MATH 308 Lecture 14

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Lecture Notes

Non-Homogeneous Equations

$y^{″}-3y^{\prime }+2y=1$

$y_h=c_1{\mathrm{e}}^{2x}+c_2{\mathrm{e}}^x$ is solution to homogeneous problem $y^{″}-3y^{\prime }+2y=0$

Let's look for a constant function solution to the nonhomogeneous function: $y(t)=\frac{1}{2}$

Guess particular solution $y_p=\frac{1}{2}+y_h$ by looking at original diff EQ.

$y^{″}-3y^{\prime }+2y={y_h}^{″}-3{y_h}^{\prime }+2y_h+1=0+1$

The general solution to the non-homogeneous function is $y=\frac{1}{2}+c_1{\mathrm{e}}^{2x}+c_2{\mathrm{e}}^x$

Exercise 2

$y^{″}-5y^{\prime }+6y={\mathrm{e}}^x$

Corresponding homogeneous function is $y^{″}-5y^{\prime }+6y=0$, and solution is $y_h=c_1{\mathrm{e}}^{2x}+c_2{\mathrm{e}}^{3x}$.

We'll guess the particular solution is a constant undetermined coefficient multiplied into ${\mathrm{e}}^x$:

$\begin{array}{rl}C{e}^x-5C{e}^x+6C{e}^x& =e^x\\ C& =\frac{1}{2}\end{array}$

So $y_p=\frac{1}{2}{\mathrm{e}}^x$ is a particular solution to the non-homogeneous differential equation.

Combining the general solution to the homogeneous equation, we get

$y=\frac{1}{2}{\mathrm{e}}^x+c_1{\mathrm{e}}^{2x}+c_2{\mathrm{e}}^{3x}$

Exercie 3

$y^{″}-4y^{\prime }+3y={\mathrm{e}}^x$

General solution to $y^{″}-4y^{\prime }+3y=0$ is $y_h=c_1{\mathrm{e}}^x+c_2{\mathrm{e}}^{3x}$

Possible particular solution is of form $y=C{\mathrm{e}}^x$, so

$\begin{array}{rl}C{\mathrm{e}}^x-4C{\mathrm{e}}^x+3C{\mathrm{e}}^x& ={\mathrm{e}}^x\\ 0& ={\mathrm{e}}^x\end{array}$

Bad guess. This isn't surprising since $c_1{\mathrm{e}}^x$ is a solution to the homogeneous differential equation, so of course it will equal 0.

Note: To make good guesses for equations of form $y^{″}+A{y}^{\prime }+By={\mathrm{e}}^{Cx}$, multiply it by $x$ so that $x$ has a degree one greater than the homogeneous solution

Let's try $y=Cx{\mathrm{e}}^x$.

$\begin{array}{rl}2C{\mathrm{e}}^x+Cx{\mathrm{e}}^x-4C{\mathrm{e}}^x-4Cx{\mathrm{e}}^x+3Cx{\mathrm{e}}^x& ={\mathrm{e}}^x\\ -2C{\mathrm{e}}^x& ={\mathrm{e}}^x\\ C& =-\frac{1}{2}\end{array}$

So $y=-\frac{1}{2}x{\mathrm{e}}^x$ is a particular solution to the non-homogeneous differential equation.

The general solution is $y=-\frac{1}{2}x{\mathrm{e}}^x+c_1{\mathrm{e}}^x+c_2{\mathrm{e}}^{3x}$

Exercise 5

$y^{″}-3y^{\prime }+2y=2x+3$

Solution to homogeneous differential equation is $y_h=c_1{\mathrm{e}}^{2x}+c_2{\mathrm{e}}^x$

Particular solution $y_p$ is a polynomial function of degree 1: $y_p=ax+b$ since the RHS is a polynomial of degree 1.

$\begin{array}{rl}{y}_p& =ax+b\\ {y_p}^{\prime }& =a\\ {y_p}^{″}& =0\\ y^{″}-3y^{\prime }+2y=0-3a+2ax+2b& =2x+3\\ 2a& =22b& =6\end{array}$

So our particular solution is $y_p=x+3$

Exercise 6

Find a particular solution to $y^{″}-3y^{\prime }+2y=20\sin (2x)$

Solution to homogeneous function is $y_h=c_1{\mathrm{e}}^{2x}+c_2{\mathrm{e}}^x$

Try particular solution $y_p=A\sin 2x+B\cos 2x$

$\begin{array}{rl}{y}_p& =A\sin 2x+B\cos 2x\\ {y_p}^{\prime }& =2A\cos 2x-2B\sin 2x\\ {y_p}^{″}& =-4A\sin 2x-2B\cos 2x\\ y^{″}-3y+2y& =-4A\sin 2x-2B\cos x-6A\cos 2x+6B\sin x+2A\sin 2x+2B\cos 2x=20\sin 2x\\ (-4A+6B+2A)\sin 2x+(-4B-6A+2B)\cos 2x& =20\sin 2x\end{array}$

We're left with the system of equations

$\{\begin{array}{l}-2A+6B=20\\ -2B+6A=0\end{array}$

We find that $A=-1$ and $B=3$.

Therefore our general solution to the nonhomogeneous equation is $y=-\sin 2x+3\cos 2x+c_1{\mathrm{e}}^{2x}+c_2{\mathrm{e}}^x$

Making Good Guesses

Given a linear second order differential operator with constant coefficients

$a{y}^{″}+b{y}^{\prime }+cy=g(x)$

We have the following cases for $g(x)$:

Exponentials

$g(x)={\mathrm{e}}^{\alpha x}$

A particular solution to the differential equation is in the form

$y=k{x}^s{\mathrm{e}}^{\alpha x}$

• $s=0$ if ${\mathrm{e}}^{\alpha x}$ is not a solution to the corresponding homogeneous problem.
• $s=1$ if ${\mathrm{e}}^{\alpha x}$ is a solution to the corresponding homogeneous problem.
• $s=2$ if ${\mathrm{e}}^{\alpha x}$ and $x{\mathrm{e}}^{\alpha x}$ are solutions to the corresponding homogeneous problem.

Polynomials

$g(x)=P_n(x)=\sum _{j=0}^n{\alpha }_j{x}^j={\alpha }_n{x}^n+{\alpha }_{n-1}{x}^{n-1}+\dots +{\alpha }_{1}x+{\alpha }_0$

Where $P_n$ is a polynomial function of degree $n$, then a particular solution is in the form $y=x^s{p}_n(x)$ where

• $p_n$ is a polynomial function of degree $n$
• $s=1$ if the constant functions are solutions to the corresponding homogeneous differential equation
• $s=0$ otherwise.

Trigonometric

$g(x)=p\cos \alpha x+q\sin \alpha x$

A particular solution is in the form

$y=x^s(A\cos \alpha x+B\sin \alpha x)$

• $s=1$ if $\cos \alpha x$ is solution to the homogeneous solution
• $s=0$ otherwise.