Differential Equations: Separable Equations

In summary, to solve the equation dy/dx = x/(y^2√(1+x)), the equation was separated into y^2 dy = dx/√(1+x). The left side was integrated to get (1/3)y^3 + C. The right side was integrated using the substitution u^2 = 1+x, resulting in 2∫(u^2-1)du. The final solution is (1/3)y^3 + 2u - 2ln(u) + C.
  • #1
KallKoll
4
0

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!
 
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  • #2
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!

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

For the RHS, try ##u^2 = 1 + x##.
 
  • #3
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!


You dropped an x from the numerator on the right side.
 
  • #4
Yes that should be, thank you. How does the u^2 work?
 
  • #5
KallKoll said:
Yes that should be, thank you. How does the u^2 work?

Did you try it? Post what you have. Latex formatted please.
 
  • #6
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.

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
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.

[tex]
\frac{x}{\sqrt{1 +x}} = \frac{u^2 - 1}{u} = u - \frac1u \\
[/tex]
 
  • #8
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] 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
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]

Thank you so much!
 

What is a separable equation?

A separable equation is a type of differential equation where the variables can be separated and solved individually. This means that the equation can be written in the form of dy/dx = g(x)h(y), where g(x) only depends on x and h(y) only depends on y.

How do you solve a separable equation?

To solve a separable equation, you need to separate the variables and integrate both sides. This means integrating the function g(x) with respect to x and h(y) with respect to y. Once both integrals are solved, you can combine them to find the solution to the original equation.

What is a particular solution in the context of separable equations?

A particular solution is a specific solution to a separable equation that satisfies the initial conditions of the problem. This means that the particular solution will satisfy the equation and any given values for the dependent and independent variables.

What is the role of initial conditions in solving separable equations?

Initial conditions are necessary in solving separable equations because they provide specific values for the dependent and independent variables. These conditions are used to find the particular solution to the equation, as the general solution may have multiple possible solutions.

Are there any applications of separable equations in real life?

Yes, separable equations have many applications in various fields of science and engineering. They are commonly used in physics, chemistry, economics, and biology to model natural phenomena and predict outcomes. For example, separable equations are used in population growth models, radioactive decay, and chemical reactions.

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