# Initial Value Problem

• September 5th 2011, 06:32 PM
VitaX
Initial Value Problem
Find the solution to the initial value problem in explicit form.

$y' = xy^3(1 + x^2)^{-\frac{1}{2}}$ ; $y(0) = 1$

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

$-\frac{1}{2y^2} = (1 + x^2)^{\frac{1}{2}} + C$

$-2y^2 = \frac{1}{1 + x^2)^{\frac{1}{2}}} + C$

$-y = \left(\frac{1}{2(1 + x^2)^{\frac{1}{2}}}\right)^{\frac{1}{2}} + C$

$y = - \frac{1}{\left(2(1 + x^2)^{\frac{1}{2}}\right)^{\frac{1}{2}}} \right) + C$

I think I screwed up somewhere because when I go to solve for C I get $C = 1 + \frac{1}{2^{\frac{1}{2}}}$.

The Book's answer in explicit form is $y = \left[3 - 2(1 + x^2)^{\frac{1}{2}}\right]^{-\frac{1}{2}}$
• September 5th 2011, 06:41 PM
SammyS
Re: Initial Value Problem
Quote:

Originally Posted by VitaX
Find the solution to the initial value problem in explicit form.

$y' = xy^3(1 + x^2)^{-\frac{1}{2}}$ ; $y(0) = 1$

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

$-\frac{1}{2y^2} = (1 + x^2)^{\frac{1}{2}} + C$

You could solve for C at this point.

The next equation is incorrect.
Quote:

$-2y^2 = \frac{1}{(1 + x^2)^{\frac{1}{2}}} + C$
It should be:
$-2y^2 = \frac{1}{(1 + x^2)^{\frac{1}{2}}+ C}$

Quote:

$-y = \left(\frac{1}{2(1+x^2)^{\frac{1}{2}}} \right)^{\frac{1}{2}} + C$

$y = -\left(\frac{1}{2(1+x^2)^{\frac{1}{2}}} \right)^{\frac{1}{2}} + C$

I think I screwed up somewhere because when I go to solve for C I get $C = 1 + \frac{1}{2^{\frac{1}{2}}}$.

The Book's answer in explicit form is $y = \left[3 - 2(1 + x^2)^{\frac{1}{2}}\right]^{-\frac{1}{2}}$
• September 5th 2011, 06:55 PM
VitaX
Re: Initial Value Problem
Why does C go in the denominator? I thought that since you are taking the reciprocal of both sides that $\frac{1}{C}$ will just be another constant, C, so I left it alone. It's these types of things that are confusing me, the work is easy but knowing when C gets changed or not is a bit confusing here. Why exactly is C effected here if there is no variable effecting it?
• September 5th 2011, 08:11 PM
SammyS
Re: Initial Value Problem
It's pretty basic algebra !

$\frac{1}{a+b}\ne\frac{1}{a}+\frac{1}{b}$
• September 5th 2011, 09:09 PM
VitaX
Re: Initial Value Problem
Yeah I screwed up on that part.

My main question is this though, after isolating $y$ does it become this: $y = \left [ \frac{-1}{2(1+x^2)^{\frac{1}{2}} + C}\right]^{\frac{1}{2}}$ or this: $y = \left [ \frac{-1}{2(1+x^2)^{\frac{1}{2}} + 2C}\right]^{\frac{1}{2}}$.

I know this is hard to explain but I'll show a link to why I'm a bit confused with the subject of C.

Problem 7 Page 4

If you go to page 4 and problem 7, you can see where he raises both sides to $e$ and $e^C$ simply becomes C. So does $2C$ become $C$ here or no? I only ask that because both $e$ and $2$ are constants, so isn't that the reason why in that problem it is written as $C$? The math is easy, but working with the C is confusing me with situations like this.
• September 6th 2011, 05:29 PM
SammyS
Re: Initial Value Problem
He uses C (upper case) as the original constant and c (lower case) as the 'derived' constant. Therefore, he has c = e^C, where ^ denotes the exponentiation operation.

In your case, you just need to be consistent about which way you use the constant, C, both when you plug-in the initial conditions to solve for C and then when you substitute that value of C back into your expression for y.

You should get the same answer for y either way.