- #1

ddoctor

- 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

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- Thread starter ddoctor
- Start date

- #1

ddoctor

- 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

shyboy

- 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]

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