Author Topic: TT1 Problem 1 (afternoon)  (Read 2402 times)

Victor Ivrii

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TT1 Problem 1 (afternoon)
« on: October 16, 2018, 05:33:52 AM »
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 $y(1)=1$.

Jiabei Bi

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Re: TT1 Problem 1 (afternoon)
« Reply #1 on: October 16, 2018, 09:41:48 AM »
here is my solution

Zhiya Lou

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Re: TT1 Problem 1 (afternoon)
« Reply #2 on: October 16, 2018, 09:50:37 AM »
$M_y= -2yx-2x^2$
$N_x= -4xy -3x^2$

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 $\ln 1 - 1 - 1 = c, \therefore c = -2$ $$\ln x - xy^2 - x^2 y = -2$$
« Last Edit: October 16, 2018, 09:56:22 AM by Zhiya Lou »

Victor Ivrii

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Re: TT1 Problem 1 (afternoon)
« Reply #3 on: October 18, 2018, 04:50:21 AM »
Everybody got it right: Jiabei solved, Zhiya typed (but please, don't use \mbox etc, this is a bad habit). You can use \text
« Last Edit: October 18, 2018, 04:52:55 AM by Victor Ivrii »