Using substitution in differential eq

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
vipertongn
Messages
97
Reaction score
0
xdy/dx+y=1/y^2:using substitution in differential eq

Homework Statement


solve using substitution
xdy/dx+y=1/y^2


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:
Physics news on Phys.org
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:

[tex]\frac 1 3 x u' + u = 1[/tex]

Once you put that in correct form and find the integrating factor, you shouldn't have any ex terms.
 
The DE is separable, which is something you should check for at the start in problems like this.

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