Differential Equations: Separable Equations

[KallKoll]

Homework Statement

Solve the equation

dy/dx = x/(y^2√(1+x))

Homework Equations

The Attempt at a Solution

I separated them:
y^2 dy = dx/√(1+x)

I then integrated the dy side, I got (1/3)y^3 + C. I am stuck at integrating the dx side. Thanks in advance!

 

[Curious3141]

Homework Statement

Solve the equation

dy/dx = x/(y^2√(1+x))

Homework Equations

The Attempt at a Solution

I separated them:
y^2 dy = dx/√(1+x)

I then integrated the dy side, I got (1/3)y^3 + C. I am stuck at integrating the dx side. Thanks in advance!

Shouldn't that be $\int y^2dy = \int \frac{x}{\sqrt{1+x}}dx$?

For the RHS, try $u^2 = 1 + x$.

 

[SammyS]

Homework Statement

Solve the equation

dy/dx = x/(y^2√(1+x))

Homework Equations

The Attempt at a Solution

I separated them:
y^2 dy = dx/√(1+x)

I then integrated the dy side, I got (1/3)y^3 + C. I am stuck at integrating the dx side. Thanks in advance!

You dropped an x from the numerator on the right side.

 

[KallKoll]

Yes that should be, thank you. How does the u^2 work?

 

[Curious3141]

Yes that should be, thank you. How does the u^2 work?

Did you try it? Post what you have. Latex formatted please.

 

[KallKoll]

Shouldn't that be $\int y^2dy = \int \frac{x}{\sqrt{1+x}}dx$?

For the RHS, try $u^2 = 1 + x$.

Did you try it? Post what you have. Latex formatted please.

Well, I set $\sqrt{1+x}$ = u. Then I set u² = 1 + x. But I don't understand how to get du from that or how that helps with getting rid of the x on top.

 

[pasmith]

Well, I set $\sqrt{1+x}$ = u. Then I set u² = 1 + x. But I don't understand how to get du from that or how that helps with getting rid of the x on top.

$$\frac{x}{\sqrt{1 +x}} = \frac{u^2 - 1}{u} = u - \frac1u \\$$

 

[HallsofIvy]

Well, I set $\sqrt{1+x}$ = u. Then I set u² = 1 + x. But I don't understand how to get du from that or how that helps with getting rid of the x on top.

From $u^2= 1+ x$, $x= u^2- 1$ and $2u du= dx$ so
$$\int \frac{xdx}{\sqrt{1+ x}}= \int \frac{(u^2- 1)(2udu)}{u}= 2\int (u^2- 1) du$$

 

[KallKoll]

From $u^2= 1+ x$, $x= u^2- 1$ and $2u du= dx$ so
$$\int \frac{xdx}{\sqrt{1+ x}}= \int \frac{(u^2- 1)(2udu)}{u}= 2\int (u^2- 1) du$$

Thank you so much!