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)

Question

asked Jun 10, 2020 125k views

1 Answer

Ligowsky · Answer 1 · 2020-06-14T01:51:41+0000

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
, 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.