Find the general solution of the given differential equation
$$\frac{y^2}{2}-2ye^t+(y-e^t)\frac{dy}{dt}=0$$
So we are learning about exact equations and the first step our professor told us to do is check the derivatives of m and n:
$$M: \frac{y^2}{2}-2ye^t $$
$$ N: y-e^t $$
Now, derivatives:
$$M: \frac{\partial \phi}{\partial y}=y-2e^t$$
$$N: \frac{\partial \phi}{\partial t}=e^t$$
but they are not equal, so I'm not sure what to do next or am I not able to do it.
$\endgroup$2 Answers
$\begingroup$Rearrange the ODE we have
$$\frac{y^2}{2}+yy'=2ye^t+y'e^t$$
Multiplying both sides by $e^t$:
$$e^t\frac{y^2}{2}+e^tyy'=2ye^{2t}+y'e^{2t}\quad (*)$$
Note that
$$LHS=\frac{d}{dt}\left(e^t\frac{y^2}{2}\right),\quad RHS=\frac{d}{dt}(e^{2t}y)$$
Integrating both sides of $(*)$ gives
$$e^ty^2=2e^{2t}y+C\Rightarrow y^2-2e^ty-Ce^{-t}=0$$
Solve for $y$ gives
$$y=\frac{-(-2e^t)\pm\sqrt{4e^{2t}+4Ce^{-t}}}{2}=e^t\pm\sqrt{e^{2t}+Ce^{-t}}$$
$\endgroup$ 3 $\begingroup$Notice that
$$ \frac{1}{N} \left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial t}\right) = \frac{(y- 2e^t) - (-e^t)}{y - e^t} = 1 $$
doesn't depend on $y$
So we can define our integrating factor to be
$$ \mu(t) = \exp\left(\int \frac{1}{N} \left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial t}\right) dt\right) = \exp\left(\int dt\right) = e^t $$
Multiplying the equation by this integrating factor to get
$$ \left(\frac{y^2e^t}{2} - 2ye^{2t}\right)dt + (ye^t - e^{2t})\,dy = 0 $$
You can check to see that this equation is exact
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