- #1
naspek
- 181
- 0
hey there..
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?
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?