# Toronto Math Forum

## MAT244-2014F => MAT244 Math--Tests => FE => Topic started by: Victor Ivrii on December 08, 2014, 04:12:22 PM

Title: FE1
Post by: Victor Ivrii on December 08, 2014, 04:12:22 PM
Solve the initial value problem
\begin{equation*}
\left(3x + 2y\right) dx + \left(x + \frac{6 y^2}{x}\right) dy = 0\, , \qquad  y(0) = 3\, .
\end{equation*}

Solution
As $M=3x+2y$, $N=x+ \frac{6 y^2}{x}$, $M_y-N_x= 2- \bigl(1-\frac{6y^2}{x^2}\bigr)$ and $(M_y-N_x )/N= 1/x$ is a function of $x$ only. So we can find integrating factor $\mu =\mu(x)$ from $\mu'/\mu = 1/x \implies \ln \mu =\ln x$ (modulo constant factor) and $\mu=x$. Therefore
$$3x^2 + 2xy) dx + (x^2 + 6 y^2) dy = 0.$$
Then
$$U_x= 3x^2+2xy ,\qquad U_y=x^2+6y^2$$
where the first equation implies that
$$U= x^3 + x^2y+ \phi(y)$$
and plugging to the second equation we see that
$$\phi'=6y^2\implies y=2y^3.$$
Then
$$U:=x^3+x^2y+2y^3 =C$$
is a general solution and finding $C=54$ from initial condition we arrive to
$$U:=x^3+x^2y+2y^3 =54.$$
Title: Re: FE1
Post by: Sang Wu on December 14, 2014, 06:57:05 PM
Hi prof, I think if we substitute x = 0, y = 3 into U, we should get C = 54 instead of 18.
Title: Re: FE1
Post by: Victor Ivrii on December 14, 2014, 07:01:50 PM
Yes, corrected