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Differential eq

  • Thread starter aaronfue
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  • #1
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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=xey/x

2. The attempt at a solution

I've gotten this far:

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

[itex]\frac{dy}{dx}[/itex]=[itex]\frac{-x^2-y^2}{x^2-xy}[/itex]

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

2c1(x+y)(1+y)=xey/x([itex]\frac{xy'-y}{x}[/itex]) + ey/x

Now....I believe that I can solve for y' in: 2c1(x+y)(1+y)=xey/x([itex]\frac{xy'-y}{x}[/itex]) + ey/x?
 

Answers and Replies

  • #2
ehild
Homework Helper
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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=xey/x

2. The attempt at a solution

I've gotten this far:

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

[itex]\frac{dy}{dx}[/itex]=[itex]\frac{-x^2-y^2}{x^2-xy}[/itex]

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

2c1(x+y)(1+y)=xey/x([itex]\frac{xy'-y}{x}[/itex]) + ey/x

Now....I believe that I can solve for y' in: 2c1(x+y)(1+y)=xey/x([itex]\frac{xy'-y}{x}[/itex]) + ey/x?
Your method is correct, but there is a sign error in the equation in red, it should be

[itex]\frac{dy}{dx}[/itex]=[itex]\frac{-x^2+y^2}{x^2-xy}[/itex]

and there are mistakes also in the blue equation:

2c1(x+y)(1+y')=xey/x([itex]\frac{xy'-y}{x^2}[/itex]) + ey/x

ehild
 
  • #3
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Your method is correct, but there is a sign error in the equation in red, it should be

[itex]\frac{dy}{dx}[/itex]=[itex]\frac{-x^2+y^2}{x^2-xy}[/itex]

and there are mistakes also in the blue equation:

2c1(x+y)(1+y')=xey/x([itex]\frac{xy'-y}{x^2}[/itex]) + ey/x

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

Thanks for catching that.
 
Last edited:

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