The question:

Use the substitution to solve

given that y = 1 at x = 1.

My attempt:

=

=

=

=

Substituting initial value gives C = 3/2

Is this correct? I don't have answers for practice exams. :(

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- Jan 4th 2011, 07:38 PMGlitchDifferential equation with substitution
**The question:**

Use the substitution to solve

given that y = 1 at x = 1.

**My attempt:**

=

=

=

=

Substituting initial value gives C = 3/2

Is this correct? I don't have answers for practice exams. :( - Jan 4th 2011, 07:41 PMdwsmith
- Jan 4th 2011, 07:44 PMJhevon
No, that's not correct. You tried to solve the problem with 3 variables. You integrated both sides with respect to x while treating v as the variable too. absurdity :p

You must substitute that for dy/dx into your differential equation then proceed via separation of variables. - Jan 4th 2011, 07:45 PMProve It
No, it's not. You need to get a DE that does not have any reference to the variable .

.

Make the substitution . Then and and the DE becomes

where

. - Jan 4th 2011, 07:46 PMRandom Variable
which implies by the chain rule (and product rule).

- Jan 4th 2011, 07:46 PMChris L T521
I'm sorry to say that this is not correct, because you're integrating with respect to the wrong variable (v instead of x).

First apply the substitution that's given to you. . So now we need to rewrite :

. Since , we end up with

Thus, the ODE becomes . This is now a separable equation.

Can you proceed?

EDIT: Man, am I slow!! - Jan 4th 2011, 07:49 PMnehme007
You are integrating with respect to v when you should be integrating with respect to x. You need to make the substitution and solve the resulting equation for v using separation of variables. Once you have v you can get y since y = vx.

- Jan 4th 2011, 11:00 PMGlitch
Thanks guys!

With the 3C in the solution, is it fair to just write 'C' since it's still an unknown constant? - Jan 5th 2011, 03:42 AMProve It