Question

Find an integrating factor of the form u(x, y)=P(x)Q(y) for the following equation and solve it.

Answer 1

\[(\alpha*y+\gamma*x*y)dx+(\beta*x+\delta*x*y)dy=0\]

Answer 2

@SithsAndGiggles

Answer 3

Are \(\alpha,\beta,\gamma,\delta\) constants?

Answer 4

If they are constants, working through the method I used last time gets an integrating factor of \(u(x,y)=\dfrac{1}{xy}\).

Answer 5

I did use your method but I got stucked in the middle of finding the integrating factor.

Answer 6

\[F(x)(\beta*x+\delta*x*y)-G(y)(\alpha*y+\gamma*x*y)\]

Answer 7

Okay, if you recall we have this equation:
\[(1)\quad\quad F(x)\underbrace{(\beta x+\delta xy)}_{N(x,y)}-G(y)\underbrace{(\alpha y+\gamma xy)}_{M(x,y)}=M_y-N_x\]
where the differential equation was
\[M(x,y)~dx+N(x,y)~dy=0\]
\(F(x)\) and \(G(y)\) are equivalent to \(\dfrac{u_x}{u}\) and \(\dfrac{u_y}{u}\), respectively, and \(M_y\) and \(N_x\) denote the partial derivatives with respect to the subscript variables.

Equation (1) becomes
\[\begin{align*}F(x)(\beta x+\delta xy)-G(y)(\alpha y+\gamma xy)&=(\alpha+\gamma x)-(\beta+\delta y)\\\\
F(x)(\beta x+\delta xy)&=G(y)(\alpha y+\gamma xy)+(\alpha+\gamma x)-(\beta+\delta y)\\\\
F(x)&=\frac{G(y)(\alpha y+\gamma xy)+(\alpha+\gamma x)-(\beta+\delta y)}{x(\beta +\delta y)}\end{align*}\]
(Note: Solving for \(G(y)\) should get the same result.)

As before, since \(F(x)\) is supposed to be a function of \(x\) only, we want to be able to remove any factors of \(y\). To do this, we'll say the numerator is equal to a constant times the function of \(y\) in the denominator.
\[G(y)(\alpha y+\gamma xy)+(\alpha+\gamma x)-(\beta+\delta y)=C(\beta+\delta y)\]
If we let \(C=-1\), we have a few terms disappear and we can solve for \(G(y)\):
\[\begin{align*}G(y)(\alpha y+\gamma xy)+(\alpha+\gamma x)\color{red}{-(\beta+\delta y)}&=\color{red}{-\beta-\delta y}\\\\
G(y)y(\alpha +\gamma x)+(\alpha+\gamma x)&=0\\\\
(G(y)y+1)(\alpha +\gamma x)&=0\\\\
G(y)y+1&=0\\\\
G(y)&=-\frac{1}{y}\end{align*}\]
Substituting this into the \(F(x)\) equation gives
\[\begin{align*}F(x)&=\frac{-\dfrac{1}{y}(\alpha y+\gamma xy)+(\alpha+\gamma x)-(\beta+\delta y)}{x(\beta +\delta y)}\\\\
F(x)&=\frac{-\alpha -\gamma x+(\alpha+\gamma x)-(\beta+\delta y)}{x(\beta +\delta y)}\\\\
F(x)&=-\frac{\beta+\delta y}{x(\beta +\delta y)}\\\\
F(x)&=-\frac{1}{x}\end{align*}\]
In terms of the notation you're given, \(F(x)=P(x)\) and \(G(y)=Q(y)\).

The integrating factor is then
\[\log u(x,y)=-\int\frac{dx}{x}-\int\frac{dy}{y}~~\implies~~u(x,y)=\frac{1}{xy}\]

Answer 8

The answer I got is \[\frac{ (\alpha*y+\gamma*x*y)x+(\beta*x+\delta*x*y)y }{ xy }=C\]

Answer 9

But the book's answer is \[\left| x \right|^{\alpha}*\left| y \right|^{\beta}*e ^{\gamma*x}*e ^{\delta*y}=C\]

Answer 10

Which is the right answer?

Answer 11

@SithsAndGiggles

Answer 12

Distributing the IF gives
\[\left(\frac{\alpha}{x}+\gamma\right)dx+\left(\frac{\beta}{y}+\delta\right)dy=0\]
Partials:
\[\frac{\partial}{\partial y}\left(\frac{\alpha}{x}+\gamma\right)=0\\
\frac{\partial}{\partial x}\left(\frac{\beta}{y}+\delta\right)=0\]
Solving for \(\Psi\):
\[\begin{align*}\Psi_x&=\frac{\alpha}{x}+\gamma\\\\
\Psi &=\int\left(\frac{\alpha}{x}+\gamma\right)~dx\\\\
&=\alpha\ln|x|+\gamma x+f(y)\\\\
\Psi_y&=f'(y)\\\\
\frac{\beta}{y}+\delta&=f'(y)\\\\
f(y)&=\beta \ln|y|+\delta y+C
\end{align*}\]
which makes the solution
\[\Psi=\alpha\ln|x|+\gamma x+\beta \ln|y|+\delta y=C\]
Taking the exponential of both sides will give you the solution from your textbook.

Answer 13

@SithsAndGiggles , my computer won't show what you typed since processing math is 0%. Can you please take a snapshot of what you just typed?

Answer 14

Thank you so much, @SithsAndGiggles !!! I'm so sorry that my internet is super slow sometimes.

Answer 15

You're welcome! And don't worry about it, technology tends to be a inconvenient sometimes.