Re: Initial Value Problem

Quote:

Originally Posted by

**VitaX** Find the solution to the initial value problem in explicit form.

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

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

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

You could solve for C at this point.

The next equation is incorrect. Quote:

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

It should be:

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

Quote:

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

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

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

The Book's answer in explicit form is $\displaystyle y = \left[3 - 2(1 + x^2)^{\frac{1}{2}}\right]^{-\frac{1}{2}}$

Re: Initial Value Problem

Why does C go in the denominator? I thought that since you are taking the reciprocal of both sides that $\displaystyle \frac{1}{C}$ 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?

Re: Initial Value Problem

It's pretty basic algebra !

$\displaystyle \frac{1}{a+b}\ne\frac{1}{a}+\frac{1}{b}$

Re: Initial Value Problem

Yeah I screwed up on that part.

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

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 $\displaystyle e$ and $\displaystyle e^C$ simply becomes C. So does $\displaystyle 2C$ become $\displaystyle C$ here or no? I only ask that because both $\displaystyle e$ and $\displaystyle 2$ are constants, so isn't that the reason why in that problem it is written as $\displaystyle C$? The math is easy, but working with the C is confusing me with situations like this.

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.