# Differential Equations: Separable Equations

## Homework Statement

Solve the equation

dy/dx = x/(y^2√(1+x))

## 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!

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Curious3141
Homework Helper

## Homework Statement

Solve the equation

dy/dx = x/(y^2√(1+x))

## 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
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Solve the equation

dy/dx = x/(y^2√(1+x))

## 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.

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

Curious3141
Homework Helper
Yes that should be, thank you. How does the u^2 work?
Did you try it? Post what you have. Latex formatted please.

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
Homework Helper
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
Homework Helper
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
$$\int \frac{xdx}{\sqrt{1+ x}}= \int \frac{(u^2- 1)(2udu)}{u}= 2\int (u^2- 1) du$$

1 person
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!