1st order nonhomogeneous dif'l e'n

  • Thread starter rbwang1225
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  • #1
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f'+x*f=x^(2m+1), m is an integer

I have no idea for finding particular solution.

Any help would be appreciated.
 
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  • #2
HallsofIvy
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Well, that is a linear equation so it should be fairly easy to find an integrating factor.

That is, you want to find a function, u(x), such that multiplying by it makes the left side an "exact derivative":
[tex]u(x)\frac{df}{dx}+ xu(x)f(x)= \frac{d(u(x)f}{dx}[/tex]

By the product rule
[tex]\frac{d(uf)}{dx}= u\frac{df}{dx}+ \frac{du}{dx}f[/tex]

Comparing those, we must have
[tex]\frac{du}{dx}= xu[/tex]
or
[tex]\frac{du}{u}= x dx[/tex]

Integrate that to find u(x), then multiply the entire differential equation by u(x) to get an exact equation.

By the way, you really need to use parenthese to clarify. Is that right hand side
(x^2)(m+1) or x^(2m)+ 1 or x^(2m+1)?

(I expect it is the last since that exponent being odd is important in integrating.)
 
  • #3
tiny-tim
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hi rbwang1225! :smile:

(try using the X2 icon just above the Reply box :wink:)
f'+x*f=x^2m+1, m is an integer

I have no idea for finding particular solution.

Any help would be appreciated.

is that f' + x*f = x2m+1 or f' + x*f = x2m + 1 ? :confused:

either way, you have a polynomial on the RHS, so the obvious guess for a particular solution is … ? :wink:
 
  • #4
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rbwang1225
The particular solution is

f*(x)=x^(2m)-2*m*x^(2m-2)+(2^2)*m*(m-1)*x^(2m-4)-...+((-1)^m)*(2^m)*m!
 

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