### Author Topic: Q2: TUT0801  (Read 810 times)

#### Coollight

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##### Q2: TUT0801
« on: October 04, 2019, 04:40:30 PM »
Show the given equation is not exact but becomes exact when multiplied by the given integrating factor, then solve the equation

\begin{align*}
\  x^2 y^3 + x(1+y^2)y' = 0, \ \mu(x, y) = \frac{1}{xy^3}\\
\end{align*}
Solution:
\begin{align*}
\ (x^2 y^3)dx + x(1+y^2)dy &= 0 \\
\ M = x^2 y^3 \ &and \ N = x(1+y^2) \\
\ M_{y} = 3x^2y^2 \ &and \ N_{x} = 1+y^2\\
\ Since \ M_{y} \neq N_{x}, \ the \ given \ equation \ is \ not \ exact. \\
\\
When \ multiplied \ by \ \mu(x, y) = \frac{1}{xy^3} \ on \ both \ sides, \ then \\
\ \frac{1}{xy^3}(x^2 y^3)dx + \frac{1}{xy^3}x(1+y^2)dy &= 0 \\
\ x dx + (y^{-3} + y^{-1})dy &= 0  \ is \ exact. \\
\ Then \ there \ exists \ a \ function \ \phi(x,y) \ s.t. \ \phi_{x} = M = x \ &and \ \phi_{y} = N = y^{-3} + y^{-1} \\
\ \phi = \int M dx = \int x dx &= \frac{1}{2}x^2 + h(y) \\
\ \phi_{y} = h(y)' \equiv N &= y^{-3} + y^{-1} \\
\ h(y)' &= y^{-3} + y^{-1} \\
\ h(y) &= -\frac{1}{2}y^{-2} + \ln|y| \\
\ \phi &= \frac{1}{2}x^2 + -\frac{1}{2}y^{-2} + \ln|y| \\
So \ we \ will \ have, \\
\ \frac{1}{2}x^2 -\frac{1}{2}y^{-2} + \ln|y| &= C \ as \ the \ general \ solution
\end{align*}