### Author Topic: P1-Morning  (Read 2329 times)

#### Victor Ivrii ##### P1-Morning
« 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 $y(1)=1$.

#### Vivian Ngo

• Full Member
•   • Posts: 23
• Karma: 4 ##### Re: P1-Morning
« Reply #1 on: February 15, 2018, 05:48:52 PM »
*Typed solutions to come* (I have class until 9pm today) *Typed solutions to come*

#### Vivian Ngo

• Full Member
•   • Posts: 23
• Karma: 4 ##### Re: P1-Morning
« Reply #2 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, $\phi= x^4y^2 + x^2\ln x = 1$

« Last Edit: February 21, 2018, 09:10:24 AM by Victor Ivrii »