# IVP and interval of validity

• Jun 22nd 2018, 03:53 AM
Vinod
IVP and interval of validity
I want to solve the following IVP and also want to find interval of validity of the solution.
$y'=\frac{xy^3}{\sqrt{1+x^2}} y(0)=-1$

Solution:
First separate and then integrate both sides.$y^{-3}dy=x(1+x^2)^{0.5}dx$

$\int y^{-3}dy=\int x(1+x^2)^{0.5} dx$

$\frac{-1}{2y^2}=\frac{\sqrt{(1+x^2)^3}}{3} +c$

Applying initial condition to get the value of c.

$\frac{-1}{2}=\frac{\sqrt{1}}{3}+c. c=\frac{-5}{6}$

The implicit solution is then

$\frac{-1}{2y^2}=\frac{\sqrt{(1+x^2)^3}}{3}-\frac56$

Now let's solve for y(x).

$\frac{1}{y^2}=\frac{-5}{3}-\frac23 \sqrt{(1+x^2)^3}$

$y^2=\frac{3}{5-2\sqrt{(1+x^2)^3}}$

$y(x)=\pm\frac{\sqrt{3}}{\sqrt{5-2\sqrt{(1+x^2)^3}}}$

Reapplying the initial condition shows us that the "-" is the correct sign. The explicit solution is then,$y(x)=\frac{-\sqrt{3}}{\sqrt{5-2\sqrt{(1+x^2)^3}}}$

I am not giving full computation for the interval of validity.
$x^2=(\frac{25}{4})^{0.3333333}-1$

Hence -0.917614<x<0.917614 is the interval of validity of the solution.

But answer provided to me do not match with my calculated answer. Would any one find here what is wrong with my calculations?
• Jun 22nd 2018, 08:31 AM
SlipEternal
Re: IVP and interval of validity
Looks correct to me. What answer was provided to you?
• Jun 22nd 2018, 08:44 AM
Vinod
Re: IVP and interval of validity
Quote:

Originally Posted by SlipEternal
Looks correct to me. What answer was provided to you?

Hello,
The explicit solution is $y(x)=\frac{1}{\sqrt{3-2\sqrt{(1+x^2)}}}$

And the interval of validity of solution is $\frac{-\sqrt{5}}{2}<x<\frac{\sqrt{5}}{2}$

Now I am putting this problem in wolfram alpha to verify the answer.
• Jun 22nd 2018, 09:24 AM
SlipEternal
Re: IVP and interval of validity
Oh, you forgot a negative exponent before integrating $x(1+x^2)^{-1/2}$