Toronto Math Forum
MAT244--2018F => MAT244--Tests => Term Test 1 => Topic started by: Victor Ivrii on October 16, 2018, 05:33:52 AM
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ind integrating factor and then a general solution of ODE
\begin{equation*}
\bigl(1-xy^2-2x^2y\bigr) - \bigl(2x^2y+x^3\bigr) y'=0.
\end{equation*}
Also, find a solution satisfying $y(1)=1$.
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here is my solution
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$M_y= -2yx-2x^2$
$N_x= -4xy -3x^2$
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 $\ln 1 - 1 - 1 = c, \therefore c = -2$ $$\ln x - xy^2 - x^2 y = -2$$
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Everybody got it right: Jiabei solved, Zhiya typed (but please, don't use \mbox etc, this is a bad habit). You can use \text