What is the issue with the two methods used to solve the ODE dy/dx=x(1-y)?

In summary, the conversation discusses solving the equation dy/dx=x(1-y) using both linear and separable methods. The two methods yield similar but slightly different answers, leading to confusion about which method is correct. The conversation also includes a brief explanation and correction of a typo in the separable method.
  • #1
badtwistoffate
81
0
Got the eqn dy/dx=x(1-y) and it can be solved both linear and separable methods.(Linear method being using a integrating factor) Problem I am having is that with this two methods i get two different (yet similar answers) and was wondering if you can see my problem with this two methods I am using.

Integrating Factor method:
y'+xy=x, u'(x)=e^(x^2/2)

[e^(x^2/2)y]'=x*e^(x^2/2)

e^(x^2/2)y=integral(x*e^(x^2/2)), do u substitution, get...

e^(x^2/2)y=e^(x^2/2)+c

y=1+c/e^(x^2/2) or y=1+c*e^(-x^2/2)

Separable method:
dy/(1-y)=x dx, integrate both sides

-ln(1-y)=e^(x^2/2)+C, raise both sides to e.

1/(1-y)=K*e^(x^2/2)+C, rearrange to get y=.

y=1-1/K*e^(x^2/2)

so we get two different answers with these methods, where is the problem lieing or are both wrong?
 
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  • #2
badtwistoffate said:
Separable method:
dy/(1-y)=x dx, integrate both sides

-ln(1-y)=e^(x^2/2)+C, raise both sides to e.

1/(1-y)=K*e^(x^2/2)+C, rearrange to get y=.

y=1-1/K*e^(x^2/2)
You got a bit sloppy near the end. Some mistakes are just typo's I think.

[tex]-\ln(1-y)=\frac{1}{2}x^2+C[/tex]
[tex]1-y=K\exp(-\frac{1}{2}x^2)[/tex]
[tex]y=1-K\exp(-\frac{1}{2}x^2)[/tex]

So it's the same.
 

1. What is an Integrating Factor in an ODE?

An integrating factor is a function that is multiplied to both sides of an ordinary differential equation (ODE) to make it easier to solve. It is used to convert a non-exact ODE into an exact one by introducing an additional term that helps to integrate the equation.

2. How do I determine the integrating factor for an ODE?

To determine the integrating factor for an ODE, you can use the method of undetermined coefficients or the method of variation of parameters. The method of undetermined coefficients involves finding a function that satisfies a certain differential equation, while the method of variation of parameters involves finding a function that satisfies a system of equations.

3. Why is an integrating factor needed in ODEs?

An integrating factor is needed in ODEs because it helps to transform a non-exact equation into an exact one, making it easier to solve. It also helps to simplify the process of solving higher-order ODEs by reducing them to first-order equations.

4. Are there any limitations to using an integrating factor in ODEs?

While an integrating factor is a useful tool in solving ODEs, it may not work for all types of equations. In some cases, it may not be possible to find an integrating factor that can transform a non-exact equation into an exact one. Additionally, the use of integrating factors may result in more complex solutions that are difficult to interpret.

5. Can an integrating factor be used for systems of ODEs?

Yes, an integrating factor can be used for systems of ODEs. In this case, the integrating factor is a matrix function that is multiplied to both sides of the system to make it easier to solve. It helps to reduce the system to a set of first-order equations, which can then be solved using standard techniques.

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