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Solve the Following Initial Value Problem: 2XYY'+Y^2-4X^3=0. where Y(1)=2

The answer is y= sqrt((x^4+3)/x)

1 Answer

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2xyy'+y^2-4x^3=0

Let
z(x)=y(x)^2, so that
z'(x)=2y(x)y'(x) (which appears in the first term on the left side):


xz'+z=4x^3

This ODE is linear in
z, and we don't have to find any integrating factor because the left side is already the derivative of a product:


(xz)'=4x^3\implies xz=x^4+C\implies z=\frac{x^4+C}x


\implies y(x)=\sqrt{\frac{x^4+C}x}

With
y(1)=2, we get


2=√(1+C)\implies C=3

so the solution is as given in your post.

User Ligowsky
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