forked from Sasserisop/MATH201
41 lines
2.0 KiB
Markdown
41 lines
2.0 KiB
Markdown
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# Reduction of order
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#reduction_of_order
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Consider the equation:
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$$y''+p(x)y'+q(x)y=f(x)$$this equation (1) does not have a general solution procedure always.
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But, if $y_{1}(x)$ solves the homogenous counterpart: $y''+p(x)y'+q(x)y=0$
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then we can find the general solution to the non homogenous equation (1) by guessing it in the form: $y(x)=v(x)y_{1}(x)$
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let's calculate the derivatives wrt. x:
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$y'=v'y_{1}+vy_{1}'$
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$y''=v''y_{1}+2v'y_{1}'+vy_{1}''$
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plugging in:
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$(v''y_{1}+2v_{1}'y_{1}'+y_{1}''v)+p(x)(v'y_{1}+vy_{1}')+q(x)vy_{1}=f(x)$
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$v\cancelto{ 0 }{ (y_{1}''+p(x)y_{1}'+q(x)y_{1}) }+v''y_{1}+(2y_{1}'+p(x)y_{1})v'=f(x)$
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$y_{1}v''+(2y_{1}'+p(x)y_{1})=f(x)$
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$v''+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)v'=\frac{f(x)}{y_{1}}$
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substitute $v'=u$
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$u'+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)u=\frac{f(x)}{y_{1}}$<- This is now a linear first order equation #de_L_type2
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This can be solved with prior tools now, We compute the integrating factor $\mu$
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$\mu=e^{\int(2y_{1}'/y_{1}+p)dx}=e^{\ln(y_{1}^2)}e^{\int P(x) \, dx}=y_{1}^2\cdot e^{\int p(x) \, dx}$
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From there, continue on as you would with any linear first order equation.
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Isn't this nice? some kind of magic. We made some guesses and we arrived somewhere.
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#ex #second_order_nonhomogenous #reduction_of_order
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Find the general solution to the equation:
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$$y''+4xy'+(4x^2+2)y=8e^{-x(x+2)}$$
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if $y_{1}(x)=e^{-x^2}$ is one solution.
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> Ouch look at those x terms. And the exponent on the RHS. This isn't even in Cauchy Euler form!
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we guess the general solution will be in the form: $y(x)=v(x)y_{1}=v(x)e^{-x^2}$
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substitute: $v'=u$
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plug into formula derived above:
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$u'+\left( \frac{2y_{1}'}{y_{1}}+4x \right)u=\frac{8{e^{-x^2}e^{-2x}}}{e^{-x^2}}$ <-(note: $p(x)=4x$)
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$u'+\underbrace{ \left( \frac{2{e^{-x^2}(-2x)}}{e^{-x^2}}+4x \right) }_{ =0 }u=8e^{-2x}$
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$u'=8e^{-2x}$
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> Lucky us! This is just a separable equation. No need to treat it like a linear equation.
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$u=-4e^{-2x}+c_{1}$
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$v'=u=-4e^{-2x}+c_{1}$
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$v(x)=2e^{-2x}+c_{1}x+c_{2}$
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general solution:
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$$y(x)=v(x)y_{1}(x)=(2e^{-2x}+c_{1}x+c_{2})e^{-x^2}$$
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We are done.
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