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

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In summary, the given differential equation is (x2-y2)dx + (x2-xy)dy=0 and the task is to verify if the function c1(x+y)2=xey/x is a solution. The method used includes solving for \frac{dy}{dx} in the equation and using implicit differentiation on the function. There was a sign error in the equation and mistakes in the blue equation, which have been corrected. The final equation is 2c1(x+y)(1+y')=xey/x(\frac{xy'-y}{x2}) + ey/x.
  • #1
aaronfue
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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?
 
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  • #2
aaronfue said:

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
ehild said:
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:

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

1. How do you verify a solution for a differential equation?

To verify a solution for a differential equation, you need to substitute the solution function into the original equation and check if it satisfies the equation. If the solution function satisfies the equation, then it is a valid solution for the differential equation.

2. What is the process for verifying a solution for a differential equation?

The process for verifying a solution for a differential equation involves substituting the solution function into the original equation, simplifying the equation, and checking if the resulting equation is true. If the equation is true, then the solution is verified.

3. Can a differential equation have more than one solution?

Yes, a differential equation can have more than one solution. In fact, most differential equations have an infinite number of solutions. However, not all solutions may be valid or physically meaningful.

4. How do you know if a solution is physically meaningful?

A solution is physically meaningful if it satisfies any physical constraints or boundary conditions that are given in the problem. For example, if the differential equation represents the motion of a particle, the solution must satisfy the initial position and velocity of the particle.

5. What is the purpose of verifying a solution for a differential equation?

The purpose of verifying a solution for a differential equation is to ensure that the solution is correct and satisfies the given equation. This is important in applications of differential equations, as an incorrect solution can lead to incorrect predictions or interpretations.

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