Verifying a Solution for a Diff. Equation: (x2-y2)dx + (x2-xy)dy=0

[aaronfue]

Homework Statement

Given: (x²-y²)dx + (x²-xy)dy=0,

Verify that the following function is a solution for the given differential equation:

c₁(x+y)²=xe^y/x

2. The attempt at a solution

I've gotten this far:

1st - I solved for \frac{dy}{dx} in the given equation.

\frac{dy}{dx}=\frac{-x^2-y^2}{x^2-xy}

2nd - I used implicit differentiation on the function and got:

2c₁(x+y)(1+y)=xe^y/x(\frac{xy'-y}{x}) + e^y/x

Now...I believe that I can solve for y' in: 2c₁(x+y)(1+y)=xe^y/x(\frac{xy'-y}{x}) + e^y/x?

 

[ehild]

Homework Statement

Given: (x²-y²)dx + (x²-xy)dy=0,

Verify that the following function is a solution for the given differential equation:

c₁(x+y)²=xe^y/x

2. The attempt at a solution

I've gotten this far:

1st - I solved for \frac{dy}{dx} in the given equation.

\frac{dy}{dx}=\frac{-x^2-y^2}{x^2-xy}

2nd - I used implicit differentiation on the function and got:

2c₁(x+y)(1+y)=xe^y/x(\frac{xy'-y}{x}) + e^y/x

Now...I believe that I can solve for y' in: 2c₁(x+y)(1+y)=xe^y/x(\frac{xy'-y}{x}) + e^y/x?

Your method is correct, but there is a sign error in the equation in red, it should be

\frac{dy}{dx}=\frac{-x^2+y^2}{x^2-xy}

and there are mistakes also in the blue equation:

2c₁(x+y)(1+y')=xe^y/x(\frac{xy'-y}{x^2}) + e^y/x

ehild

 

[aaronfue]

Your method is correct, but there is a sign error in the equation in red, it should be

\frac{dy}{dx}=\frac{-x^2+y^2}{x^2-xy}

and there are mistakes also in the blue equation:

2c₁(x+y)(1+y')=xe^y/x(\frac{xy'-y}{x^2}) + e^y/x

ehild

My mistake. In the original equation it was supposed to be (x² + y²)dx. So my original \frac{dy}{dx}=\frac{-x^2-y^2}{x^2-xy} equation was okay. And you are correct about the blue equation. I completely forgot the y'.

Thanks for catching that.