1. ## 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!

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

At this point could you not try to integrate both sides, left hand side by substitution and right hand side by parts?

3. i am not quite sure how to do it without differentiating...

4. We need to consider the following

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

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

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

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?

5. 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?

6. You can have e such that $e^{-a} = \frac{1}{e^{a}}$