forked from Sasserisop/MATH201
23 lines
1.7 KiB
Markdown
23 lines
1.7 KiB
Markdown
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# Variation of parameters
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$ay''+by'+cy=f(t)$
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1) $y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}(t)$ <- h is homogenous, ie: $f(t)=0$
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Lagrange proposed this method to find the particular solution $y_{p}$:
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$y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ <- btw $y_{1}$ and $y_{2}$ are often called a fundamental pair.
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we put $y_p$ into the equation and make it equal to the RHS. To do so, find the derivatives first:
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$y'_{p}=v_{1}'y_{1}+v_{1}y_{1}'+v_{2}'y_{2}+v_{2}y_{2}'$
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to avoid second derivatives in the equation and problems with uniqueness, Lagrange imposed:
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1) $v_{1}'y_{1}+v_{2}'y_{2}=0$ this simplifies our work down the road as well.
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$y'_{p}=\cancel{ v_{1}'y_{1} }+v_{1}y_{1}'+\cancel{ v_{2}'y_{2} }+v_{2}y_{2}'$
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so $y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}''$
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now we plug into the second order equation:
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$a(v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}'')+b(v_{1}y_{1}'+v_{2}y_{2}')+c(v_{1}y_{1}+v_{2}y_{2})=f(t)$
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$v_{1}(\cancel{ ay_{1}''+ by_{1}' +cy_{1} })+v_{2}(\cancel{ ay_{2}''+ by_{2}'+cy_{2} })+a(v_{1}'y_{1}'+v_{2}'y_{2}')$
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Both $y_{1}$ and $y_{2}$ are solutions to the homogenous counterpart. So the first two terms above equal to zero.
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2) $v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a}$
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we now have a system of two equations:
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$\det \begin{pmatrix}y_{1} & y_{2} \\y_{1}'& y_{2}'\end{pmatrix}$ = Wronskian = $W[y_{1},y_{2}]\ne 0$
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by definition $y_1$ and $y_2$ are linearly independent solutions so the above can never be 0!
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$v_{1}'=-\frac{{f(t)y_{2}t}}{aW[y_{1},y_{2}]}$; $v_{2}'=\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]}$ <- integrate both sizes to get $v_{1}$ and $v_{2}$. When integrating, you don't need to add a generic constant.
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Finally, your solution is: $y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)$ where you get $y_{1},y_{2}$ from your homogenous solution.
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