# Using substitution in differential eq

1. Aug 15, 2010

### vipertongn

xdy/dx+y=1/y^2:using substitution in differential eq

1. The problem statement, all variables and given/known data
solve using substitution
xdy/dx+y=1/y^2

3. The attempt at a solution
Thanks to the people who've help me thus far. here's a bernulli problem that I'm having. I change this problem around to...
dy/dx=y^3/xy^2
xy^2dy=y^3dx

using u sub.
u=y^3
du=3y^2dy

substituted problem

1/3xdu=udx
du/dx=3xu
du/dx-3xu=0

then I get e^(integral -3x)=e^(-3x^2/2)

Here's where I'm stuck

e^(-3x^2/2)u=integral 0*e^(-3x^2/2)

doesn't that just have c? which later becomes
u=ce^(3x^2/2)
However, that's not the solution of the equation which is
y^3=1+cx^-3

Can someone explain why?

Last edited: Aug 15, 2010
2. Aug 15, 2010

### LCKurtz

There are three terms in your original DE, so you have dropped one. Also you need to be careful whether your x is in the numerator or denominator as you work. Use parentheses when there is doubt. After your substitution your DE should look like this:

$$\frac 1 3 x u' + u = 1$$

Once you put that in correct form and find the integrating factor, you shouldn't have any ex terms.

3. Aug 16, 2010

### ehild

ehild

4. Aug 16, 2010

### Staff: Mentor

The DE is separable, which is something you should check for at the start in problems like this.

The original equation is equivalent to
$$\frac{y^2 dy}{1 - y^3} = \frac{dx}{x}$$