$y″+P(x)y\prime +Q(x)y=f(x).$$

The complementary function (from 4.3) is $y_c=c_1{y}_1+c_2{y}_2$, where $c_1$ and $c_2$ are constants.

Now consider $y_p=u_1(x)y_1(x)+u_2(x)y_2(x)$, with the constants replaced by functions. Differentiate it twice:

$$\begin{array}{cc}\hfill {\displaystyle y_p\prime }& {\displaystyle =u_1\prime y_1+u_1{y}_1\prime +u_2\prime y_2+u_2{y}_2\prime }\hfill \\ \hfill {\displaystyle y_p″}& {\displaystyle =u_1″y_1+2u_1\prime y_1\prime +u_1{y}_1″+u_2″y_2+2u_2\prime y_2\prime +u_2{y}_2″}\hfill \end{array}$$

Substituting $y″$, $P(x)y\prime$, and $Q(x)y$ into $y″+P(x)y\prime +Q(x)y=f(x)$, the DE becomes

$$\begin{array}{c}{\displaystyle u_1″y_1+2u_1\prime y_1\prime +u_1{y}_1″+u_2″y_2+2u_2\prime y_2\prime +u_2{y}_2″}\hfill \\ {\displaystyle +P{u}_1\prime y_1+P{u}_1{y}_1\prime +P{u}_2\prime y_2+P{u}_2{y}_2\prime }\hfill \\ {\displaystyle +Q{u}_1{y}_1+Q{u}_2{y}_2=f(x).}\hfill \end{array}$$

Since $y_1$ and $y_2$ are solutions of the homogeneous DE, the terms with $u_1$ itself (not its derivatives) add up to zero:

$$u_1(y_1″+P{y}_1\prime +Q{y}_1)=0.$$

Likewise for $y_2$: the $u_2$ terms add up to zero. What remains is

$$u_1″y_1+2u_1\prime y_1\prime +u_2″y_2+2u_2\prime y_2\prime +P(u_1\prime y_1+u_2\prime y_2)=f(x).$$

Split $2u_1\prime y_1\prime$ into $u_1\prime y_1\prime +u_1\prime y_1\prime$. The piece $u_1″y_1+u_1\prime y_1\prime$ is the derivative ${\displaystyle \frac{d}{dx}(u_1\prime y_1)}$ by the Product Rule, and in the same way $u_2″y_2+u_2\prime y_2\prime$ is ${\displaystyle \frac{d}{dx}(u_2\prime y_2)}$. The equation becomes

$${\displaystyle \frac{d}{dx}(u_1\prime y_1+u_2\prime y_2)+u_1\prime y_1\prime +u_2\prime y_2\prime +P(u_1\prime y_1+u_2\prime y_2)=f(x).}$$

If we require $u_1\prime y_1+u_2\prime y_2=0$, the first and last groups are zero, so $u_1\prime y_1\prime +u_2\prime y_2\prime =f(x)$. Here we are considering a system with unknown functions $u_1\prime$ and $u_2\prime$:

$$\begin{array}{cc}\hfill {\displaystyle u_1\prime y_1+u_2\prime y_2}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle u_1\prime y_1\prime +u_2\prime y_2\prime }& {\displaystyle =f(x)}\hfill \end{array}$$

Here the unknowns are $u_1\prime$ and $u_2\prime$, so we form

$$W=\left|\begin{array}{cc}{y}_1& y_2\\ y_1\prime & y_2\prime \end{array}\right|,W_1=\left|\begin{array}{cc}0& y_2\\ f(x)& y_2\prime \end{array}\right|,W_2=\left|\begin{array}{cc}{y}_1& 0\\ y_1\prime & f(x)\end{array}\right|.$$

$W_1$ and $W_2$ use $f(x)$, the right side of the standard form, so the DE must be in standard form before forming them.

4.6b

$$W=\left|\begin{array}{cc}{y}_1& y_2\\ y_1\prime & y_2\prime \end{array}\right|,W_1=\left|\begin{array}{cc}0& y_2\\ f(x)& y_2\prime \end{array}\right|,W_2=\left|\begin{array}{cc}{y}_1& 0\\ y_1\prime & f(x)\end{array}\right|$$

Expanding the determinants, $W_1=-f(x)y_2$ and $W_2=f(x)y_1$.

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle u_1\prime }& {\displaystyle =\frac{W_1}{W}⟹u_1=\int \frac{W_1}{W}dx}\hfill \\ \hfill {\displaystyle u_2\prime }& {\displaystyle =\frac{W_2}{W}⟹u_2=\int \frac{W_2}{W}dx}\hfill \end{array}}$$

$y_p=u_1{y}_1+u_2{y}_2$ and $y=y_c+y_p$

Example

Solve the differential equation by variation of parameters.

$$y″+y=\sec \theta \tan \theta$$

Solution

The auxiliary equation is $m^2+1=0$, so $m=\pm i$ and

$$y_c=c_1\cos \theta +c_2\sin \theta ,y_1=\cos \theta ,y_2=\sin \theta .$$

The Wronskian and the other two determinants:

$$W=\left|\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right|={\cos }^2\theta +{\sin }^2\theta =1$$

$$W_1=\left|\begin{array}{cc}0& \sin \theta \\ \sec \theta \tan \theta & \cos \theta \end{array}\right|=-{\tan }^2\theta ,W_2=\left|\begin{array}{cc}\cos \theta & 0\\ -\sin \theta & \sec \theta \tan \theta \end{array}\right|=\tan \theta$$

Then

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle u_1}& {\displaystyle =\int (-{\tan }^2\theta )d\theta =\int (1-{\sec }^2\theta )d\theta }\hfill \\ \hfill {\displaystyle }& {\displaystyle =\theta -\tan \theta .}\hfill \end{array}}$$

Tip: No $+C$ is needed here: it would only add $C{y}_1$ to the solution, which we already have in $y_c$.

$${\displaystyle u_2=\int \tan \theta d\theta =-\int \frac{-\sin \theta }{\cos \theta }d\theta =-\ln |\cos \theta |=\ln |\sec \theta |}$$

For this integral, let $u=\cos \theta$, so $du=-\sin \theta d\theta$.

So the particular solution is

$$\begin{array}{cc}\hfill {\displaystyle y_p=u_1{y}_1+u_2{y}_2}& {\displaystyle =(\theta -\tan \theta )\cos \theta +\sin \theta \ln |\sec \theta |}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\theta \cos \theta -\sin \theta +\sin \theta \ln |\sec \theta |.}\hfill \end{array}$$

The term $-\sin \theta$ is a multiple of $y_2=\sin \theta$, so it combines with $c_2\sin \theta$ in $y_c$:

$$y=c_1\cos \theta +c_2\sin \theta +\theta \cos \theta -\sin \theta +\sin \theta \ln |\sec \theta |$$

4.6c

Example

Solve the initial-value problem by variation of parameters.

$$2y″+y\prime -y=x+1,y(0)=1,y\prime (0)=0$$

Solution

Divide by 2 to put the DE in standard form before forming $W$, $W_1$, and $W_2$:

$${\displaystyle y″+\frac{1}{2}y\prime -\frac{1}{2}y=\frac{x+1}{2},f(x)=\frac{x+1}{2}}$$

The auxiliary equation of the original DE is $2m^2+m-1=0$, so $(2m-1)(m+1)=0$ and $m=1/2,-1$:

$$y_c=c_1{e}^{-x}+c_2{e}^{x/2},y_1=e^{-x},y_2=e^{x/2}$$

$${\displaystyle W=\left|\begin{array}{cc}{e}^{-x}& e^{x/2}\\ -e^{-x}& \frac{1}{2}{e}^{x/2}\end{array}\right|=\frac{1}{2}{e}^{-x/2}+e^{-x/2}=\frac{3}{2}{e}^{-x/2}}$$

$${\displaystyle W_1=\left|\begin{array}{cc}0& e^{x/2}\\ \frac{x+1}{2}& \frac{1}{2}{e}^{x/2}\end{array}\right|=-e^{x/2}\left(\frac{x+1}{2}\right),W_2=\left|\begin{array}{cc}{e}^{-x}& 0\\ -e^{-x}& \frac{x+1}{2}\end{array}\right|=e^{-x}\left(\frac{x+1}{2}\right)}$$

For $u_1$, integrate by parts with $u=x+1$, $du=dx$, $dv=e^{x}dx$, $v=e^x$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle u_1}& {\displaystyle =\frac{1}{3}\int \frac{-e^{x/2}(x+1)}{e^{-x/2}}dx=-\frac{1}{3}\int e^x(x+1)dx}\hfill \\ \hfill {\displaystyle }& {\displaystyle =-\frac{1}{3}(x+1)e^x+\frac{1}{3}{e}^x=-\frac{1}{3}x{e}^x}\hfill \end{array}}$$

For $u_2$, integrate by parts with $u=x+1$, $du=dx$, $dv=e^{-x/2}dx$, $v=-2e^{-x/2}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle u_2}& {\displaystyle =\frac{1}{3}\int \frac{e^{-x}(x+1)}{e^{-x/2}}dx=\frac{1}{3}\int e^{-x/2}(x+1)dx}\hfill \\ \hfill {\displaystyle }& {\displaystyle =-\frac{2}{3}(x+1)e^{-x/2}-\frac{4}{3}{e}^{-x/2}=-\frac{2}{3}x{e}^{-x/2}-2e^{-x/2}}\hfill \end{array}}$$

Then ${\displaystyle y_p=u_1{y}_1+u_2{y}_2=-\frac{1}{3}x-\frac{2}{3}x-2=-x-2}$, so

$$y=c_1{e}^{-x}+c_2{e}^{x/2}-x-2.$$

Apply the initial conditions, using ${\displaystyle y\prime =-c_1{e}^{-x}+\frac{1}{2}{c}_2{e}^{x/2}-1}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle y(0)=1⟹1}& {\displaystyle =c_1+c_2-2⟹c_1+c_2=3}\hfill \\ \hfill {\displaystyle y\prime (0)=0⟹0}& {\displaystyle =-c_1+\frac{1}{2}{c}_2-1⟹-c_1+\frac{1}{2}{c}_2=1}\hfill \end{array}}$$

Adding the two equations gives ${\displaystyle \frac{3}{2}{c}_2=4}$, so ${\displaystyle c_2=\frac{8}{3}}$ and ${\displaystyle c_1=3-\frac{8}{3}=\frac{1}{3}}$.

Example

Solve the differential equation by variation of parameters.

$${\displaystyle y″-4y=\frac{e^{2x}}{x}}$$

Solution

The auxiliary equation is $m^2-4=0$, so $m=\pm 2$:

$$y_c=c_1{e}^{2x}+c_2{e}^{-2x},y_1=e^{2x},y_2=e^{-2x}$$

$$W=\left|\begin{array}{cc}{e}^{2x}& e^{-2x}\\ 2e^{2x}& -2e^{-2x}\end{array}\right|=-2-2=-4$$

$${\displaystyle W_1=\left|\begin{array}{cc}0& e^{-2x}\\ \frac{e^{2x}}{x}& -2e^{-2x}\end{array}\right|=-\frac{1}{x},W_2=\left|\begin{array}{cc}{e}^{2x}& 0\\ 2e^{2x}& \frac{e^{2x}}{x}\end{array}\right|=\frac{e^{4x}}{x}}$$

$${\displaystyle u_1=\int \frac{W_1}{W}dx=\frac{1}{4}\int \frac{1}{x}dx=\frac{1}{4}\ln |x|}$$

$${\displaystyle u_2=\int \frac{W_2}{W}dx=-\frac{1}{4}\int \frac{e^{4x}}{x}dx}$$

The integrand ${\displaystyle \frac{e^{4x}}{x}}$ has no elementary antiderivative, so $u_2$ stays an integral, written with a variable upper limit:

$${\displaystyle u_2=-\frac{1}{4}{\int }_{x_0}^x\frac{e^{4t}}{t}dt}$$

4.6d

Example

Solve the given third-order differential equation by variation of parameters.

$$y‴+4y\prime =\sec 2x$$

Solution

The auxiliary equation is $m^3+4m=0$, so $m(m^2+4)=0$ and $m=0,\pm 2i$:

$$y_c=c_1+c_2\cos 2x+c_3\sin 2x$$

Expanding the Wronskian down the first column (only its top entry is nonzero):

$$W=\left|\begin{array}{ccc}1& \cos 2x& \sin 2x\\ 0& -2\sin 2x& 2\cos 2x\\ 0& -4\cos 2x& -4\sin 2x\end{array}\right|=1\left|\begin{array}{cc}-2\sin 2x& 2\cos 2x\\ -4\cos 2x& -4\sin 2x\end{array}\right|=8$$

For $W_k$, replace column $k$ of $W$ with $0,0,\sec 2x$:

$$W_1=\left|\begin{array}{ccc}0& \cos 2x& \sin 2x\\ 0& -2\sin 2x& 2\cos 2x\\ \sec 2x& -4\cos 2x& -4\sin 2x\end{array}\right|=\sec 2x\left|\begin{array}{cc}\cos 2x& \sin 2x\\ -2\sin 2x& 2\cos 2x\end{array}\right|=2\sec 2x$$

$$W_2=\left|\begin{array}{ccc}1& 0& \sin 2x\\ 0& 0& 2\cos 2x\\ 0& \sec 2x& -4\sin 2x\end{array}\right|=\left|\begin{array}{cc}0& 2\cos 2x\\ \sec 2x& -4\sin 2x\end{array}\right|=-2$$

$$W_3=\left|\begin{array}{ccc}1& \cos 2x& 0\\ 0& -2\sin 2x& 0\\ 0& -4\cos 2x& \sec 2x\end{array}\right|=\left|\begin{array}{cc}-2\sin 2x& 0\\ -4\cos 2x& \sec 2x\end{array}\right|=-2\tan 2x$$

Then $u_k\prime =W_k/W$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle u_1}& {\displaystyle =\int \frac{1}{4}\sec 2xdx=\frac{1}{8}\ln |\sec 2x+\tan 2x|}\hfill \\ \hfill {\displaystyle u_2}& {\displaystyle =\int (-\frac{1}{4})dx=-\frac{1}{4}x}\hfill \\ \hfill {\displaystyle u_3}& {\displaystyle =\int (-\frac{1}{4}\tan 2x)dx=\frac{1}{8}\ln |\cos 2x|}\hfill \end{array}}$$

With $y_p=u_1·1+u_2\cos 2x+u_3\sin 2x$, and on an interval where $\cos 2x\ne 0$: