MATH201/content/Reduction of order (lec 11).md

# 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_linear
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.
## What you need to remember:
#remember
I know memorizing formulas robs the richness of mathematics, but that is just the nature of test taking imo. If you want to minimize the amount of work to the lowest possible level, this would be the fastest algorithm (that I know of), it's a little heavy on memorization:
1) $y''+p(x)y'+q(x)y=f(x)$
2) $u'+\left( \frac{2y_{1}'}{y_{1}}+p(x) \right)u=\frac{f(x)}{y_{1}}$ <- Notice, if the coefficient for the $u$ term is $0$, you can treat the equation as a separable equation to minimize computation (integrate both sides to get u, then move on to step 5). Otherwise, move on to step 3.
3) $\mu(x)=y_{1}^2\cdot e^{\int p(x) \, dx}$ <- where $y_{1}$ is one of your homogenous solutions.
4) $u(x)=\frac{1}{\mu(x)}\int \mu(x) \frac{f(x)}{y_{1}} dx$ <-don't confuse $q(x)$ with $\frac{f(x)}{y_{1}}$
5) $y(x)=y_{1}(x)\int u(x) dx$ is your general solution.

Notice the gotcha, we have three nested integrals. If you notice any of the integrals is difficult to compute, you might have to give up and use another method like #mouc or #voparam . No method is better than the other. For example, solving $y''+y'+y=xe^x$ is trivial using #mouc but practically impossible using #reduction_of_order .

# Reduction of order example
#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.

integrating both sides:
$u=-4e^{-2x}+c_{1}$
$v'=u=-4e^{-2x}+c_{1}$
integrating both sides again:
$v(x)=2e^{-2x}+c_{1}x+c_{2}$
general solution:
$$y(x)=v(x)y_{1}(x)=(2e^{-2x}+c_{1}x+c_{2})e^{-x^2}$$
We are done.