MATH 308 Lecture 15

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Lecture Notes

Remark for section 3.5

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

$y_{h}=c_{1}\,\mathrm {e} ^{2x}+c_{2}\,\mathrm {e} ^{x}$

Particular Solution $y_{p}=A\,\sin {2x}+B\,\cos {2x}+a\,x+b+C\,x\,\mathrm {e} ^{x}$

Section 3.6: Variation of Parameters

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

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:

$c_{1}'\,y_{1}+c_{2}'\,y_{2}=0$

Using the above examples, we let $y_{1}=\mathrm {e} ^{-x}$ and $y_{2}=\mathrm {e} ^{2x}$:

${\begin{aligned}y&=c_{1}\,\mathrm {e} ^{-x}+c_{2}\,\mathrm {e} ^{2x}\\y'&=c_{1}'\,\mathrm {e} ^{-x}-c_{1}\,\mathrm {e} ^{-x}+c_{2}'\,\mathrm {e} ^{2x}+2c_{2}\,\mathrm {e} ^{2x}\\&=\left(c_{1}'\,\mathrm {e} ^{-x}+c_{2}'\,\mathrm {e} ^{2x}\right)-c_{1}\,\mathrm {e} ^{-x}+2c_{2}\,\mathrm {e} ^{2x}\\&=-c_{1}\,\mathrm {e} ^{-x}+2c_{2}\,\mathrm {e} ^{2x}\\y''&=c_{1}\,\mathrm {e} ^{-x}-c_{1}'\,\mathrm {e} ^{-x}+2c_{2}'\,\mathrm {e} ^{2x}+4c_{2}\,\mathrm {e} ^{2x}\end{aligned}}$

Simplification: We were able to simplify $y'$ with the restriction $c_{1}'\,\mathrm {e} ^{-x}+c_{2}'\,\mathrm {e} ^{2x}=0$.

Now we can plug our functions into the differential equation:

${\begin{aligned}\mathrm {e} ^{3x}&=y''-y'-2y\\&=(c_{1}\,\mathrm {e} ^{-x}-c_{1}'\,\mathrm {e} ^{-x}+2c_{2}'\,\mathrm {e} ^{2x}+4c_{2}\,\mathrm {e} ^{2x})-(-c_{1}\,\mathrm {e} ^{-x}+2c_{2}\,\mathrm {e} ^{2x})-2(c_{1}\,\mathrm {e} ^{-x}+c_{2}\,\mathrm {e} ^{2x})\\&={\cancel {c_{1}\,\mathrm {e} ^{-x}}}-c_{1}'\,\mathrm {e} ^{-x}+2c_{2}'\,\mathrm {e} ^{2x}+{\cancel {4c_{2}\,\mathrm {e} ^{2x}}}+{\cancel {c_{1}\,\mathrm {e} ^{-x}}}{\cancel {-2c_{2}\,\mathrm {e} ^{2x}}}{\cancel {-2c_{1}\,\mathrm {e} ^{-x}}}{\cancel {-2c_{2}\,\mathrm {e} ^{2x}}}\\&=-c_{1}'\,\mathrm {e} ^{-x}+2c_{2}'\,\mathrm {e} ^{2x}\end{aligned}}$

Use restriction as second equation in system:

${\begin{cases}-c_{1}'\,\mathrm {e} ^{-x}+2c_{2}'\,\mathrm {e} ^{2x}=\mathrm {e} ^{3x}\\c_{1}'\,\mathrm {e} ^{-x}+c_{2}'\,\mathrm {e} ^{2x}=0\end{cases}}$

We solve for $c_{1}'$ and $c_{2}'$:

${\begin{aligned}c_{1}'&=-{\frac {1}{3}}\,\mathrm {e} ^{4x}&c_{2}'&={\frac {1}{3}}\,\mathrm {e} ^{x}\\c_{1}&=-{\frac {1}{12}}\,\mathrm {e} ^{4x}+A&c_{2}&={\frac {1}{3}}\,\mathrm {e} ^{x}+B\end{aligned}}$

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

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

Method for Solving Second Order ODEs

To determine a particular solution to $y''+p(x)\,y'+q(x)\,y=g(x)$

Find $y_{1}$ and $y_{2}$ for corresponding homogeneous equation $y''+p(x)\,y'+q(x)\,y=0$

Solve the following system for $c_{1}(x)$ and $c_{2}(x)$

${\begin{cases}c_{1}'\,y_{1}+c_{2}'\,y_{2}=0\\c_{1}'\,y_{1}'+c_{2}'\,y_{2}'=g(x)\end{cases}}$

or

${\begin{aligned}c_{1}(x)&=-\int _{x_{0}}^{x}{\frac {y_{2}(s)\,g(s)}{W(y_{1},y_{2})(s)}}\,\mathrm {d} s&c_{2}(x)&=\int _{x_{0}}^{x}{\frac {y_{1}(s)\,g(s)}{W(y_{1},y_{2})(s)}}\,\mathrm {d} s\end{aligned}}$

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 $c_{2}$

${\begin{aligned}c_{1}(x)&=-\int \sin {x}\,{\frac {1}{\sin {x}}}\,\mathrm {d} x\\&=-x+A\\c_{2}(x)&=\int \cos {x}\,{\frac {1}{\sin {x}}}\,\mathrm {d} x\\&=\ln {(\sin {x})}+B\end{aligned}}$

General solution is

${\begin{aligned}y&=\left(-x+A\right)\,\cos {x}+\left(\ln {(\sin {x})}+B\right)\sin {x}\\&=-x\,\cos {x}+A\,\cos {x}+\ln {(\sin {x})}\,\sin {x}+B\,\sin {x}\end{aligned}}$