Is This Differential Equation Misplaced in the Precalculus Forum?

[fan_103]

1.Solve this differential equation



2.(x^2+y^2)y'=xy



3. Thanks

 

[Mark44]

What have you tried? The rules in this forum state that you must show some effort before we'll help you out.

 

[fan_103]

Ok mate I have tried it
(x^2+y^2)dy/dx=xy

Divide by $$y*(x^2+y^2)$$ : 1/y dy/dx= $$x/(x^2+y^2)$$.

$$\int {1/y}$$ dy= 1/2$$\int {x/(x^2+y^2)}$$

My answer is y=A $$\sqrt{x^2+y^2}$$.

 

[willem2]

I don't see how you get that answer from the equation above it.
There are various classes of differential equations that can be solved.
This one is in one of those classes.
It's probably more instructive to search for which class it is in yourself.

 

[fan_103]

Can anyone help me please its very important!

 

[HallsofIvy]

Ok mate I have tried it
(x^2+y^2)dy/dx=xy

Divide by $$y*(x^2+y^2)$$ : 1/y dy/dx= $$x/(x^2+y^2)$$.

$$\int {1/y}$$ dy= 1/2$$\int {x/(x^2+y^2)}$$

You don't have a "dx" on the right

My answer is y=A $$\sqrt{x^2+y^2}$$.

That may seem like a technical point but it would have reminded you that you are integrating on the right with respect to x and y is some (unknown) function of x, not a constant. This is NOT a separable equation- that is you cannot get only x on one side of the equation and only y on the other.

It is, however, a "homogeneous" first order equation. If you divide both sides of the equation by $x^2$, you get
$$(1+ \left(\frac{y}{x}\right)^2)dy= \frac{y}{x}dx$$
Let u= y/x. Don't for get to replace dy: y= ux so dy= xdu+ udx.

Why was this posted under "Precalculus"?

 

[fan_103]

I forgot to write the dx on the right.
Thanks a lot mate!Really appreciate ur effort!

Why was this posted under "Precalculus"? Where should I post these kind of questions...

 

[Mark44]

I forgot to write the dx on the right.
Thanks a lot mate!Really appreciate ur effort!

Why was this posted under "Precalculus"? Where should I post these kind of questions...

In Calculus and Beyond...