# Reduction of order
#reduction_of_order 
Consider the equation:
$$y''+p(x)y'+q(x)y=f(x)$$this equation (1) does not have a general solution procedure always.
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.