Initial Value Problems

• Sep 15th 2009, 02:19 PM
Smac
Initial Value Problems
1. y(dy/dx) = (y^3 - 1) (x/y) e^(ax^2) ; y(0) = -1
a > 0 is fixed constant

I solved to split up the x and y components and got:
(y^2)/(y^3 -1) dy = (xe^ax^2) dx
n(y) = m(x) (using equation above i made this one)
m(x) - n(x) = 0

then dm/dx - dn/dy = C where C=constant
soo...
3Ln(y^3 -1) - (e^(ax^2))(1- 1/4a^2) = C

.... solve for y....

y = cube rt (e^(1/3)((e^(ax^2))(x-1/4a^2)+C)

sorry for all the "( )" i know it makes it hard to read.

then when i plugged in y(0) = -1 i couldnt get it to work because i ended with...

-1 = e^-(1/12a^2) + C

and the e function cant be negative right? help!
• Sep 15th 2009, 02:38 PM
pickslides
$\displaystyle \frac{y^2}{y^3-1} ~dy = xe^{ax^2}~dx$

At this point could you not try to integrate both sides, left hand side by substitution and right hand side by parts?
• Sep 15th 2009, 03:07 PM
Smac
i am not quite sure how to do it without differentiating...
• Sep 15th 2009, 03:33 PM
pickslides
We need to consider the following

$\displaystyle \int \frac{y^2}{y^3-1} ~dy = \int xe^{ax^2}~dx$

To solve this we need to do the opposite to differeniating.

Do you know of this $\displaystyle = \int x^{n}~dx= \frac{x^{n+1}}{n+1}+c$ ?

This is a simple idea on how to solve such an equation, but in your case it's not quite this simple. Have you heard of integration by parts or by substitution?
• Sep 15th 2009, 04:50 PM
Smac
that is what i did. but i was just verifying that the rest was good as well. that i set everything up right. its just hard to tell lol. i used substitution on the 'y' side and by parts on the 'x' side. but i got -1 = e to a power, and the e exponential can never be negative, right?
• Sep 15th 2009, 04:59 PM
pickslides
You can have e such that $\displaystyle e^{-a} = \frac{1}{e^{a}}$