[SOLVED] seperable equation

**nameck** · March 24th 2010, 07:01 AM

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 + C = √x^2 – 16 + C
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?

**Prove It** · March 24th 2010, 02:59 PM

Originally Posted by nameck

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 + C = √x^2 – 16 + C
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?

You have gotten it to the step

$\int{2y\,dy} = \int{\frac{x}{\sqrt{x^2 - 16}}\,dx}$ correctly.

$y^2 + C_1 = \frac{1}{2}\int{2x(x^2 - 16)^{-\frac{1}{2}}\,dx}$

$y^2 + C_1 = \frac{1}{2}\int{u^{-\frac{1}{2}}\,du}$ after making the substitution $u = x^2 - 16$

$y^2 + C_1 = u^{\frac{1}{2}} + C_2$

$y^2 + C_1 = \sqrt{x^2 - 16} + C_2$

$y^2 = \sqrt{x^2 - 16} + C$ , where $C = C_2 - C_1$

Now use your initial condition to find $C$

$y(5) = 2$

so $2^2 = \sqrt{5^2 - 16} + C$

$4 = 3 + C$

$C = 1$ .

Therefore the solution is:

$y^2 = \sqrt{x^2 - 16} + 1$

or if you like

$y = \pm \sqrt{\sqrt{x^2 - 16} + 1}$ .

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