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 $W(y_{1},y_{2})\neq 0$.

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 $(x-1)\,y''-x\,y'+y=0$ and solution $y_{1}=\mathrm {e} ^{x}$, we find the second solution as follows:

${\begin{aligned}y_{2}&=\lambda \,\mathrm {e} ^{x}\\y_{2}'&=\lambda '\,\mathrm {e} ^{x}+\lambda \,\mathrm {e} ^{x}\\y_{2}''&=\lambda ''\,\mathrm {e} ^{x}+2\lambda '\,\mathrm {e} ^{x}+\lambda \,\mathrm {e} ^{x}\\0&=(x-1)\,\left(\lambda ''+2\lambda '+\lambda \right)\,\mathrm {e} ^{x}-x\,(\lambda +\lambda ')\,\mathrm {e} ^{x}+\lambda \,\mathrm {e} ^{x}\\0&=(x-1)\,\lambda ''+(x-2)\,\lambda '+{\cancel {0\lambda }}\\\mu &=\mathrm {e} ^{\int {\frac {x-2}{x-1}}\,\mathrm {d} x}=\mathrm {e} ^{\int 1-{\frac {1}{x-1}}\,\mathrm {d} x}={\frac {\mathrm {e} ^{x}}{x-1}}\\{\frac {\mathrm {d} }{\mathrm {d} x}}\left(\lambda '\,{\frac {\mathrm {e} ^{x}}{x-1}}\right)&=0\\\lambda '\,{\frac {\mathrm {e} ^{x}}{x-1}}&=C\\\lambda '&=C\,(x-1)\,\mathrm {e} ^{-x}\\\lambda &=-C\,((x+1)\,\mathrm {e} ^{-x}-\mathrm {e} ^{-x})\end{aligned}}$

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:

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

Laplace Transforms

Review starting with MATH 308 Lecture 17.