- #1
KillerZ
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Homework Statement
Solve the given differential equation by using appropriate substitution.
[tex]3(1+t^{2})\frac{dy}{dx} = 2ty(y^{3} - 1)[/tex]
Homework Equations
[tex]y = u^{\frac{1}{1-n}}[/tex]
The Attempt at a Solution
[tex]\frac{dy}{dt} = \frac{2ty^{4} - 2ty}{3 + 3t^{2}}[/tex]
[tex]\frac{dy}{dt} + \frac{2ty}{3 + 3t^{2}} = \frac{2ty^{4}}{3 + 3t^{2}}[/tex]
[tex]-\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}[/tex]
[tex]-\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}}[/tex]
[tex]-\frac{1}{3}\frac{du}{dt} + (\frac{2ty}{3 + 3t^{2}})u = \frac{2ty^{4}}{3 + 3t^{2}}[/tex]
[tex]\frac{du}{dt} - (\frac{2ty}{1 + t^{2}})u = -\frac{2ty^{4}}{1 + t^{2}}[/tex]
Integrating factor:
[tex]I(t) = e^{-\int\frac{2t}{1+t^{2}}dt}[/tex]
use substitution:
[tex]e^{ln|u|^{-1}}[/tex]
[tex]I(t) = \frac{1}{1+t^{2}}[/tex]
[tex]y(t) = \frac{1}{\frac{1}{1+t^{2}}}[-\int\frac{2t}{(1+t^{2})(1+t^{2})}dt + c][/tex]
[tex]y(t) = 1+t^{2}[\frac{1}{1+t^{2}} + c][/tex]
[tex]y(t) = 1 + c + ct^{2}[/tex]