Using substitution in differential eq

In summary: Integrating both sides, we get -\frac{1}{3} \ln (1 - y^3) = \ln x + C or \ln (x^3) + \ln (1 - y^3) = C or \ln (x^3 (1 - y^3)) = C or x^3 (1 - y^3) = CIn summary, the given differential equation can be solved by using substitution and then integrating both sides. The resulting equation can then be solved for the solution.
  • #1
vipertongn
98
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?
 
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  • #2
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
And what about substituting u=xy?

ehild
 
  • #4
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]
 

FAQ: Using substitution in differential eq

How do I know when to use substitution in solving a differential equation?

Substitution is often used when a differential equation cannot be solved by other methods such as separation of variables or integration. It is also useful when dealing with non-linear equations.

What is the process of substitution in solving a differential equation?

The process of substitution involves replacing one variable with another in order to simplify the equation. This can be done by making a substitution of the form u = g(x) or by using trigonometric identities.

Can substitution be used in all types of differential equations?

No, substitution may not always be applicable or effective in solving a differential equation. It is most commonly used for first-order linear or non-linear equations, and may not be useful for higher-order or more complex equations.

Are there any limitations or challenges when using substitution in solving differential equations?

One limitation of substitution is that it may only yield a particular solution, rather than the general solution of the equation. This method may also be more time-consuming and require more algebraic manipulation compared to other methods.

Can I use substitution in combination with other methods to solve a differential equation?

Yes, substitution can be used in combination with other methods such as separation of variables or integration to solve a differential equation. It can also be used in conjunction with techniques like power series or Laplace transforms for more complex equations.

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