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Taking the integral of xe^ax^2

  • #1

Homework Statement


dy/dx=x*e^(ax^2)

solve the differential equation



Homework Equations


integral of e^x=e^x


The Attempt at a Solution


im not really sure how to do it when there are two variables in the exponent? i tried several things like u=x^2 1/2du=xdx then 1/2*int[e^a*u] results, but i can't take the integral without a different substitution since int[e^a*u] is not = to e^a*u.

the answer is 1/2a * e^ax^2 + C.. any methods to integrate this? several methods would be best
 

Answers and Replies

  • #2
Mentallic
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Think about how you compute [tex]\frac{d}{dx}e^{f(x)}[/tex]
 
  • #3
349
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a is a constant, keep that in mind . . .
 
  • #4
I solved it with a as a constant and got the right answer, but . . .

How do i know a is a constant? I can't just assume that can I? What is the proof for a being a constant
 
Last edited:
  • #5
Mentallic
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I solved it with a as a constant and got the right answer, but . . .

How do i know a is a constant? I can't just assume that can I? What is the proof for a being a constant
Because a is assumed to be a constant in these situations, just as how [itex]\pi[/itex] nearly always represents the irrational number, but in some other cases, it can be mean something completely different, such as representing a product.

And if it wasn't a constant, then you can't solve the problem. So which do you think is more likely? :tongue:
 
  • #6
HallsofIvy
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Was the question just "integrate [itex]xe^{ax}[/itex]" or was it to find [itex]\int xe^{ax}dx[/itex]?
 
  • #7
That makes sense but it still bugs me to just assume it is a constant :(

And to the other person, the question was

dy/dx=x*e^(ax^2)

solve the differential equation, so yes it would be integral of x*e^ax^2 dx (except you forgot the squared on top of the x in e's exponent)

thanks for everybody's help
 
  • #8
epenguin
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The situation is no different than if you were told to solve dy/dx = a .
 
  • #9
Mentallic
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Was the question just "integrate [itex]xe^{ax}[/itex]" or was it to find [itex]\int xe^{ax}dx[/itex]?
What's the difference?
 
  • #10
HallsofIvy
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Was the question just "integrate [itex]xe^{ax^2}[/itex]" or was it to find [itex]\int xe^{ax^2}dx[/itex]?
What's the difference?
The second, [itex]\int xe^{ax^2} dx[/itex], makes it explicit that the 'variable of integration' is x ("dx") while the first, "integrate [itex]xe^{ax^2}[/itex]", is ambiguous. Of course, it is a standard convention that such things as "x", "y", "z" are used as variables while such things as "a", "b", "c", etc. are used to denote constants.

If it were [itex]\int xe^{ax^2}da[/itex], it would be [itex](1/x)\int e^{ax^2}+ C[/itex].
 
  • #11
Mentallic
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The second, [itex]\int xe^{ax^2} dx[/itex], makes it explicit that the 'variable of integration' is x ("dx") while the first, "integrate [itex]xe^{ax^2}[/itex]", is ambiguous. Of course, it is a standard convention that such things as "x", "y", "z" are used as variables while such things as "a", "b", "c", etc. are used to denote constants.
Yes, but without those standard conventions then we could just as well argue that a=f(x) in the integrand, just as the OP has done.
 

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