Differential Equations?

Question

I am stuck on a variation of parameters problem. 

y'' - y = coshx

I can find yc which is yc = c1e^x + c2e^(-x)

But I am having difficulties with the yp. I am ending up with:

u1 = (hsinhx-coshx) / [(2h^2+2)e^x]

u2 = [(coshhx+hsinhx)e^x] / (2h^2 + 2)

and when i do u1y1 + u2y2 i get:

yp = (hsinhx) / (h^2+1)

The correct answer for yp is suppose to be:

yp = (1/2)xsinhx

with a final solution of y = c1e^x + c2e^(-x) + (1/2)xsinhx

Any help would be greatly appreciated.

schmiso · Accepted Answer

Your complementary solution is correct, but i think you misinterpreted the function on RHS of  DE.
It is not a cosine (cos(h·x)). It is a hyperbolic cosine cosh(x).

For the solution it is better to use definition of  the hyperbolic cosine:
cosh(x) = (1/2)·(e^x + e^-x)
The hyperbolic sine, which occurs in the solution, is defined by:
sinh(x) = (1/2)·(e^x + e^-x)

Let
y1= e^x
y2= e^-x

The Wronskian of the fundamental system is:
W = y1·y2' - y2·y1' = e^x·(-e^-x) - e^x ·e^-x = -2

The functions u1 and u2 are
u1 = -∫ y2·f(x)/W dx = ∫ e^-x·((1/2)·(e^x + e^-x)) /(-2) dx
= (1/4)· ∫ 1 + e^-2x dx = (1/4)·x - (1/8)·e^-2x + c1
u2 = ∫ y1·f(x)/W dx = ∫ e^x·((1/2)·(e^x + e^-x)) /(-2) dx
= -(1/4)· ∫ e^2x + 1 dx = -(1/8)·e^2x - (1/4)·x  + c2

=>
y =  y1·u1 + y2·u2
= e^x · ((1/4)·x - (1/8)·e^-2x + c1) + e^-x · (-(1/8)·e^2x - (1/4)·x  + c2)
= (c1 - (1/8))·e^x + (c2 - (1/8))·e^-x + (1/4)·x·(e^x - e^-x)
= C1·e^x + C2·e^-x + (1/2)·x·sinh(x)

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Differential Equations?

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