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Initial value problem: $(1-x^2)\dfrac{dy}{dx}=2xy$, $y(0)=5$. Differential equations is one of my toughest subjects. Could someone help with solving

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This is separable. – Ishan Banerjee Mar 6 '13 at 12:00
So what do I need to do? Integrate – user58315 Mar 6 '13 at 12:07
The thing is you have the differentiation of $(1-x^2)$ in the numerator on the RHS once you separate the variables to LHS and RHS. – lsp Mar 6 '13 at 12:25
up vote 3 down vote accepted

Getting all '$x$' variables on side and all '$y$' variables on the other side, we get :

$(dy/y) = 2x.dx/(1-x^2)$ ----(1)

Let $1-x^2 = p;$

$ \implies -2x.dx = dp$ ----(2)

$(2)$ in $(1)$ $\implies $

$(dy/y) = (-dp/p)$

Integrating on both sides we get :$ ln(y) + ln(p) = C \implies **ln(y) + ln(1-x^2) = C**$

To make the answer look more simpler, we can take the constant after integration as $ln(C)$ instead on just $C$. This gives the following equation:

$ ln(y) + ln(1-x^2) = ln(C)$

$ \implies ln(y) = ln(C) - ln(1-x^2) \implies ln(y) = ln(C/(1-x^2)) $

$ \implies y = C/(1-x^2)$

Hope the answer is clear !

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Separating the variables, $\frac{dy}{y}=\frac{2x}{1-x^2}$.Integrating, $\log{|y|}+\log(|1-x^2|)=C$ for some constant $C$. I hope you can take over from here.

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$$(1-x^2)\frac{dy}{dx}-2xy=0\implies (1-x^2)\frac{d(y)}{dx}+y\frac{d(1-x^2)}{dx}=0$$

Remembering product rule, it can be written as $$\frac{d(y(1-x^2))}{dx}=0\implies y(1-x^2)=c$$

Then plug the initial condition to get constant $c$.

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