- #1

- 118

- 0

f'+x*f=x^(2m+1), m is an integer

I have no idea for finding particular solution.

Any help would be appreciated.

I have no idea for finding particular solution.

Any help would be appreciated.

Last edited:

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter rbwang1225
- Start date

- #1

- 118

- 0

I have no idea for finding particular solution.

Any help would be appreciated.

Last edited:

- #2

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 964

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

Science Advisor

Homework Helper

- 25,832

- 251

(try using the X

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 = x

either way, you have a

- #4

- 3

- 0

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!

Share: