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