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

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The integral in the linear differential equation cannot be evaluated in closed form, as it involves the integral of x^2e^(x^2), which lacks a finite expression using elementary functions. The discussion suggests that there may be a typo in the original problem statement, with possible corrections indicating different forms of the differential equation. Despite the inability to express the solution in closed form, a valid solution exists, represented as y = e^(-x^2)(∫x^2e^(x^2)dx + C). The participants emphasize the importance of verifying the original equation to ensure accurate evaluation.
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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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?
 
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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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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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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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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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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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.
 

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