Using substitution in differential eq

  • Thread starter vipertongn
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  • #1
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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:

Answers and Replies

  • #2
LCKurtz
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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.
 
  • #3
ehild
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And what about substituting u=xy?

ehild
 
  • #4
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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]
 

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