Variation of Parameters?

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.

acafrao341 · Accepted Answer

Where does the h come in?

The complementary solutions, exp x and exp(-x), can be rewritten as cosh x and sinh x, since these are just linear combinations of exponential functions.  This will make the algebra easier.  Their derivatives are sinh x and cosh x, respectively.  The functions u1 and u2 are found by solving the system of equations

     (cosh x)u1' + (sinh x)u2' = 0
     (sinh x)u1' + (cosh x)u2' = cosh x.

This yields u1' = -cosh x sinh x / (cosh^2 x + sinh^2 x)
     = -sinh 2x/2cosh 2x = (-1/2) tanh 2x
     u2' = cosh^2 x / (cosh^2 x + sinh^2 x)
     = (1 + cosh 2x) /2cosh 2x
     = (1/2) (sech 2x + 1)

Now integrate u1' and u2', to get

     u1 = (-1/4) log cosh 2x
     u2 = (1/2) arctan exp(2x).

The particular solution is then

     yp = u1 y1 + u2 y2
     = (-1/4) (log cosh 2x) cosh x + (1/2) (arctan exp(2x)) sinh x.

The general solution is

     y = c1 cosh x + c2 sinh x - (1/4) (log cosh 2x) cosh x + (1/2) (arctan exp(2x)) sinh x.

If there are initial conditions, c1 and c2 may now be found.

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Variation of Parameters?

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