# Solving a Non-Exact ODE: $(y-2x^2y)dx +xdy = 0$

• MHB
• NotaMathPerson
In summary, the given ODE is in exact form, but when tested for exactness, it fails. The integrating factor found using the technique of $\frac{M_y-N_x}{N}$ is $u(x)=-2x$, but this still results in a non-exact ODE. However, when treating the ODE as separable and using the integrating factor $\mu(x)=e^{-x^2}$, the ODE becomes exact.
NotaMathPerson
Solve the ode

$$(y-2x^2y)dx +xdy = 0$$

The equation is in exact form $$Q(x,y)dx+ P(x,y)dy =0$$

When I test for exactness it fails. Then I used the technique $$\frac{M_y-N_x}{N}$$

I get

$u(x)=-2x$ as my integrating factor.

But I end still end up with a non-exact d.e why is that?

$$(-2xy+4x^3y)dx-2x^2dy=0$$

Last edited:
It's separable. Factor out the $y$ on the $dx$ term.

Ackbach said:
It's separable. Factor out the $y$ on the $dx$ term.

Yes i did that already. What I am after is the answer as to why the method i used above did not work.

You need to exponentiate the integral of $(M_y-N_x)/N$. That is, in general,
$$u(x)=\exp\left[\int\left(\frac{M_y-N_x}{N}\right)dx\right].$$
I get $u(x)=e^{-x^2}$. See if you agree.

NotaMathPerson said:
Solve the ode

$$(y-2x^2y)dx +xdy = 0$$

The equation is in exact form $$Q(x,y)dx+ P(x,y)dy =0$$

When I test for exactness it fails. Then I used the technique $$\frac{M_y-N_x}{N}$$

I get

$u(x)=-2x$ as my integrating factor.

But I end still end up with a non-exact d.e why is that?

$$(-2xy+4x^3y)dx-2x^2dy=0$$

I agree with Ackbach that treating the given ODE as separable is the way to go here. However, if you wish to compute an integrating factor, you should use:

$$\displaystyle \mu(x)=\exp\left(\int -2x\,dx\right)=e^{-x^2}$$

$$\displaystyle \left(e^{-x^2}y\left(1-2x^2\right)\right)dx+\left(xe^{-x^2}\right)dy=0$$

And you can see by inspection this ODE is exact. :)

## 1. What is a non-exact ODE?

A non-exact ODE is a type of ordinary differential equation that cannot be solved using the standard methods of integrating factors or separation of variables. It often involves terms that are not directly dependent on the dependent variable, making it difficult to manipulate into a solvable form.

## 2. How do you solve a non-exact ODE?

To solve a non-exact ODE, you can use various methods such as the method of integrating factors, the method of undetermined coefficients, or the method of variation of parameters. These methods involve manipulating the equation and using known solutions to find a solution to the non-exact ODE.

## 3. What is the role of the integrating factor in solving a non-exact ODE?

The integrating factor is a function that is multiplied to both sides of the non-exact ODE to transform it into an exact ODE. This allows for the use of standard methods to solve the equation and obtain a general solution.

## 4. Can a non-exact ODE have multiple solutions?

Yes, a non-exact ODE can have multiple solutions. The general solution to a non-exact ODE usually includes an arbitrary constant, and different values of this constant can lead to different solutions to the equation.

## 5. What are the applications of solving non-exact ODEs?

Non-exact ODEs are commonly used in physics, engineering, and other scientific fields to model various phenomena such as population growth, chemical reactions, and heat transfer. Being able to solve these equations allows for a better understanding and prediction of the behavior of these systems.

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