MAT244-2014F > TT1

TT1-problem 1

(1/1)

Victor Ivrii:
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:
\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  $\mu = e^{2y} / y$

e^{2y} + (2xe^{2y} - 1/y ) yâ€²= 0

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,

\Psi (x, y) = xe^{2y} - ln|y| = C

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

The solution is

xe^{2y} - ln|y| = e^{-4} - ln 2

Bogdan Scaunasu:
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 $y = -2$:
$$(1) e ^ {2 (-2)} - \ln |-2| + C = 0 \\ e ^ {-4} - \ln 2 + C = 0 \\ \boxed{C = \ln 2 - e ^ {-4}}$$
$$\boxed{x e ^ {2 y} - \ln |y| = e ^ {-4} - \ln 2}$$ ** fixed as per Professor Ivrii's remark
Bogdan: observe that in virtue of initial condition we are interested in negative $y$.