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Solve the initial value problem?

Solve the initial value problem: y'=2x/y, y(0)=2.

dy/dx=2x/y

y dy=2x dx

y^2/2=x^2+c

Now what?

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1 Answer

y^2/2=x^2+c

Plug in y(0) = 2,

2^2 / 2 = 0^2 + c

c = 2

Answer: y = sqrt[2(x^2 + 2)]