MATH 308 Lecture 14

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Lecture Notes

Non-Homogeneous Equations

$y''-3y'+2y=1$

$y_{h}=c_{1}\,\mathrm {e} ^{2x}+c_{2}\,\mathrm {e} ^{x}$ is solution to homogeneous problem $y''-3y'+2y=0$

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

Guess particular solution $y_{p}={\tfrac {1}{2}}+y_{h}$ by looking at original diff EQ.

$y''-3y'+2y=y_{h}''-3y_{h}'+2y_{h}+1=0+1$

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

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

Corresponding homogeneous function is $y''-5y'+6y=0$, 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 $\mathrm {e} ^{x}$:

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

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

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

General solution to $y''-4y'+3y=0$ 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

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

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 $y''+A\,y'+B\,y{=}\mathrm {e} ^{C\,x}$, multiply it by $x$ so that $x$ has a degree one greater than the homogeneous solution

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

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

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

$y''-3y'+2y=2x+3$

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: $y_{p}=a\,x+b$ since the RHS is a polynomial of degree 1.

${\begin{aligned}y_{p}&=a\,x+b\\y_{p}'&=a\\y_{p}''&=0\\y''-3y'+2y=0-3a+2a\,x+2\,b&=2x+3\\2a&=22b&=6\end{aligned}}$

So our particular solution is $y_{p}=x+3$

Exercise 6

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

Solution to homogeneous function is $y_{h}=c_{1}\,\mathrm {e} ^{2x}+c_{2}\,\mathrm {e} ^{x}$

Try particular solution $y_{p}=A\sin {2x}+B\cos {2x}$

${\begin{aligned}y_{p}&=A\sin {2x}+B\cos {2x}\\y_{p}'&=2A\cos {2x}-2B\sin {2x}\\y_{p}''&=-4A\sin {2x}-2B\cos {2x}\\y''-3y+2y&=-4A\sin {2x}-2B\cos {x}-6A\cos {2x}+6B\sin {x}+2A\sin {2x}+2B\cos {2x}=20\sin {2x}\\(-4A+6B+2A)\sin {2x}+(-4B-6A+2B)\cos {2x}&=20\sin {2x}\end{aligned}}$

We're left with the system of equations

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

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

$a\,y''+b\,y'+c\,y=g(x)\,\!$

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

$y=x^{s}\left(A\,\cos {\alpha \,x}+B\,\sin {\alpha \,x}\right)\,\!$

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