MATH 308 Lecture 13

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Lecture Notes

Wronskian

$W(y_{1},y_{2})=y_{1}\,y_{2}'-y_{2}\,y_{1}'$

$W(y_{1},y_{2})\neq 0$ iff $y_{1}$ and $y_{2}$ form a fundamental set of solutions. Any solution is of form $y=c_{1}\,y_{1}+c_{2}\,y_{2}$

Abel's Theorem

$W(y_{1},y_{2})(t)=C\,\mathrm {e} ^{-\int p(t)\,\mathrm {d} t}$

Exercise 14

$y''+p(x)\,y'+q(t)\,y=0$

$y_{1}$ and $y_{2}$ are two solutions, and they reach the maximum at the same point $t_{0}$. Are they a fundamental set of solutions?

Evaluate $W(y_{1},y_{2})(t_{0})=y_{1}(t_{0})\,y_{2}'(t_{0})-y_{2}(t_{0})\,y_{1}'(t_{0})=y_{1}(t_{0})\cdot 0-y_{2}(t_{0})\cdot 0=0$.

Therefore, since $W(y_{1},y_{2})=0$, $y_{1}$ and $y_{2}$ cannot be a fundamental set of solutions.

Chapter 3.4: Reduction of Order method

Given linear, homogeneous differential equation and solution $y_{1}$:

Find solutions of the form $y(x)=\lambda (x)\,y_{1}(x)$

Differentiate $y$ as many times as needed:

${\begin{aligned}y'&=\lambda '\,y_{1}+\lambda \,y_{1}'\\y''&=\lambda ''\,y_{1}+2\lambda '\,y_{1}+\lambda \,y_{1}''\\y^{(n)}&=\ldots \end{aligned}}$

Plug the derivatives into the original differential equation. The final term should always be $0\lambda$.

Solve the reduced order differential equation for $\lambda '$ and plug back into original $y$.

Exercise 15

$y'''-2y''-5y'+6y=0$

We are told that $e^{x}$ is a solution, but we want to find more solutions of the form $y=\lambda (x)\,\mathrm {e} ^{x}$, where the $\mathrm {e} ^{x}$ is the solution.

${\begin{aligned}y'(x)&=\lambda '(x)\,\mathrm {e} ^{x}+\lambda (x)\,\mathrm {e} ^{x}\\y''(x)&=\lambda ''(x)\,\mathrm {e} ^{x}+2\lambda '(x)\,\mathrm {e} ^{x}+\lambda (x)\mathrm {e} ^{x}\\y'''(x)&=\lambda '''(x)\,\mathrm {e} ^{x}+3\lambda ''(x)\,\mathrm {e} ^{x},+3\lambda '(x)\,\mathrm {e} ^{x}+\lambda (x)\,\mathrm {e} ^{x}\end{aligned}}$

Therefore, our differential equation becomes

$\lambda '''(x)\,\mathrm {e} ^{x}+3\lambda ''(x)\,\mathrm {e} ^{x}+3\lambda '(x)\,\mathrm {e} ^{x}+\lambda (x)\,\mathrm {e} ^{x}+2\left(\lambda ''(x)\,\mathrm {e} ^{x}+2\lambda '(x)\,\mathrm {e} ^{x}+\lambda (x)\mathrm {e} ^{x}\right)-5\left(\lambda '(x)\,\mathrm {e} ^{x}+\lambda (x)\,\mathrm {e} ^{x}\right)+5\lambda (x)\,\mathrm {e} ^{x}=0$

This can be reduced to

${\begin{aligned}\lambda '''(x)\,\mathrm {e} ^{x}+\lambda ''\,\mathrm {e} ^{x}-6\lambda '(x)\,\mathrm {e} ^{x}+0\lambda &=0\\\lambda '''(x)+\lambda ''(x)-6\lambda '(x)&=0\end{aligned}}$

The $0\lambda$ is always the case. Notice that $\lambda '(x)$ is a solution to a second order differential equation. Let $y(x)=\lambda '(x)$:

$y''+y'-6y=0$

Characteristic equation is $r^{2}+r-6=0$, roots are $r\in \{-3,2\}$.

${\begin{aligned}\lambda '&=c_{1}\,\mathrm {e} ^{-3t}+c_{2}\,\mathrm {e} ^{2t}\\\lambda &=-{\frac {c_{1}}{3}}\,\mathrm {e} ^{-3t}+{\frac {c_{2}}{2}}\,\mathrm {e} ^{2x}+C\end{aligned}}$

Therefore, our solution is

$y=\lambda (x)\,\mathrm {e} ^{x}=\left(-{\frac {c_{1}}{3}}\right)\,\mathrm {e} ^{-2x}+\left({\frac {c_{2}}{2}}\right)\,\mathrm {e} ^{3x}+C\,\mathrm {e} ^{x}$

Remark

$y''-4y'+4y=0$

Characteristic equation has root $r=2$ with multiplicity 2.

$y_{1}=\mathrm {e} ^{2x}$

Find second solution $y_{2}=\lambda (x)\,\mathrm {e} ^{2x}$

${\begin{aligned}y_{2}'&=\lambda '(x)\,\mathrm {e} ^{2x}+2\lambda '(x)\,\mathrm {e} ^{2x}\\y_{2}''&=\lambda ''(x)\,\mathrm {e} ^{2x}+4\lambda '(x)\,\mathrm {e} ^{2x}+4\lambda (x)\,\mathrm {e} ^{2x}\end{aligned}}$

${\begin{aligned}0&=y_{2}''-4y_{2}'+y_{2}\\&=\left(\lambda ''(x)\,\mathrm {e} ^{2x}+4\lambda '(x)\,\mathrm {e} ^{2x}+4\lambda (x)\,\mathrm {e} ^{2x}\right)+4\left(\lambda '(x)\,\mathrm {e} ^{2x}+2\lambda '(x)\,\mathrm {e} ^{2x}\right)+4\lambda (x)\,\mathrm {e} ^{2x}\end{aligned}}$

Therefore $\lambda ''(x)\,\mathrm {e} ^{2x}=0$, giving $\lambda (x)=a\,x+b$