Use the Reduction of Order method to solve $$(2-x)y'''+(2x-3)y''-xy'+y=0$$ (such that $x<2$) using $u(x)=e^x$.
How do you use the method for $3$rd order... I have seen it been used on $2$nd but not higher. Please help.
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$\begingroup$It's basically the same idea: write $y(x) = Y(x) u(x)$ and substitute in to the differential equation. If $u(x)$ is a solution, the terms in $Y(x)$ will cancel out and you're left with a DE of lower order in $Y'$. In this particular case it's even nicer because the $Y'$ terms also cancel.
$\endgroup$ 3 $\begingroup$Don't forget $y=x$ is also a particular solution of the ODE.
Let $y=xu$ ,
Then $y'=xu'+u$
$y''=xu''+u'+u'=xu''+2u'$
$y'''=xu'''+u''+2u''=xu'''+3u''$
$\therefore(2-x)(xu'''+3u'')+(2x-3)(xu''+2u')-x(xu'+u)+xu=0$
$x(2-x)u'''+3(2-x)u''+x(2x-3)u''+2(2x-3)u'-x^2u'-xu+xu=0$
$x(2-x)u'''+2(x^2-3x+3)u''-(x^2-4x+6)u'=0$
Let $v=u'$ ,
Then $v'=u''$
$v''=u'''$
$\therefore x(2-x)v''+2(x^2-3x+3)v'-(x^2-4x+6)v=0$
Note that $v=e^x$ is a particular solution of the ODE.
Let $v=e^xw$ ,
Then $v'=e^xw'+e^xw$
$v''=e^xw''+e^xw'+e^xw'+e^xw=e^xw''+2e^xw'+e^xw$
$\therefore x(2-x)(e^xw''+2e^xw'+e^xw)+2(x^2-3x+3)(e^xw'+e^xw)-(x^2-4x+6)e^xw=0$
$x(2-x)(w''+2w'+w)+2(x^2-3x+3)(w'+w)-(x^2-4x+6)w=0$
$x(2-x)w''+2x(2-x)w'+x(2-x)w+2(x^2-3x+3)w'+2(x^2-3x+3)w-(x^2-4x+6)w=0$
$x(2-x)w''-2(x-3)w'=0$
$x(2-x)w''=2(x-3)w'$
$\dfrac{w''}{w'}=\dfrac{2(x-3)}{x(2-x)}$
$\int\dfrac{w''}{w'}=\int\dfrac{2(x-3)}{x(2-x)}dx$
$\int\dfrac{w''}{w'}=-\int\left(\dfrac{2}{x}+\dfrac{6}{2-x}\right)dx$
$\ln w'=-2\ln x-6\ln(2-x)+c_1$
$\ln w'=-\ln(x^2(2-x)^6)+c_1$
$w'=-\dfrac{C_1}{x^2(2-x)^6}$
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