What are some approaches to solving differential equations like the one below?
$(1-x^2)y''+2xy'-2y = 0$
For instance, using power series would be one way, but what other methods exist?
Approaches to solving $(1-x^2)y''+2xy'-2y = 0$.
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ordinary-differential-equations
2 Answers
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$$(1-x^2)\frac{d^2y}{dx^2}+2x\frac{dy}{dx}-2y = 0$$ ODE of homogeneous kind. The usual change of function is $y(x)=xu(x)$
$y'=xu'+u \quad;\quad y''=xu''+2u'\quad\to\quad (1-x^2)(xu''+2u')+2x(xu'+u)-2xu=0$ $$x(1-x^2)u''+2u'=0$$ $$\frac{u''}{u'}=\frac{2}{x(x^2-1)}=-\frac{2}{x}+\frac{2x}{x^2-1}$$ $$u'=c_1\frac{x^2-1}{x^2}$$ $$u=c_1(x+\frac{1}{x})+c_2$$ $$y(x)=c_1(x^2+1)+c_2x$$
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HINT
Put $1-x^2 = t$ and differentiate and put back in your ODE