# Solve $x^2 (a - bx) {d^2y \over dx^2} - x (5a - 4bx) {dy \over dx} + 3(2a - bx)y = 6a^2$

What do I substitute here $$x^2 (a - bx) {d^2y \over dx^2} - x (5a - 4bx) {dy \over dx} + 3(2a - bx)y = 6a^2$$ to get it into this form? $$u^2 {dv^2 \over du^2} + P_1 u {dv \over du} + P_2 v = F(u)$$ The solution (according to answer sheet): $y(a-bx) = Ax^2 + Bx^3 + C$

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The "solution" does not appear to be a solution. Perhaps there's a typo somewhere ... –  user26872 Jul 28 '12 at 18:00
That looks somehow similar to an Euler Cauchy equation. I'll give it a try. –  Pedro Tamaroff Jul 28 '12 at 19:16
@PeterTamaroff Yup ... it's Euler Cauchy equation. The problem is how to reduce it into standard form. –  Santosh Linkha Jul 28 '12 at 19:17
Actually it isn't really a Euler Cauchy equation. A EC eqn. is of the form $$(a_0 x^n D^n+\cdots a_{n-1}xD+a_n)y=F(x)$$ –  Pedro Tamaroff Jul 28 '12 at 19:27
In fact this belongs to an ODE of the form eqworld.ipmnet.ru/en/solutions/ode/ode0226.pdf. –  doraemonpaul Jul 28 '12 at 22:20

When either $a$ or $b$ but not both equal to $0$ , it is clear that $x^2(a-bx)\dfrac{d^2y}{dx^2}-x(5a-4bx)\dfrac{dy}{dx}+3(2a-bx)y=6a^2$ will immediately reduce to and ODE of the form $x^2\dfrac{d^2y}{dx^2}+P_1x\dfrac{dy}{dx}+P_2y=F(x)$ , i.e. Euler Cauchy equation.

So the remaining problem is the case that $a\neq0$ and $b\neq0$ .

$x^2(a-bx)\dfrac{d^2y}{dx^2}-x(5a-4bx)\dfrac{dy}{dx}+3(2a-bx)y=6a^2$

$x^2\left(\dfrac{bx}{a}-1\right)\dfrac{d^2y}{dx^2}-x\left(\dfrac{4bx}{a}-5\right)\dfrac{dy}{dx}+\left(\dfrac{3bx}{a}-6\right)y=-6a$

Which belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0226.pdf.

According to the site, we can deduce the standard method of solving this ODE should be like this:

Let $u=\dfrac{bx}{a}$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=\dfrac{b}{a}\dfrac{dy}{du}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{b}{a}\dfrac{dy}{du}\right)=\dfrac{b}{a}\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)=\dfrac{b}{a}\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}=\dfrac{b}{a}\dfrac{d^2y}{du^2}\dfrac{b}{a}=\dfrac{b^2}{a^2}\dfrac{d^2y}{du^2}$

$\therefore\left(\dfrac{au}{b}\right)^2(u-1)\dfrac{b^2}{a^2}\dfrac{d^2y}{du^2}-\dfrac{au}{b}(4u-5)\dfrac{b}{a}\dfrac{dy}{du}+(3u-6)y=-6a$

$u^2(u-1)\dfrac{d^2y}{du^2}-u(4u-5)\dfrac{dy}{du}+(3u-6)y=-6a$

Let $y=u^nv$ ,

Then $\dfrac{dy}{du}=u^n\dfrac{dv}{du}+nu^{n-1}v$

$\dfrac{d^2y}{du^2}=u^n\dfrac{d^2v}{du^2}+nu^{n-1}\dfrac{dv}{du}+nu^{n-1}\dfrac{dv}{du}+n(n-1)u^{n-2}v=u^n\dfrac{d^2v}{du^2}+2nu^{n-1}\dfrac{dv}{du}+n(n-1)u^{n-2}v$

$\therefore u^2(u-1)\left(u^n\dfrac{d^2v}{du^2}+2nu^{n-1}\dfrac{dv}{du}+n(n-1)u^{n-2}v\right)-u(4u-5)\left(u^n\dfrac{dv}{du}+nu^{n-1}v\right)+(3u-6)u^nv=-6a$

$(u-1)\left(u^{n+2}\dfrac{d^2v}{du^2}+2nu^{n+1}\dfrac{dv}{du}+n(n-1)u^nv\right)-(4u-5)\left(u^{n+1}\dfrac{dv}{du}+nu^nv\right)+(3u-6)u^nv=-6a$

$(u-1)\left(u^2\dfrac{d^2v}{du^2}+2nu\dfrac{dv}{du}+n(n-1)v\right)-(4u-5)\left(u\dfrac{dv}{du}+nv\right)+(3u-6)v=-6a$

$u^2(u-1)\dfrac{d^2v}{du^2}+2nu(u-1)\dfrac{dv}{du}+n(n-1)(u-1)v-u(4u-5)\dfrac{dv}{du}-n(4u-5)v+(3u-6)v=-6a$

$u^2(u-1)\dfrac{d^2v}{du^2}+u((2n-4)u-2n+5)\dfrac{dv}{du}+((n^2-5n+3)u-(n^2-6n+6))v=-6a$

In order to simplify the ODE to becomes a Gaussian hypergeometric equation, we should take $n^2-6n+6=0$ , i.e. $n=3\pm\sqrt{3}$

Choose $n=3+\sqrt{3}$ , the ODE becomes

$u^2(u-1)\dfrac{d^2v}{du^2}+u((2+2\sqrt{3})u-1-2\sqrt{3})\dfrac{dv}{du}+\sqrt{3}uv=-6a$

$u(u-1)\dfrac{d^2v}{du^2}+((2+2\sqrt{3})u-1-2\sqrt{3})\dfrac{dv}{du}+\sqrt{3}v=-\dfrac{6a}{u}$

In fact this belongs to a Gaussian hypergeometric equation with parameters $\alpha=\dfrac{1+2\sqrt{3}+\sqrt{13}}{2}$ , $\beta=\dfrac{1+2\sqrt{3}-\sqrt{13}}{2}$ and $\gamma=1+2\sqrt{3}$ .

So the complementary solution is $y_c=C_1x^{3+\sqrt{3}}{}_2F_1\left(\dfrac{1+2\sqrt{3}+\sqrt{13}}{2},\dfrac{1+2\sqrt{3}-\sqrt{13}}{2};1+2\sqrt{3};\dfrac{bx}{a}\right)+C_2x^{3-\sqrt{3}}{}_2F_1\left(\dfrac{1-2\sqrt{3}-\sqrt{13}}{2},\dfrac{1-2\sqrt{3}+\sqrt{13}}{2};1-2\sqrt{3};\dfrac{bx}{a}\right)$

But the form of the particular solution is too ugly and should be unavoidable to find it by Variation of parameters.

I think the book should make deep introspection about the reliability of EqWorld.

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I will post a partial answer, as I might have made a sign error or been taken astray at some point. The method amounts to finding some sort of integrating factor for the equation. It seems reasonable to start with the change of the dependent variable: $$u=a-bx$$ $$x=-\frac{1}{b}\left(u-a\right)$$ $$\frac{dy}{dx}=-b\frac{dy}{du}$$ $$\frac{d^{2}y}{dx^{2}}=-b^{2}\frac{d^{2}y}{du^{2}}$$ Rewrite the equation: $$\left(u-a\right)^{2}u\frac{d^{2}y}{du}+\left(u-a\right)\left(4u+a\right)\frac{dy}{du}+3\left(u+a\right)y=6a^{3}$$ Split the $\frac{dy}{du}$ term and regroup: $$\left(u-a\right)^{2}u\frac{d^{2}y}{du}+\left(u-a\right)\left(u+a\right)\frac{dy}{du}+3\left(u-a\right)u\frac{dy}{du}+3\left(u+a\right)y=6a^{3}$$ $$\left(u-a\right)\left[\left(u-a\right)u\frac{d^{2}y}{du^{2}}+\left(u+a\right)\frac{dy}{du}\right]+3\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]=6a^{3}$$ Differentiating the second square bracket: $$\frac{d}{du}\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]=\left(2u-a\right)\frac{dy}{du}+\left(u-a\right)u\frac{d^{2}y}{du^{2}}+y+\left(u+a\right)y\qquad(1)$$ We can express the first one as follows: $$\left(u-a\right)u\frac{d^{2}y}{du^{2}}+\left(u+a\right)\frac{dy}{du}=\frac{d}{du}\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]-\left(2u-a\right)\frac{dy}{du}-y$$ Inserting in the equation: $$\left(u-a\right)\frac{d}{du}\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]-\left(u-a\right)\left[\left(2u-a\right)\frac{dy}{du}+y\right]+3\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]=6a^{3}$$ $$\left(u-a\right)\frac{d}{du}\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]+\left(u^{2}-a^{2}\right)\frac{dy}{du}+2ay=6a^{3}$$

Now using the product rule we obtain: $$\left(u-a\right)\frac{d}{du}\left[\left(u-a\right)u\frac{dy} {du}+\left(u+a\right)y\right]=\frac{d}{du}\left[\left(u-a\right)^{2}u\frac{dy}{du}+\left(u^{2}-a^{2}\right)y\right]-\left(u-a\right)u\frac{dy}{du}-\left(u+a\right)y$$ The original equation is then expressed as follows: $$\frac{d}{du}\left[\left(u-a\right)^{2}u\frac{dy}{du}+\left(u^{2}-a^{2}\right)y\right]+\left(u-a\right)\left[a\frac{dy}{du}-y\right]=6a^{3}$$

Now it might have made sense in (1) to look for $y$ in some specific form, say $y=rs$ and then impose suitable conditions on the factors.

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