- #1

- 9

- 0

Could someone 'remind me' why the equation u" + u = km/L^2 has the solution of the form u(theta) = km/L^2 + C cos(theta - theta(o)).

Any references would be appreciated.

Thanks

Dave

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 ddoctor
- Start date

- #1

- 9

- 0

Could someone 'remind me' why the equation u" + u = km/L^2 has the solution of the form u(theta) = km/L^2 + C cos(theta - theta(o)).

Any references would be appreciated.

Thanks

Dave

- #2

- 137

- 0

[tex]u^{''}+u=0[/tex]

has solution

[tex]u_0=C\times cos(\theta-\theta_0)[/tex]

on the other hand, your equation has a particular solution

[tex]u_p=km/L^2=const[/tex]

so the general solution is

[tex]u=u_0+u_p[/tex]

Share: