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Ordinary differentiation equations (ODE)

  • Thread starter naspek
  • Start date
  • #1
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?
 

Answers and Replies

  • #2
33,518
5,196
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
In the step above, where's your constant of integration?
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?
 
  • #3
181
0
ok...
2yy’ = x / √x^2 – 16
2y dy/dx = x / √x^2 – 16
∫2y dy = ∫ x / √x^2 – 16 dx
y^2 + C = √x^2 – 16 + C
 
  • #4
33,518
5,196
ok...
2yy’ = x / √x^2 – 16
2y dy/dx = x / √x^2 – 16
∫2y dy = ∫ x / √x^2 – 16 dx
y^2 + C = √x^2 – 16 + C
It wouldn't be the same constant on both sides. If it were, you could subtract C from both sides to get back to what you had.
You could do this:
y^2 + A = √(x^2 – 16) + B

Note that I added parentheses, which are necessary here.
Or you could just put the constant on one side as

y^2 = √(x^2 – 16) + C, where C = B - A. This is how it's usually done.

Now use your initial condition, y(5) = 2, to find C.
 

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