What is the Correct Approach for Solving This Differential Equation?

[aaronfue]

Homework Statement

y = 2xy' + y(y')² ; y² = c₁(x + c₁/4)

Homework Equations

So far I've gotten the second equation to be: y = (c₁x + c₁²/4)^1/2

I was then going to take the derivative of that equation and plug them into the first equation after setting it to zero.

Is that the right way to handle this particular equation?

 

[Mark44]

Homework Statement

y = 2xy' + y(y')² ; y² = c₁(x + c₁/4)

Your problem statement is incomplete - it doesn't say what you need to do. Are you supposed to show that the equation with y² is a solution of the diff. equation?

Also, your equation for y² should be y² = c₁(x + c₂/4)

Homework Equations

So far I've gotten the second equation to be: y = (c₁x + c₁²/4)^1/2

I was then going to take the derivative of that equation and plug them into the first equation after setting it to zero.

"after setting it to zero" - ??

Your equation for y should be written as y = ±(c₁(x + c₂/4))^1/2, since the original equation determines two values for y: one positive and one negative.

Substitute your expressions for y and y' into the differential equation. If the solution is correct, you'll get an equation that is identically true.

Is that the right way to handle this particular equation?

 

[aaronfue]

Mark44,

You are correct, the problem asked to verify that the indicated function is a solution of the given differential equation.

- My initial equation was correct. There is no c₂ in the equation. (see attached document from textbook)

- "after setting it to zero" : I was referring to y = 2xy' + y(y')^2 --> 0 = 2xy' + y(y')^2 - y

Thanks for your reply.

 

[Mark44]

Mark44,

You are correct, the problem asked to verify that the indicated function is a solution of the given differential equation.

- My initial equation was correct. There is no c₂ in the equation. (see attached document from textbook)

That is possibly a typo in the book. The right side of that equation is the same as c₁x + c₂, where c₂ = (1/4)c₁².

- "after setting it to zero" : I was referring to y = 2xy' + y(y')^2 --> 0 = 2xy' + y(y')^2 - y

OK, that wasn't clear to me. You could work with the equation as-is, without moving y to the other side. It doesn't make much difference either way.