Differential Equations: Separable Equations

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  • #1
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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!
 

Answers and Replies

  • #2
Curious3141
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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##.
 
  • #3
SammyS
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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.
 
  • #4
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Yes that should be, thank you. How does the u^2 work?
 
  • #5
Curious3141
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Yes that should be, thank you. How does the u^2 work?
Did you try it? Post what you have. Latex formatted please.
 
  • #6
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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 [itex]\sqrt{1+x}[/itex] = 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.
 
  • #7
pasmith
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Well, I set [itex]\sqrt{1+x}[/itex] = 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.
[tex]
\frac{x}{\sqrt{1 +x}} = \frac{u^2 - 1}{u} = u - \frac1u \\
[/tex]
 
  • #8
HallsofIvy
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Well, I set [itex]\sqrt{1+x}[/itex] = 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 [itex]u^2= 1+ x[/itex], [itex]x= u^2- 1[/itex] and [itex]2u du= dx[/itex] so
[tex]\int \frac{xdx}{\sqrt{1+ x}}= \int \frac{(u^2- 1)(2udu)}{u}= 2\int (u^2- 1) du[/tex]
 
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  • #9
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From [itex]u^2= 1+ x[/itex], [itex]x= u^2- 1[/itex] and [itex]2u du= dx[/itex] so
[tex]\int \frac{xdx}{\sqrt{1+ x}}= \int \frac{(u^2- 1)(2udu)}{u}= 2\int (u^2- 1) du[/tex]
Thank you so much!
 

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