Toronto Math Forum
MAT2442014F => MAT244 MathTests => FE => Topic started 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 $M=3x+2y$, $N=x+ \frac{6 y^2}{x}$, $M_yN_x= 2 \bigl(1\frac{6y^2}{x^2}\bigr)$ and $(M_yN_x )/N= 1/x$ 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
$$
U:=x^3+x^2y+2y^3 =54.
$$

Hi prof, I think if we substitute x = 0, y = 3 into U, we should get C = 54 instead of 18.

Yes, corrected