Differential equations problem

[sahen]

solve the following ivp
xy' - y = 3xe^2y/x
y(1)=-1

how can i get rid of e ? does anybody help me ?
thanks in advance.

 

[CompuChip]

I assume that e is just the Euler number 2.7...
Why would you want to get rid of it? And why then don't you ask: "how can I get rid of 3?"

 

[Defennder]

It's hard to read what you wrote. Do you mean: $$xy' - y = \frac{3xe^2y}{x}$$?

 

[sahen]

It should be xy' - y = 3xe$$^{2y/x}$$
I guess i need to study more thanks for your help.

 

[CompuChip]

Ah, so the equation is
[tex] x y' - y = 3 x \exp\left[ \frac{2y}{x} \right]<br /> $$... that makes the problem significantly more complex <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /><br /> I'm not even sure there is an exact solution.$$[/tex]

 

[crystalplane]

For the IVP problem, you should find I(X)
you may get I(x)=e^x dx
then you multiply I(X) on both sides and you can solve the problem i guess

 

[Hurkyl]

... that makes the problem significantly more complex .

On the contrary, it suggests an obvious thing to try. And due to good fortune^*, it works.

Really, this is one of those problems that (at least for the beginner) should fall into the category of "this looks complicated -- there is only one thing I could possibly do, and I just have to hope it works".


*: Okay, fine, it's more likely that it was rigged to work.

 

[Defennder]

It takes a substitution to make things a lot easier as Hurkyl said.