- #1

naspek

- 181

- 0

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 of

**y(5) = 2**?