Toronto Math Forum
MAT2442018F => MAT244Tests => Term Test 1 => Topic started by: Victor Ivrii on October 16, 2018, 05:33:52 AM

ind integrating factor and then a general solution of ODE
\begin{equation*}
\bigl(1xy^22x^2y\bigr)  \bigl(2x^2y+x^3\bigr) y'=0.
\end{equation*}
Also, find a solution satisfying $y(1)=1$.

here is my solution

$M_y= 2yx2x^2$
$N_x= 4xy 3x^2$
M$_y$ $\neq$ N$_x$ so the original equation is not exact
Let $\mu$ only depends on x
$\mu'(x)$ = $\frac{M_y  N_x}{N}$ $\mu(x)$
$\mu'(x)$ = $\frac{2xy 2x^2 + 4xy + 3x^2 }{x(2xy + x^2)}$ $\mu(x)$
$\mu'(x)$ = $\frac{1}{x}$ $\mu(x)$
$\mu(x)$ = $\frac{1}{x}$
multiply $\mu(x)$ to both side:
($\frac{1}{x}$  $y^2$ 2xy) (2xy + $x^2$)$y'$ = 0 (Now M$_y$ = N$_x$, equation is exact)
There exist $\psi(x,y)$ st
$$ \psi(x, y) = \int{\frac{1}{x}  y^2  2xy} \mbox{d}x = \ln x  xy^2 x^2y + h(y) $$
$$ \frac{\partial \psi}{\partial y} = 2yx  x^2 + h'(x) = 2xy  x^2 $$
$$ \mbox{Therefore, } h'(x) \Longrightarrow h(x) = c $$
$$ \ln x  xy^2  x^2y = c $$Given $$y(1) = 1$$ when $$x = 1, y = 1$$ such that $\ln 1  1  1 = c, \therefore c = 2$ $$\ln x  xy^2  x^2 y = 2$$

Everybody got it right: Jiabei solved, Zhiya typed (but please, don't use \mbox etc, this is a bad habit). You can use \text