Solving a differential equation

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Homework Statement


Solve
(1+bx)y''(x)-ay(x)=0

Homework Equations


The Attempt at a Solution



I'm used to solving homogeneous linear ODE's where you form a characteristic equation and solve that way, here there is the function of x (1+bx) so how does that change things?
 
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Would dividing both sides by 1+bx help?
 
Ok so if I did that then what? I can define a characteristic equation such that

r^2-\frac{a}{1+bx}=0

and r=\pm\sqrt{\frac{a}{1+bx}}

where b^2-4ac = 4a(1+bx) > 0

so a solution is y=ce^{rx} but that doesn't satisfy the ODE so its not correct?
 
You can't use constant coefficient methods on a DE like this with variable coefficients. Perhaps there is a clever substitution that will help, or maybe not. Problems like this are typically solved with series solutions, especially if you know ##a## and ##b##. Where did this equation come from? If it's from a text, the recent material may give a hint how to solve it.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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