Author Topic: TUT0302  (Read 293 times)

syc425

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TUT0302
« on: October 04, 2019, 02:02:37 PM »
Find an integrating factor and solve the given equation.
$1+(\frac{x}{y}-sin(y))y'=0$

$M=1$
$N=(\frac{x}{y}-sin(y))$
$R=\frac{My-Nx}{M}$
$R=\frac{0-(\frac{1}{y}-0)}{1}$
$R=-\frac{1}{y}$
$\mu=e^{-\int -\frac{1}{y}dy}$
$\mu=e^{lny}$
$\mu=y$

Multiply $\mu$ to both side:
$y+(x-ysin(y))y'=0$
$My=1$
$Nx=1$
It is exact now, so $\mu=y$ is the integrating factor.

Integrate M with respect to x and we get:
$xy+h(y)$

Take derivative on both side with respect to $y$ and we get:
$x+h'(y)$

$h'(y)=-ysiny$
$h(y)=-\int ysiny$