Author Topic: TT1-problem 1  (Read 1468 times)

Victor Ivrii

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TT1-problem 1
« on: October 09, 2014, 01:57:52 AM »
Find integrating factor and then a general solution of ODE
\begin{equation*}
y + (2xy - e^{-2y})y' = 0 \ .
\end{equation*}
Also, find a solution satisfying $y(1)=-2$ .

Roro Sihui Yap

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Re: TT1-problem 1
« Reply #1 on: October 09, 2014, 12:09:00 PM »
\begin{equation*}
y + (2xy - e^{-2y})y' = 0 \ .
\end{equation*}
Let $M(x,y) = y$, $N(x,y) =2xy - e^{-2y}$. Then $M_y(x,y) = 1$,  $N_x(x,y) = 2y$.

The equation is not exact. Lets find an integrating factor to make it exact.
$d\mu/dy = (N_x  - M_y)\mu / M\implies
d\mu/dy = (2y - 1)\mu / y \implies
d\mu/ \mu  = (2 - (1/y))dy\implies
\ln \mu = 2y - lny \implies
\mu = e^{2y - lny} \implies
\mu = e^{2y} / y $

Now multiply the equation by  $\mu = e^{2y} / y$
\begin{equation}
e^{2y} + (2xe^{2y} - 1/y ) y′= 0
\end{equation}
Now  $M(x,y) = e^{2y}$,  $N(x,y) =(2xe^{2y} - 1/y )$. 

There is a  $\Psi(x, y)$ such that:
\begin{gather}
 \Psi _x(x, y) = M(x,y) = e^{2y}\\
\Psi _y (x, y) = N(x,y) = (2xe^{2y} - 1/y )
\end{gather}

Integrating (2)  we have $\Psi (x, y) = xe^{2y} + f(y)$. Using this, differentiate to get
$\Psi_y (x, y) = 2xe^{2y} + f'(y) $. 

Compare this with (3): $f'(y) = -1/y$. $Meaning f(y) = -ln|y|$
So, 
\begin{equation}
\Psi (x, y) = xe^{2y} - ln|y| = C
\end{equation}

When x = 1, y = -2
$e^{-4} - ln |-2| = C $

The solution is
\begin{equation}
xe^{2y} - ln|y| = e^{-4} - ln 2
\end{equation}
« Last Edit: October 09, 2014, 12:11:30 PM by Roro Sihui Yap »

Bogdan Scaunasu

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Re: TT1-problem 1
« Reply #2 on: October 09, 2014, 12:24:11 PM »
Let $M = y$ and $N = 2xy-e^{-2y}$. Then:
$$
\frac{\partial M}{\partial y} = 1 \neq 2y = \frac{\partial N}{\partial x}
$$
Therefore we must introduce an integrating factor $\mu$ such that:
$$
\frac{\partial \mu M}{\partial y} = \frac{\partial \mu}{\partial y} M + \mu \frac{\partial M}{\partial y} \\
\frac{\partial \mu N}{\partial x} = \frac{\partial \mu}{\partial x} N + \mu \frac{\partial N}{\partial x}
$$
Assume $\mu_x = 0$ so as to make calculations easier. Then:
$$
\frac{\partial \mu M}{\partial y} = \frac{\partial \mu N}{\partial x} \\
\frac{\partial \mu}{\partial y} M + \mu \frac{\partial M}{\partial y} = \mu \frac{\partial N}{\partial x} \\
\mu \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} = \frac{\partial \mu}{\partial y} \\
\frac{\partial \mu}{\partial y} = \frac{2y - 1}{y} \mu = (2 - y^{-1}) \mu \\
\int \frac{\partial \mu}{\mu} = \int (2 - y^{-1}) dy \\
\ln \mu = 2y - \ln y + ln C \\
\boxed{\mu = C y^{-1} e ^ {2y}}
$$
Find some function $\Psi$ such that:
\begin{aligned}
\Psi_x & = \mu M = e ^ {-2y} \\
\Psi_y & = \mu N = 2 x e ^ {2 y} - y ^ {-1} \\
\end{aligned}
Then integrating $\Psi_x$:
\begin{aligned}
\int \Psi_x dx & = \int e ^ {-2y} dx \\
& = e ^ {-2y} \int dx \\
\Psi & = xe ^ {-2y} + V
\end{aligned}
Then integrating $\Psi_y$:
\begin{aligned}
\int \Psi_y dy & = \int (2 x e ^ {2 y} - y ^ {-1}) dy \\
& = 2 x \int e ^ {2 y} dy - \int y ^ {-1} dy \\
\Psi & = x e ^ {2 y} - \ln y + U
\end{aligned}
Therefore we have:
$$
\Psi = x e ^ {2 y} - \ln y + C = 0
$$
To solve the IVP $y(1) = -2$, we substitute $x = 1$ and $y = -2$:
$$
(1) e ^ {2 (-2)} - \ln |-2| + C = 0 \\
e ^ {-4} - \ln 2 + C = 0 \\
\boxed{C = \ln 2 - e ^ {-4}}
$$
Therefore the answer is:
$$
\boxed{x e ^ {2 y} - \ln |y| = e ^ {-4} - \ln 2}
$$
** fixed as per Professor Ivrii's remark
« Last Edit: October 09, 2014, 04:31:23 PM by Bogdan Scaunasu »

Victor Ivrii

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Re: TT1-problem 1
« Reply #3 on: October 09, 2014, 02:48:47 PM »
Bogdan: observe that in virtue of initial condition we are interested in negative $y$.