The discussion revolves around solving a separable differential equation of the form dy/dx = x/(y^2√(1+x)). Participants are exploring the integration of both sides after separating the variables.
Several participants have provided guidance on the integration process and substitution techniques. There is an ongoing exploration of how to handle the integration of the right-hand side, with various interpretations being considered.
Some participants express confusion about the substitution method and how to derive the necessary differential elements. There is a recognition of missing steps in the integration process, particularly regarding the manipulation of variables.
KallKoll said: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!
KallKoll said: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!
KallKoll said:Yes that should be, thank you. How does the u^2 work?
Curious3141 said:Shouldn't that be ##\int y^2dy = \int \frac{x}{\sqrt{1+x}}dx##?
For the RHS, try ##u^2 = 1 + x##.
Curious3141 said:Did you try it? Post what you have. Latex formatted please.
KallKoll said: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] soKallKoll said: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.
HallsofIvy said: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]