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i've got unsolved question here..

Question:

Solve the initial value problem.

2yy’ = x / √x^2 – 16 ; y(5) = 2

2yy’ = x / √x^2 – 16

2y dy/dx = x / √x^2 – 16

∫2y dy = ∫ x / √x^2 – 16 dx

y^2 = √x^2 – 16

y = (√x^2 – 16)^1/2

my answer was wrong. the answer is y = ((√x^2 – 16)+ 1)^1/2

how am i going to get the correct answer..?

and.. what is the use ofy(5) = 2?

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# Homework Help: Ordinary differentiation equations (ODE)

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