MATH 308 Lecture 23

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Lecture Notes

Exam Review

Fundamental Sets of Solutions

Recall that $y_1$ and $y_2$ are a fundamental set of solutions if $W(y_1,y_2)\ne 0$.

Reduction of Order

Given a differential equation and one solution $y_1$, we want to find a second solution in the form $y_2=y_1\lambda$

For example, given $(x-1)y^{″}-x{y}^{\prime }+y=0$ and solution $y_1={\mathrm{e}}^x$, we find the second solution as follows:

$\begin{array}{rl}{y}_2& =\lambda {\mathrm{e}}^x\\ {y_2}^{\prime }& ={\lambda }^{\prime }{\mathrm{e}}^x+\lambda {\mathrm{e}}^x\\ {y_2}^{″}& ={\lambda }^{″}{\mathrm{e}}^x+2{\lambda }^{\prime }{\mathrm{e}}^x+\lambda {\mathrm{e}}^x\\ 0& =(x-1)({\lambda }^{″}+2{\lambda }^{\prime }+\lambda ){\mathrm{e}}^x-x(\lambda +{\lambda }^{\prime }){\mathrm{e}}^x+\lambda {\mathrm{e}}^x\\ 0& =(x-1){\lambda }^{″}+(x-2){\lambda }^{\prime }+\cancel{0\lambda }\\ \mu & ={\mathrm{e}}^{\int \frac{x-2}{x-1}\mathrm{d}x}={\mathrm{e}}^{\int 1-\frac{1}{x-1}\mathrm{d}x}=\frac{{\mathrm{e}}^x}{x-1}\\ \frac{\mathrm{d}}{\mathrm{d}x}\left({\lambda }^{\prime }\frac{{\mathrm{e}}^x}{x-1}\right)& =0\\ {\lambda }^{\prime }\frac{{\mathrm{e}}^x}{x-1}& =C\\ {\lambda }^{\prime }& =C(x-1){\mathrm{e}}^{-x}\\ \lambda & =-C((x+1){\mathrm{e}}^{-x}-{\mathrm{e}}^{-x})\end{array}$

Variation of Parameters

Recall that for non-homogeneous differential equations, a particular solution may be found by solving for the homogeneous solution, and then using the corresponding $y_1$ and $y_2$ in the formulae:

$\begin{array}{rl}{c}_1& =-\underset{x_0}{\overset{x}{\int }}\frac{y_{2}g}{W(y_1,y_2)}\mathrm{d}x\\ c_2& =\underset{x_0}{\overset{x}{\int }}\frac{y_{1}g}{W(y_1,y_2)}\mathrm{d}x\end{array}$

Laplace Transforms

Review starting with MATH 308 Lecture 17.