Can you help me evaluate the integral in this linear differential equation?

In summary: In this case we cannot express the integral as a closed form so all we can do is write the integral in the solution just as it is, so the solution would be $$y=e^{-x^2}\left ( \int x^2e^{x^2} dx+C\right )$$ which is a solution not in closed form. So a solution exists , but simply we cannot write the solution in closed form.
  • #1
Butterfly41398
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Homework Statement
Fing the general solution
Relevant Equations
Integrating factor = exp{integral[p(x)]}
I tried it but I don't know how to evaluate the integral on the last equation. Help.
 

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  • #2
Butterfly41398 said:
Homework Statement:: Fing the general solution
Relevant Equations:: Integrating factor = exp{integral[p(x)]}

I tried it but I don't know how to evaluate the integral on the last equation. Help.
Can you post a photo of the problem as given in your book?
 
  • #3
It seems to me what you did is almost correct (you forgot the integration constant though so that equation should be $$ye^{x^2}=\int x^2e^{x^2}dx+C$$)
However the above integral doesn't have a closed form. Perhaps the original equation is $$\frac{dy}{dx}=x^3-2xy$$? (or even $$\frac{dy}{dx}=x-2xy$$) cause if it is so then the integral will be $$\int x^3 e^{x^2} dx$$ (or $$\int xe^{x^2} dx$$) and will have a closed form.
 
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  • #4
Delta2 said:
However the above integral doesn't have a closed form.
Which is why I asked to see the original problem, in case it's different from what is being worked on here.
 
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  • #5
Mark44 said:
Can you post a photo of the problem as given in your book?
That's really what is given on the book sir.
 

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  • #6
Delta2 said:
It seems to me what you did is almost correct (you forgot the integration constant though so that equation should be $$ye^{x^2}=\int x^2e^{x^2}dx+C$$)
However the above integral doesn't have a closed form. Perhaps the original equation is $$\frac{dy}{dx}=x^3-2xy$$? (or even $$\frac{dy}{dx}=x-2xy$$) cause if it is so then the integral will be $$\int x^3 e^{x^2} dx$$ (or $$\int xe^{x^2} dx$$) and will have a closed form.
I've attached the original pocture sir. It's really x^2. So does that mean the differential equation can't be solved? And by means "closed form" what does it mean sir. I want to know more. Tnx in advance.
 

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  • #7
By closed form we mean a form that is finite and contains only the elementary well known functions (##x^n,\sin x, e^x,\ln x,a^x## e.t.c) for example $$y=\frac{\tan(x^2e^{x^2}+1)+\ln({x^3+1})}{2^x}$$ is a closed form (no matter how complex it might be).

In this case we cannot express the integral as a closed form so all we can do is write the integral in the solution just as it is, so the solution would be $$y=e^{-x^2}\left ( \int x^2e^{x^2} dx+C\right )$$ which is a solution not in closed form. So a solution exists , but simply we cannot write the solution in closed form.

I suspect a typo in the statement of the problem, it should be ##x## or ##x^3##, can't explain it otherwise.
 
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  • #8
Delta2 said:
I suspect a typo in the statement of the problem
I do, too, unless the intent was to give the solution in terms of the integral shown in Delta2's post.
 

Related to Can you help me evaluate the integral in this linear differential equation?

1. What is a linear differential equation?

A linear differential equation is an equation that involves a dependent variable and its derivatives with respect to one or more independent variables. The equation is considered linear if it can be written in the form of a polynomial, with the dependent variable and its derivatives having a degree of 1.

2. What is the order of a linear differential equation?

The order of a linear differential equation is the highest derivative present in the equation. For example, a first-order linear differential equation would have a maximum degree of 1, while a second-order equation would have a maximum degree of 2.

3. What is the general solution to a linear differential equation?

The general solution to a linear differential equation is a family of solutions that satisfies the equation. It contains a constant of integration, which allows for an infinite number of possible solutions. To find the specific solution, initial conditions must be given.

4. How is a linear differential equation solved?

A linear differential equation can be solved using various methods, such as separation of variables, integrating factor, or the method of undetermined coefficients. The specific method used depends on the form of the equation and the given initial conditions.

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Linear differential equations have many real-world applications, such as in physics, engineering, and economics. They are used to model systems that involve rates of change, such as population growth, chemical reactions, and electrical circuits.

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