Differential Equations, Separable, Explicit Solution

[Destroxia]

Homework Statement

Solve the differential equation, explicitly.

dy/dx = (2x)/(1+2y)

The answer given by the book is...

-1/2 + 1/2sqrt(2x - 2x^2 +4)

Homework Equations

Process for solving separable differential equations

The Attempt at a Solution

dy/dx = (2x)/(1+2y)

(1 + 2y)*dy = 2x*dx

∫(1+2y)*dy = ∫2x*dx

y + y^2 = x^2 + C

...

I seem to have solved it, implicitly, but I see no way of solving it explicitly.

 

[Mark44]

Homework Statement

Solve the differential equation, explicitly.

dy/dx = (2x)/(1+2y)

The answer given by the book is...

-1/2 + 1/2sqrt(2x - 2x^2 +4)

Homework Equations

Process for solving separable differential equations

The Attempt at a Solution

dy/dx = (2x)/(1+2y)

(1 + 2y)*dy = 2x*dx

∫(1+2y)*dy = ∫2x*dx

y + y^2 = x^2 + C

...

I seem to have solved it, implicitly, but I see no way of solving it explicitly.

Use the quadratic formula to solve for y in terms of x. Note that the quadratic formula will give you two solutions. Your book seems to have chosen one of the solutions arbitrarily.

 

[Destroxia]

Use the quadratic formula to solve for y in terms of x. Note that the quadratic formula will give you two solutions. Your book seems to have chosen one of the solutions arbitrarily.

Ughhh, I thought about that, I just never applied it... Thank you so much for your help!

P.S. Should I use the C(constant) as the C in the quadratic formula? or just leave it on the other side with the x?

 

[Mark44]

Is there an initial condition in your problem that you didn't include? The book's solution doesn't include the constant, which makes me suspect that they are using information not shown here.

 

[Destroxia]

Is there an initial condition in your problem that you didn't include? The book's solution doesn't include the constant, which makes me suspect that they are using information not shown here.

Yes, I'm sorry I forgot to include the initial condition, y(2) = 0

P.S. Never mind, I figured out how to do the quadratic with this equation!

 

[Mark44]

Also, are you sure you wrote the problem down correctly? I don't get the same solution as you showed.

 

[Destroxia]

Also, are you sure you wrote the problem down correctly? I don't get the same solution as you showed.

This is the solution I ended up attempting...dy/dx = (2x)/(1+2y)

(1 + 2y)*dy = 2x*dx

∫(1+2y)*dy = ∫2x*dx

y + y^2 = x^2 + C

(initial condition) 0 + 0 = 2^2 + C, C = -4

y^2 + y + (4 - x^2) = 0

y = (-1 +/- sqrt(1-4*(4-x^2)))/2(1)

y = (-1 +/- sqrt(x^2 - 15))/2

y = -(1/2) +/- (1/2)sqrt(x^2-15)The only problem I'm having is, how did they know to only use the one solution, instead of both of them, we didn't learn about intervals of validity or anything yet.

 

[Mark44]

You have what I got, which is different from the solution you posted. It's possible they have a typo in their answer, or even that it is just flat wrong. You can verify that your answer is correct by differentiating it to show the dy/dx = (2x)/(1 + 2y) is identically true. I checked the book's answer and it didn't satisfy the differential equation.