Integrating factors found by inspection

Flora Mae Ramirez Quiño's picture

y(2-3xy)dx-xdy=0

fitzmerl duron's picture

The differential equation $y(2-3xy)dx-xdy=0$ has the form $M(x,y)dx+N(x,y)dy=0$, which is of first order and first degree, doesn't seem to fall into a variable-separable, homogenous, or exact type of differential equation. But delving deeper into the equation, we noticed that the given differential equation could be an exact differential equation if we modify it properly.

You're interested in getting the integrating factor of $y(2-3xy)dx-xdy =0$, so here's how.....

To check that the given differential equation is an exact-type, a necessary condition $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ must be satisfied, otherwise, it would be an inexact differential equation.

So....

$$y(2-3xy)dx-xdy=0$$ $$(2y-3xy^2)dx-xdy=0$$ $$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$ $$\frac{\partial}{\partial y}(2y-3xy^2)=\frac{\partial }{\partial x}(-xdy)$$

To get $\frac{\partial M}{\partial y}$:

$$\frac{\partial M}{\partial y} =\frac{\partial }{\partial y}(2y-3xy^2)$$ $$\frac{\partial }{\partial y}(2y-3xy^2) =2-3x(2y)$$ $$\frac{\partial }{\partial y}(2y-3xy^2) =2-6xy$$ $$\frac{\partial M}{\partial y} =\frac{\partial }{\partial y}(2y-3xy^2)= 2-6xy$$

To get $\frac{\partial N}{\partial x}$:

$$\frac{\partial N}{\partial x} =\frac{\partial (-x)}{\partial x}$$ $$\frac{\partial (-x)}{\partial x} = -1$$ $$\frac{\partial N}{\partial x} =\frac{\partial (-x)}{\partial x} = -1$$

We see that $\frac{\partial}{\partial y}(2y-3xy^2) \neq \frac{\partial }{\partial x}(-x)$

Therefore, the given differential equation is an inexact differential equation, not an exact-type, so we need to convert this inexact differential equation into an exact-type. This is how to do it.

To make an inexact differential equation into an exact-type, multiply an integrating factor $I(x,y)$ to the entire differential equation. Our integrating factor would be $x^p y^q$. Solve for $p$ and $q$ to make it exact.

$$y(2-3xy) dx - xdy = 0$$ $$(2y-3xy^2)dx - xdy = 0$$ $$x^py^q((2y-3xy^2)dx - xdy = 0)$$ $$(2x^py^{q+1} - 3x^{p+1}y^{q+2})dx - (x^{p+1}y^q)dy = 0$$

Then...

$$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(2x^py^{q+1} - 3x^{p+1}y^{q+2})$$ $$\frac{\partial M}{\partial y} = (q+1)(2x^p)(y^{(q+1)-1})-(q+2)(3x^{p+1})(y^{(q+2)-1})$$ $$\frac{\partial M}{\partial y} = 2(q+1)x^py^q-3(q+2)x^{p+1}y^{q+1}$$

And...

$$\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(- (x^{p+1}y^q))$$ $$\frac{\partial N}{\partial x} = (-1)((p+1)x^{(p+1)-1}y^q)$$ $$\frac{\partial N}{\partial x} = (-1)((p+1)x^py^q)$$ $$\frac{\partial N}{\partial x} = -(p+1)x^py^q$$

Therefore...
$$\frac{\partial M}{\partial y} = 2(q+1)x^py^q-3(q+2)x^{p+1}y^{q+1}$$ $$\frac{\partial N}{\partial x} = -(p+1)x^py^q + 0$$

We now solve for $p$ and $q$:

This is a linear equation, two equations and two unknowns...
$$2(q+1) = -(p+1)$$ $$2q+2=-p-1$$ $$2q+p=-1-2$$ $$\color{red}{p+2q=-3}$$

And....

$$\color{red}{-3(q+2)=0}$$ $$-3q-6=0$$ $$\color{red}{\underline{q=-2}}$$

Then...
$$p+2q=-3$$ $$p+2(-2)=-3$$ $$\color{red}{\underline{p=1}}$$

Therefore, the integrating factor that we seek is $\color{green} {I(x,y) = x^py^q = x^1y^{-2} = \frac{x}{y^2}}$.

To verify that inexact differential equation $y(2-3xy)dx-xdy=0$ can be made exact by multiplying it by $\frac{x}{y^2}$, we do this:

$$\left( \frac{x}{y^2}(y(2-3xy)dx-xdy=0)\right)$$ $$\left( \frac{x}{y^2}((2y-3xy^2)dx-xdy=0)\right)$$ $$\left(\frac{2x}{y}-3x^2\right)dx - \frac{x^2}{y^2}dy = 0$$

Then...

$$\frac{\partial M}{\partial y} = \frac{\partial }{\partial y}\left(\frac{2x}{y}-3x^2\right)$$ $$\frac{\partial M}{\partial y} = -\frac{2x}{y^2}$$

And...

$$\frac{\partial N}{\partial x} = \frac{\partial }{\partial x}\left(\frac{2x}{y}-3x^2\right)$$ $$\frac{\partial N}{\partial x} = -\frac{2x}{y^2}$$

So...
$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$

The integrating factor of $y(2-3xy)dx-xdy=0$ is $\color{green} {\frac{x}{y^2}}$

Alternate solutions are highly encouraged.....

Nisha Alcantara's picture

2ydx-3xy2dx-xdy=0

(2ydx-3xy2dx-xdy=0)x/y2

((2yxdx-x2dy)/y2) - 3x2 = 0

Integrate
d(x2/y) - 3x2 = 0

x2/y - x3 = c

Simplify

x2(1-xy) = cy this is your general solution

I hope this helps..

From Fredierick Uy..

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