Where did the "t" come from?

Let's start over.

y' - y = x^2, y(0) = -1

Solve the homogeneous equation:

yh' - yh = 0

The characteristic equation is:

m - 1 = 0 ==> m = 1

Thus

yh(x) = Ae^{x}

Now we need a particular solution:

yp' - yp = x^2

Try

yp(x) = Bx^2 + Cx + D

yp'(x) = 2Bx + C

So

[2Bx + C] - [Bx^2 + Cx + D] = x^2

-Bx^2 + (2B - C)x + (C - D) = x^2

Thus

-B = 1

2B - C = 0

C - D = 0

Or B = -1, C = -2, and D = -2

So yp(x) = -x^2 - 2x - 2

Thus the solution to

y' - y = x^2

is

y(x) = yh(x) + yp(x) = Ae^{x} - x^2 - 2x - 2

We know that y(0) = -1, so

y(0) = Ae^{0} - (0)^2 - 2(0) - 2 = -1

A - 2 = -1

A = 1

Thus

y(x) = e^{x} - x^2 - 2x - 2

fits the bill.

-Dan