d3y/dx3+dy/dx=sin(2x). Here, i get as far as to lambda is 0,i,-i. Thereafter, I am not sure even how to write this in general form, let alone what to do with that sin thingy... can someone please help?

Also given is: y(0)=1, dy/dx(0)=0, d2y/dx2=0.

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- Mar 11th 2009, 11:07 AMfiksidiff eqn, sin(2x).
d3y/dx3+dy/dx=sin(2x). Here, i get as far as to lambda is 0,i,-i. Thereafter, I am not sure even how to write this in general form, let alone what to do with that sin thingy... can someone please help?

Also given is: y(0)=1, dy/dx(0)=0, d2y/dx2=0. - Mar 11th 2009, 11:56 AMfiksi
- Mar 12th 2009, 09:30 AMCoomast
Hello fiksi,

I assume that you can find the solution for the homogeneous equation, it is:

$\displaystyle y_h=A+B\cdot cos(x)+C\cdot sin(x)$

Now, what you need to use as particular solution is:

$\displaystyle y_p=D\cdot sin(2x)+ E\cdot cos(2x)$

Substitute this in the original DE and it will give you a system for finding D and E. (they are 0 and 1/6)

best regards,

Coomast