MATH 308 Lecture 15

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Lecture Notes

Remark for section 3.5

How would we go about solving $y^{″}-3y^{\prime }+2y=20\sin 2x+2x+3+{\mathrm{e}}^x$?

$y_h=c_1{\mathrm{e}}^{2x}+c_2{\mathrm{e}}^x$

Particular Solution $y_p=A\sin 2x+B\cos 2x+ax+b+Cx{\mathrm{e}}^x$

Section 3.6: Variation of Parameters

Find solution to $y^{″}-y^{\prime }-2y={\mathrm{e}}^{3x}$

Homogeneous solution is $y_h=c_1{\mathrm{e}}^{-x}+c_2{\mathrm{e}}^{2x}$

What about non-constant coefficients; i.e. what if the constants were actually functions of $x$:

$y=c_1(x)y_1+c_2(x)y_2$

There are infinitely many possibilities for $c_1$ and $c_2$, so we impose a restriction to limit the number of possibilities:

${c_1}^{\prime }{y}_1+{c_2}^{\prime }{y}_2=0$

Using the above examples, we let $y_1={\mathrm{e}}^{-x}$ and $y_2={\mathrm{e}}^{2x}$:

$\begin{array}{rl}y& =c_1{\mathrm{e}}^{-x}+c_2{\mathrm{e}}^{2x}\\ y^{\prime }& ={c_1}^{\prime }{\mathrm{e}}^{-x}-c_1{\mathrm{e}}^{-x}+{c_2}^{\prime }{\mathrm{e}}^{2x}+2c_2{\mathrm{e}}^{2x}\\ & =({c_1}^{\prime }{\mathrm{e}}^{-x}+{c_2}^{\prime }{\mathrm{e}}^{2x})-c_1{\mathrm{e}}^{-x}+2c_2{\mathrm{e}}^{2x}\\ & =-c_1{\mathrm{e}}^{-x}+2c_2{\mathrm{e}}^{2x}\\ y^{″}& =c_1{\mathrm{e}}^{-x}-{c_1}^{\prime }{\mathrm{e}}^{-x}+2{c_2}^{\prime }{\mathrm{e}}^{2x}+4c_2{\mathrm{e}}^{2x}\end{array}$

Simplification: We were able to simplify $y^{\prime }$ with the restriction ${c_1}^{\prime }{\mathrm{e}}^{-x}+{c_2}^{\prime }{\mathrm{e}}^{2x}=0$.

Now we can plug our functions into the differential equation:

$\begin{array}{rl}{\mathrm{e}}^{3x}& =y^{″}-y^{\prime }-2y\\ & =(c_1{\mathrm{e}}^{-x}-{c_1}^{\prime }{\mathrm{e}}^{-x}+2{c_2}^{\prime }{\mathrm{e}}^{2x}+4c_2{\mathrm{e}}^{2x})-(-c_1{\mathrm{e}}^{-x}+2c_2{\mathrm{e}}^{2x})-2(c_1{\mathrm{e}}^{-x}+c_2{\mathrm{e}}^{2x})\\ & =\cancel{c_1{\mathrm{e}}^{-x}}-{c_1}^{\prime }{\mathrm{e}}^{-x}+2{c_2}^{\prime }{\mathrm{e}}^{2x}+\cancel{4c_2{\mathrm{e}}^{2x}}+\cancel{c_1{\mathrm{e}}^{-x}}\cancel{-2c_2{\mathrm{e}}^{2x}}\cancel{-2c_1{\mathrm{e}}^{-x}}\cancel{-2c_2{\mathrm{e}}^{2x}}\\ & =-{c_1}^{\prime }{\mathrm{e}}^{-x}+2{c_2}^{\prime }{\mathrm{e}}^{2x}\end{array}$

Use restriction as second equation in system:

$\{\begin{array}{l}-{c_1}^{\prime }{\mathrm{e}}^{-x}+2{c_2}^{\prime }{\mathrm{e}}^{2x}={\mathrm{e}}^{3x}\\ {c_1}^{\prime }{\mathrm{e}}^{-x}+{c_2}^{\prime }{\mathrm{e}}^{2x}=0\end{array}$

We solve for ${c_1}^{\prime }$ and ${c_2}^{\prime }$:

$\begin{array}{rlrl}{c_1}^{\prime }& =-\frac{1}{3}{\mathrm{e}}^{4x}& {c_2}^{\prime }& =\frac{1}{3}{\mathrm{e}}^x\\ c_1& =-\frac{1}{12}{\mathrm{e}}^{4x}+A& c_2& =\frac{1}{3}{\mathrm{e}}^x+B\end{array}$

So general solution to original differential equation $y^{″}-y^{\prime }-2y={\mathrm{e}}^{3x}$ is:

$\begin{array}{rl}y& =(-\frac{1}{12}{\mathrm{e}}^{4x}+A){\mathrm{e}}^{-x}+(\frac{1}{3}{\mathrm{e}}^x+B){\mathrm{e}}^{2x}\\ & =(-\frac{1}{12}{\mathrm{e}}^{3x}+\frac{1}{3}{\mathrm{e}}^{3x})+A{\mathrm{e}}^{-x}+B{\mathrm{e}}^{2x}\\ & =\frac{1}{4}{\mathrm{e}}^{3x}+A{\mathrm{e}}^{-x}+B{\mathrm{e}}^{2x}\end{array}$

Method for Solving Second Order ODEs

To determine a particular solution to $y^{″}+p(x)y^{\prime }+q(x)y=g(x)$

Find $y_1$ and $y_2$ for corresponding homogeneous equation $y^{″}+p(x)y^{\prime }+q(x)y=0$

Solve the following system for $c_1(x)$ and $c_2(x)$

$\{\begin{array}{l}{c_1}^{\prime }{y}_1+{c_2}^{\prime }{y}_2=0\\ {c_1}^{\prime }{y_1}^{\prime }+{c_2}^{\prime }{y_2}^{\prime }=g(x)\end{array}$

or

$\begin{array}{rlrl}{c}_1(x)& =-\underset{x_0}{\overset{x}{\int }}\frac{y_2(s)g(s)}{W(y_1,y_2)(s)}\mathrm{d}s& c_2(x)& =\underset{x_0}{\overset{x}{\int }}\frac{y_1(s)g(s)}{W(y_1,y_2)(s)}\mathrm{d}s\end{array}$

A particular solution will be

$y_p(x)=c_1(x)y_1(x)+c_2(x)y_2(x)$

Exercise 7

Find general solution to $y^{″}+y=0$

Solution to homogeneous equation is $y=c_1\cos x+c_2\sin x$, where $y_1=\cos x$ and $y_2=\sin x$.

$W(y_1,y_2)={\cos }^2(x)+{\sin }^2(x)=1$

Plug in to equations for $c_1$ and $c_2$

$\begin{array}{rl}{c}_1(x)& =-\int \sin x\frac{1}{\sin x}\mathrm{d}x\\ & =-x+A\\ c_2(x)& =\int \cos x\frac{1}{\sin x}\mathrm{d}x\\ & =\ln (\sin x)+B\end{array}$

General solution is

$\begin{array}{rl}y& =(-x+A)\cos x+(\ln (\sin x)+B)\sin x\\ & =-x\cos x+A\cos x+\ln (\sin x)\sin x+B\sin x\end{array}$