Solving differential equations using substitution

In summary, the conversation is about solving a differential equation by making an appropriate substitution. The first attempt involved selecting the easiest differential to integrate, while the second attempt involved choosing the substitution for the dx differential. The conversation ends with the possibility of not being able to find a closed form solution for the equation.
  • #1
hbomb
58
0
Solve the differential equation by making an appropriate substitution

[tex]x^2dy=(xy+x^2e^\frac{y}{x})[/tex]

Here was my first attempt:

Let y=ux, dy=udx+xdu

[tex]x^2(udx+xdu)=(ux^2+x^2e^\frac{ux}{x})dx[/tex]
[tex]x^2(udx+xdu)=(ux^2+x^2e^u)dx[/tex]

[tex]x^2udx+x^3du=x^2udx+x^2dx+e^udx[/tex]

[tex]x^3du=x^2dx+e^udx[/tex]

And after that I got stuck.
My teacher said to pick the differential that would be the easiest to integrate and find the substitutions for that. So naturally that would be the dy differential.


So then I did it the other way:

Let x=vy, dx=vdy+ydu

[tex]v^2y^2dy=(y^2v+y^2v^2e^2\frac{y}{vy})(vdy+ydu)[/tex]

[tex]vy^2(vdy)=vy^2(1+ve^\frac{1}{v})(vdy+ydu)[/tex]

[tex]vdy=vdy+v^2e^\frac{1}{v}dy+ydv+yve^\frac{1}{v}dv[/tex]

[tex]-v^2e^\frac{1}{v}dy=ydv+yve^\frac{1}{v}dv[/tex]

[tex]-v^2e^\frac{1}{v}dy=y(1+ve^\frac{1}{v})dv[/tex]

[tex]\frac{1}{y}dy=(\frac{1+ve^\frac{1}{v}}{-v^2e^\frac{1}{v}})dv[/tex]

Integrating this gives me the following result:

[tex]ln|y|=-ln|v|-\frac{1}{e^\frac{1}{v}}[/tex]

substituting [tex]\frac{x}{y}[/tex] back in for v we get

[tex]ln|y|=-ln|\frac{x}{y}|-\frac{1}{e^\frac{1}{v}}[/tex]
 
Last edited:
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  • #2
is this your diff eq
[tex]x^2dy=(xy+x^2e^\frac{y}{x})[/tex]
or do you mean this
[tex]x^2 \frac{dy}{dx} =(xy+x^2 e^\frac{y}{x})[/tex]
if the first one was your question and u didnt make a typing eerror then in taht case i can't get a closed form solution the best i can come up with is
[tex] y = \frac{y}{x} + e^{\frac{y}{x}} [/tex]
but i could be wrong!
 

What is substitution in the context of solving differential equations?

Substitution is a method used to solve differential equations by replacing a variable with a new variable or expression. This allows for the equation to be simplified and easier to solve.

Why is substitution used in solving differential equations?

Substitution is used in solving differential equations because it can help reduce the complexity of the equation and make it more manageable. It can also help isolate the variable being solved for, making it easier to find a solution.

What are the steps involved in solving a differential equation using substitution?

The steps involved in solving a differential equation using substitution are: 1) Identify the variable to be substituted, 2) Choose a suitable substitution, 3) Substitute the variable in the equation, 4) Simplify the equation, 5) Solve for the new variable, 6) Substitute the solution back into the original equation to check for correctness.

Can substitution always be used to solve differential equations?

No, substitution may not always be the best or most efficient method for solving differential equations. It is important to consider other methods and choose the most appropriate one for each specific equation.

Are there any common mistakes to avoid when using substitution to solve differential equations?

One common mistake is incorrectly choosing the substitution, which can lead to an incorrect solution. It is also important to carefully substitute the variable and simplify the equation, as errors in these steps can also result in an incorrect solution.

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