Problem with differential equation

jacksonjs20
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Hi could someone please explain why

(1-x^2)y'' = 2xy'-2y

Many thanks in advance.
 
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jacksonjs20 said:
Hi could someone please explain why

(1-x^2)y'' = 2xy'-2y

Many thanks in advance.

Your question has no sens. Or the wording is incomplete.
One cannot explain why an equation is an equation.
 
Because you have not converted it to homogeneous like other people would have?
 
Solving the linear ODE : (1-x^2)y'' - 2xy'+2y = 0 is rather easy.
A first solution is obvious. Then the equation can be reduced to a first order linear ODE.
 
Thread 'Direction Fields and Isoclines'

Thread 'Direction Fields and Isoclines'

I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
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