Differential Equations-Particular Integral

[Mehta]

1. The problem statement: I have to solve the following differential equation without using variation of parameters:
2. ((d² y)/(d x²))+y= (cosh x) (cos x)
3. The Attempt at a Solution :

The auxiliary equation would become

D² + 1 = 0,where D is the differential operator.
This means D² = - 1,i.e. D= +/- i

Thus,the C. F. is
C₁ cos x + C₂ sin x
where C₁ and C₂ are constants.

Now to find the particular integral:

P. I. = (1 / (D² + 1)) (cosh x) (cos x)

= (1/ (D² + 1))(e^x+e^-x)(cos x)

=(1/ (D² + 1))(e^x)(cos x)+(1/ (D² + 1))(e^-x)(cos x)

=(e^x) (1/((D+1)²+1))(cos x) + (e^-x)(1/((D-1)²+1))(cos x)

=(e^x)(1/(D²+2D+2))(cos x)+(e^-x)(1/(D²-2D+2))(cos x)

=(e^x)(1/(-1+2D+2))(cos x)+(e^-x)(1/(-1-2D+2))(cos x)

=(e^x)(1(2D-1)/((2D+1)(2D-1)))(cos x)+(e^-x)(1(1+2D)/((1-2D)(1+2D)))(cos x)

=(e^x)((2D-1)/(4D²-1))(cos x)+(e^-x)((1+2D)/(1-4D²))(cos x)

What now to do next?

The answer is given to be ((2/5) (sin x) (sinh x))+((1/5) (cos x) (cosh x)).How do I get there?

 

[vela]

I must admit I have no idea what you're doing.

Why not use the method of undetermined coefficients to find the particular solution?