Differential Equations, Separable, Simplification of answer

[Destroxia]

Homework Statement

I believe I have solved this differential equation, yet do not know how the book arrived at it's answer...

Solve the differential equation in its explicit solution form.

The answer the book gives is...

Homework Equations

Separable Differential Equation

The Attempt at a Solution

dy/dx = x(x^2+1)/(4y^3)

(4y^3)dy = (x^3+x)dx

∫(4y^3)dy = ∫(x^3+x)dx [/B]

y^4 = 1/4x^4 + 1/2x^2 + c

(initial condition, y(0) = -1/sqrt(2))

(-1/sqrt(2))^(4) = 0 + 0 + c

C = -1/4
...

y^4 = 1/4x^4 + 1/2x^2 - 1/4

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

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I've experimented with simplifying this a bit and found a few other ways to express it, but nothing like what the book has written as the answer.

 

[RUber]

Your sign on C is wrong.

$(\frac{x^4}{4}+\frac{x^2}{2} + \frac 14 ) = (\frac{x^2}{2} + \frac 12 )^2$

 

[SammyS]

Homework Statement

I believe I have solved this differential equation, yet do not know how the book arrived at it's answer...

Solve the differential equation in its explicit solution form.

The answer the book gives is...

Homework Equations

Separable Differential Equation

The Attempt at a Solution

dy/dx = x(x^2+1)/(4y^3)

(4y^3)dy = (x^3+x)dx

∫(4y^3)dy = ∫(x^3+x)dx [/B]

y^4 = 1/4x^4 + 1/2x^2 + c

(initial condition, y(0) = -1/sqrt(2))

(-1/sqrt(2))^(4) = 0 + 0 + c

C = -1/4
...

y^4 = 1/4x^4 + 1/2x^2 - 1/4

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

-----------

I've experimented with simplifying this a bit and found a few other ways to express it, but nothing like what the book has written as the answer.

First. You made a mistake in finding C. What is (-1/sqrt(2))⁴ ? Fixing that will allow some factoring in the resulting expression.

 

[Destroxia]

Your sign on C is wrong.

$(\frac{x^4}{4}+\frac{x^2}{2} + \frac 14 ) = (\frac{x^2}{2} + \frac 12 )^2$

Oh wow, I don't think I would have seen that factor regardless. Thank you though. That helped a lot.

 

[Destroxia]

First. You made a mistake in finding C. What is (-1/sqrt(2))⁴ ? Fixing that will allow some factoring in the resulting expression.

Thank you, I understand now. In regards to the -, out front the answer from the book, I understand that comes from the square root, but how do they determine whether to go with the - or + solution. I haven't learned intervals of validity within the book yet...

 

[RUber]

The radical implies the positive. Your initial condition forces the negative choice.

 

[Dr. Courtney]

Before rejecting an answer, you should plug it into the diff eq and see if it works.