# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Term Test 1 => Topic started by: Victor Ivrii on February 15, 2018, 05:09:54 PM

Title: P1-Morning
Post by: Victor Ivrii on February 15, 2018, 05:09:54 PM
Find integrating factor and then a general solution of ODE
\begin{equation*}
(4x y^2+3\ln(x)+1)+2x^2yy'=0 \ .
\end{equation*}

Also, find a solution satisfying $y(1)=1$.
Title: Re: P1-Morning
Post by: Vivian Ngo on February 15, 2018, 05:48:52 PM
*Typed solutions to come* (I have class until 9pm today)

(https://scontent-yyz1-1.xx.fbcdn.net/v/t34.0-12/28054244_10213726492907959_247696739_n.png?oh=fd9ea2c073030ff93a9caae81614b2c5&oe=5A883A0C)

*Typed solutions to come*
Title: Re: P1-Morning
Post by: Vivian Ngo on February 16, 2018, 12:13:07 AM
$M_y = 8xy$
$N_x = 4xy$
$\implies$ The equation is not exact

$\frac{M_y-N_x}{N} = \frac{2}{x}$

$\frac{d\mu}{dx} = \frac{2}{x}\mu$

$\mu = x^2$

The integrating factor is $\mu = x^2$.

New equation: $4x^3y^2 + 3x^2 lnx + x^2 + 2x^4yy'=0$
$M_y = 8x^3y = N_x$ (The equation is exact)
$\phi_x = M$
$\phi = x^4y^2 + x^3\ln x + h(y)$
$\phi_y = 2x^4y + h'(y) = N$
$h'(y)=0$
$h(y)=C$

Thus, $\phi= x^4y^2 + x^3\ln x = C$

For particular solution passing (1,1):
$1^41^2 + 1^2\ln 1 = C$
$C=1$
Thus, $\phi= x^4y^2 + x^2\ln x = 1$