### Author Topic: MT problem 1  (Read 4409 times)

#### Victor Ivrii

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##### MT problem 1
« on: October 29, 2014, 08:56:45 PM »
If exists, find the integrating factor $\mu(x,y)\$ depending only on $\ x\$, only on $\ y\$ and on $\ x \cdot y\$  justifying your answers and then solve the ODE
\begin{equation*}
\left(3x + \frac{6}{y}\right) + \left(\frac{x^2}{y}+\frac{3y}{x}\right)y'=0.
\end{equation*}
Also, find the solution satisfying $y(1) = 2\$.

#### Roro Sihui Yap

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• Posts: 30
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##### Re: MT problem 1
« Reply #1 on: October 30, 2014, 12:10:23 AM »
\begin{equation*}
\left(3x + \frac{6}{y}\right) + \left(\frac{x^2}{y}+\frac{3y}{x}\right)y'=0.
\end{equation*}
Let $M(x,y) = 3x + \frac{6}{y} \\$ and  $N(x,y) = \frac{x^2}{y}+\frac{3y}{x}$. Then  $M_y = \frac{-6}{y^2}$,  $N_x = \frac{2x}{y}+\frac{-3y}{x^2}$

The equation is not exact. Consider $\frac{M_y - N_x}{N(x,y)}$,
\begin{equation*}
\frac{M_y - N_x}{N(x,y)} = \frac{\frac{-6}{y^2} - \frac{2x}{y} + \frac{3y}{x^2}}{\frac{x^2}{y}+\frac{3y}{x}} = \frac{-6x^2 - 2x^3y + 3y^3}{x^4y + 3y^3x}
\end{equation*}
This is not a function of $x$ only. The integrating factor depending only on $x$ does not exist

Consider $\frac{N_x - M_y}{M(x,y)}$,
\begin{equation*}
\frac{N_x - M_y}{M(x,y)} = \frac{\frac{2x}{y} - \frac{3y}{x^2} + \frac{6}{y^2}}{3x + \frac{6}{y}} = \frac{6x^2 + 2x^3y - 3y^3}{3x^3y^2 + 6x^2y}
\end{equation*}
This is not a function of $y$ only. The integrating factor depending only on $y$ does not exist.

Consider $\frac{N_x - M_y}{(x)M(x,y) - (y)N(x,y)}$,
\begin{equation*}
\frac{N_x - M_y}{(x)M - (y)N} = \frac{\frac{2x}{y} - \frac{3y}{x^2} + \frac{6}{y^2}}{3x^2 + \frac{6x}{y} - x^2 - \frac{3y^2}{x}} = \frac{6x^2 + 2x^3y - 3y^3}{6x^3y + 2x^4y^2 - 3y^4x} = \frac{1}{xy}
\end{equation*}
This is a function of $xy$ only. $\frac{d\mu}{dxy} = \frac{\mu}{xy}$. Let $u = xy$;  $\frac{d\mu}{du} = \frac{\mu}{u}$,  $\mu = u$, $\mu = xy$

Multiply the equation by $xy$
\begin{equation*}
\left(3x^2y + 6x\right) + \left(x^3+3y^2\right)y'=0.
\end{equation*}
Let $M(x,y) = 3x^2y + 6x$ and  $N(x,y) = x^3+3y^2$.

There is a  $\Psi(x, y)$ such that:
\begin{gather}
\Psi _x(x, y) = M(x,y) = 3x^2y + 6x  \label{A}\\
\Psi _y (x, y) = N(x,y) = x^3+3y^2 \label{B}\end{gather}

Integrating equation (\ref{A}), $\Psi(x, y) = x^3y + 3x^2 + f(y)$, $\Psi_y (x, y) = x^3 + f'(y)$.

Comparing with equation (\ref{B}), $f'(y) = 3y^2$ and $f(y) = y^3$

Therefore $\Psi(x, y) = x^3y + 3x^2 + y^3 = c$.
When $x=1$, $y=2$,   $2 + 3 + 8 = c$, $c = 13\implies \Psi(x, y) = x^3y + 3x^2 + y^3 = 13$
« Last Edit: October 30, 2014, 06:34:15 AM by Victor Ivrii »