Author Topic: Q1 problem 2 (day section)  (Read 789 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Q1 problem 2 (day section)
« on: September 29, 2014, 02:21:44 AM »
Please post problem and solution

Rhoda Lam

  • Newbie
  • *
  • Posts: 4
  • Karma: 5
    • View Profile
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.
\begin{equation}
(x + 2) \sin y + (x \cos y)y' = 0\label{A}\\
μ(x, y) = xe^x
\end{equation}
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}):
\begin{equation}
xe^x(x + 2) \sin y + xe^x(x \cos y)y' = 0\label{B}\\
\end{equation}
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:
\begin{equation}
x^2e^x\sin(y) = C
\end{equation}
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 »