Question

Find an integrating factor of the form u(x, y)=P(x)Q(y) for (3x^2*y^3-y^2+y)dx+(-xy+2x)dy=0.

Answer 1

@SithsAndGiggles

Answer 2

*

Answer 3

u(x,y)M dx + u(x,y)N dy=0
Now, the above equation is an exact differential equation.
\[\frac{ \delta }{ \delta y }(u(x,y) M)=\frac{ \delta }{ \delta x }(u(x,y) N)\]

Answer 4

Use this equation to find u(x,y).

Answer 5

looks the solution is going to be very long

Answer 6

Indeed.

Answer 7

it will be easy if any one of the functions P or Q is 1

Answer 8

i worked these problems assuming the IF is a function of x or y alone
but never worked a problem that requires IF to be function of both variables

Answer 9

hmm partial differential equation hehe
it hates me lol xD

Answer 10

its ODE only... we are finding IF to make it exact

Answer 11

u(x,y ) partial

Answer 12

yeah requires taking partial derivatives, but this is not a paritial differential eqn

Answer 13

sure ?

Answer 14

ugh u don't see any partial derivatives in the given eqn

Answer 15

*i

Answer 16

im not that good , but i put this question on my list to learn , tomorrow i'll type what ever i got xD

Answer 17

Using the method we've discussed before:
\[\frac{u_x}{u}N-\frac{u_y}{u}M=M_y-N_x\]
We want to find \(F(x)=\dfrac{u_x}{u}\) and \(G(y)=\dfrac{u_y}{u}\).

Partials:
\[M_y=\frac{\partial }{\partial y}[3x^2y^3-y^2+y]=9x^2y^2-2y+1\\
N_x=\frac{\partial }{\partial x}[-xy+2x]=-y+2\]
Substituting:
\[\begin{align*}F(x)(-xy+2x)\\-G(y)(3x^2y^3-y^2+y)&=9x^2y^2-2y+1-(-y+2)\\\\
G(y)&=\frac{F(x)(-xy+2x)-9x^2y^2+y+1}{3x^2y^3-y^2+y}\end{align*}\]
Let's try to eliminate the factor of \(3x^2y^3-y^2+y\) by finding a function \(C(y)\) such that
\[F(x)(-xy+2x)-9x^2y^2+y+1=C(y)(3x^2y^3-y^2+y)\]
Supposing \(C(y)=-\dfrac{3}{y}\), we have
\[\begin{align*}F(x)(-xy+2x)-9x^2y^2+y+1&=-9x^2y^2+3y-3\\\\
F(x)(-xy+2x)&=2y-4\\\\
F(x)&=\frac{2y-4}{-x(y-2)}\\\\
F(x)&=-\frac{2}{x}\end{align*}\]
Solving for \(G(y)\):
\[\begin{align*}G(y)&=\frac{-\dfrac{2}{x}(-xy+2x)-9x^2y^2+y+1}{3x^2y^3-y^2+y}\\\\
G(y)&=\frac{-9x^2y^2+3y-3}{y(3x^2y^2-y+1)}\\\\
G(y)&=\frac{-3(3x^2y^2-y+1)}{y(3x^2y^2-y+1)}\\\\
G(y)&=-\frac{3}{y}
\end{align*}\]
So the IF will be
\[\begin{align*}\log u(x,y)&=\int F(x)~dx+\int G(y)~dy\\\\
&=-2\int\dfrac{dx}{x}-3\int\frac{dy}{y}\\\\
u(x,y)&=\frac{1}{x^2y^3}\end{align*}\]

Answer 18

@Idealist10 there's a slight deviation from the approach I described the first time around. Before we let \(C\) be a constant, while here \(C\) is a function (though even when it is a constant, it's still technically a function).

The point is that this sort of "greedy" approach lets you simplify the PDE so that you can find those one-dimensional functions \(F\) and \(G\).

Answer 19

Okay, I get it. Thanks a lot!