Solving 2nd order differential equation

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Pushoam
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Homework Statement


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Homework Equations

The Attempt at a Solution



For the homogeneous equation, I have got the the root of the characteristic equation as ## e^{ix}, e^{-ix} ## .

So, the corresponding solution is ## B \sin{ x} + A \cos{ x} ## .

Then, I took the particular solution as C.

Putting C in the differential equation , I get C = -1.

So, the inhomogeneous solution is ## A \cos{ x} + B \sin{ x} – 1 ## , i.e. option(e).

Is this correct?
 

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Pushoam said:
So, the inhomogeneous solution is ## A \cos{ x} + B \sin{ x} – 1 ## , i.e. option(e).
Is this correct?
You shouldn't need to ask this. Along the lines of @scottdave's advice, whenever you get a solution to a diff. equation, you should get in the habit of checking. That way you'll know whether your answer is correct. For an initial value problem, check that 1) your solution satisfies the initial condition, and 2) your solution satisifies the differential equation.
 
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