### Author Topic: Q1 problem 2 (day section)  (Read 2266 times)

#### Victor Ivrii

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##### Q1 problem 2 (day section)
« on: September 29, 2014, 02:21:44 AM »

#### Rhoda Lam

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##### Re: Q1 problem 2 (day section)
« Reply #1 on: October 05, 2014, 09:23:47 PM »
2.6 p. 101, #22
Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.

(x + 2) \sin y + (x \cos y)y' = 0\label{A}\\
Î¼(x, y) = xe^x

Let $M(x,y) = (x+2)\sin y$, $N(x,y) = x(\cos y)$. Then $M_y(x,y) = (x+2)\cos y$, $N_x(x,y) = \cos y$.

Since $M_y(x,y) \ne N_x(x,y)$, then equation (\ref{A}) is not exact. Multiply the given integrating factor to Equation (\ref{A}):

xe^x(x + 2) \sin y + xe^x(x \cos y)y' = 0\label{B}\\

Now $M(x,y) = xe^x(x+2)\sin y$, $N(x,y) = x^2e^x(\cos y)$

Then $M_y(x,y) = xe^x(x+2)\cos y$, $N_x(x,y) = 2xe^x\cos y + x^2e^x cos y = xe^x(x+2) \cos y$. Since $M_y(x,y) = N_x(x,y)$, then Equation (\ref{B}) is exact.

There is a  $\Psi(x, y)$ such that:
\begin{gather}
\Psi _x(x, y) = M(x,y) = xe^x(x+2)\sin y\label{C}\\
\Psi _y (x, y) = N(x,y) = x^2e^x(\cos y)\label{D}
\end{gather}

Integrate Equation (\ref{C}) and we get $\Psi (x, y) =\sin(y)x^2e^x + g(y)$.

Now differentiate it to get $\Psi_y (x, y) = \cos(y)x^2e^x + g'(y) = \cos(y)x^2e^x \implies g'(y) = 0 \implies g(y) = C$

Therefore, the solution is:

x^2e^x\sin(y) = C

Observe I write \cos, \sin , \ln etc, to keep them upright rather than italic and have a proper spacing; in $\LaTeX$ they are called operators-V.I.
« Last Edit: October 05, 2014, 10:20:19 PM by Victor Ivrii »