Toronto Math Forum
MAT2442013S => MAT244 MathTests => Term Test 1 => Topic started by: Victor Ivrii on February 13, 2013, 10:36:58 PM

Find integrating factor and solve
\begin{equation*}
x\,dx +y (1+x^2+y^2)\,dy=0.
\end{equation*}

heres my solution

solution

Better quality upload

better formatted solution

I am late, but at some point, someone is going to want a coded solution that isn't rotated by 90Â° CCW.
Let $M(x,y) = x$ and let $N(x,y) = y + x^2y + y^3$. Then,
\begin{align*}
M_y(x,y) = 0, & N_x(x,y) = 2xy.
\end{align*}
So the equation is not exact, and we need an integrating factor $\mu(x,y) = \mu(y)$ to make it exact.
\begin{equation*}
\frac{d\mu(y)}{dy} = \frac{N_x  M_y}{M} \mu(y) = \frac{2xy}{x} \mu(y) = y\mu(y) \Longrightarrow \frac{d\mu(y)}{\mu(y)} = 2ydy \Longrightarrow \mu(y) = e^{y^2}
\end{equation*}
Now, we require a function $\psi(x,y)$ that satisfies $\psi_x = \mu M$ and $\psi_y = \mu N$.
\begin{equation*}
\psi_x(x,y) = xe^{y^2} \Longrightarrow \psi(x,y) = \frac{1}{2}x^2e^{y^2} + h(y)
\end{equation*}
We differentiate the result to get
\begin{equation*}
\psi_y(x,y) = x^2xe^{y^2} + h'(y) = e^{y^2}(y + x^2y + y^3) \Longleftrightarrow h'(y) = e^{y^2}(y^3 + y) \Longrightarrow h(y) = \int{(e^{y^2}(y^3 + y))}dy.
\end{equation*}
Let $u = y^2$ so that $du = 2ydy$. Then,
\begin{equation*}
h(y) = \int{(y^3e^{y^2})}dy + \int{(ye^{y^2})}dy = \frac{1}{2}\int{(ue^u)}du + \frac{1}{2}\int{e^u}du = \frac{1}{2}ue^u  \frac{1}{2}\int{e^u}du + \frac{1}{2}e^u = \frac{1}{2}ue^u  \frac{1}{2}e^u + \frac{1}{2}e^u = \frac{1}{2}y^2e^{y^2}.
\end{equation*}
Therefore, the solution is implicitly given by
\begin{equation*}
C = \frac{1}{2}e^{y^2}(x^2 + y^2).
\end{equation*}

I decided to be generous and awarded karma to all 4. Marcia definitely realized that her first scan (actually snapshot) was almost completely useless and reposted with double resolution; honestly, even her 2nd snapshot is inferior. Matthew positioned paper in the best possible way, and Yook used grayscale (better than colour; however black and white would be even better but it requires more knowledgesee my avatar for b/w) and a monsterresolution picture (but because it was grayscale file size was not much larger!)
Alexander' post is far superior (orientation is not that important, major advantage it is typed and could be easily edited and recycled so in the most strict approach (the first gets all) Matthew and Alexander would get their karma.