Hello, Harry!

We have: . y .= .C1e^{2x} + C2e^{-2x} - 3A nonhomogeneous differential equation.

A complementary solution yc and particular solution yp are given.

Find the solution satisfying the given initial conditions.

. . y′′ − 4y .= .12, . yc .= .C1e^{2x} + C2e^{-2x}, .yp = -3

. . y(0) = 10, .y'(0) = 10

We are told: .y(0) = 0

. . Hence: we have: .C1e^0 + C2e^0 - 3 .= .0 . → . C1 + C2 .= .3 .[1]

We are told: .y'(0) = 10

. . y' .= .2C1e^{2x} - 2C2e^{-2x}

. . Hence: .2C1e^0 - 2C2e^0 .= .10 . → . C1 - C2 .= .5 .[2]

Add [1] and [2]: .2C1 = 8 . → . C1 = 4

Substitute into [1]: .C2 = -1

Therefore: . y .= .4e^{2x} - e^{-2x} - 3