Solving a Diff Eq with Substitution

[KillerZ]

Homework Statement

Solve the given differential equation by using appropriate substitution.

3(1+t^{2})\frac{dy}{dx} = 2ty(y^{3} - 1)

Homework Equations

y = u^{\frac{1}{1-n}}

The Attempt at a Solution

\frac{dy}{dt} = \frac{2ty^{4} - 2ty}{3 + 3t^{2}}

\frac{dy}{dt} + \frac{2ty}{3 + 3t^{2}} = \frac{2ty^{4}}{3 + 3t^{2}}

-\frac{1}{3}u^{-\frac{4}{3}}\frac{du}{dt} + (\frac{2ty}{3 + 3t^{2}})u^{-\frac{1}{3}} = (\frac{2ty^{4}}{3 + 3t^{2}})(u^{-\frac{1}{3}})^{4}

-\frac{1}{3}\frac{du}{dt} + (\frac{2ty}{3 + 3t^{2}})\frac{u^{-\frac{1}{3}}}{u^{-\frac{4}{3}}} = \frac{2ty^{4}}{3 + 3t^{2}}

-\frac{1}{3}\frac{du}{dt} + (\frac{2ty}{3 + 3t^{2}})u = \frac{2ty^{4}}{3 + 3t^{2}}

\frac{du}{dt} - (\frac{2ty}{1 + t^{2}})u = -\frac{2ty^{4}}{1 + t^{2}}

Integrating factor:

I(t) = e^{-\int\frac{2t}{1+t^{2}}dt}

use substitution:

e^{ln|u|^{-1}}

I(t) = \frac{1}{1+t^{2}}

y(t) = \frac{1}{\frac{1}{1+t^{2}}}[-\int\frac{2t}{(1+t^{2})(1+t^{2})}dt + c]

y(t) = 1+t^{2}[\frac{1}{1+t^{2}} + c]

y(t) = 1 + c + ct^{2}

 

[LCKurtz]

For one thing, you shouldn't have t, u, and y all in the same equation. Here's an easier way to get started. Rewrite your equation:

3(1 + t²)y' + 2ty = 2ty⁴

Now multiply through by y^-4:

3(1 + t²)y^-4y' + 2ty^-3 = 2t

Now let u = y^-3, u' = -3y^-4y'

This gets you to the equation

-(1 + t²)u' + 2tu = 2t

without all the fractional exponents and resulting errors.

 

[KillerZ]

I made some typos in there here I fixed it:

Homework Statement

Solve the given differential equation by using appropriate substitution.

3(1+t^{2})\frac{dy}{dx} = 2ty(y^{3} - 1)

Homework Equations

y = u^{\frac{1}{1-n}}

The Attempt at a Solution

\frac{dy}{dt} = \frac{2ty^{4} - 2ty}{3 + 3t^{2}}

\frac{dy}{dt} + \frac{2ty}{3 + 3t^{2}} = \frac{2ty^{4}}{3 + 3t^{2}}

-\frac{1}{3}u^{-\frac{4}{3}}\frac{du}{dt} + (\frac{2t}{3 + 3t^{2}})u^{-\frac{1}{3}} = (\frac{2t}{3 + 3t^{2}})(u^{-\frac{1}{3}})^{4}

-\frac{1}{3}\frac{du}{dt} + (\frac{2t}{3 + 3t^{2}})\frac{u^{-\frac{1}{3}}}{u^{-\frac{4}{3}}} = \frac{2t}{3 + 3t^{2}}

-\frac{1}{3}\frac{du}{dt} + (\frac{2t}{3 + 3t^{2}})u = \frac{2t}{3 + 3t^{2}}

\frac{du}{dt} - (\frac{2t}{1 + t^{2}})u = -\frac{2t}{1 + t^{2}}

Integrating factor:

I(t) = e^{-\int\frac{2t}{1+t^{2}}dt}

use substitution:

e^{ln|u|^{-1}}

I(t) = \frac{1}{1+t^{2}}

y(t) = \frac{1}{\frac{1}{1+t^{2}}}[-\int\frac{2t}{(1+t^{2})(1+t^{2})}dt + c]

y(t) = 1+t^{2}[\frac{1}{1+t^{2}} + c]

y(t) = 1 + c + ct^{2}

 

[LCKurtz]

Interesting step that last one. Where did the denominator go?