U'(X) - U(X) = 0;  U(0) = 2

U''(X) - U'(X) = 0;  U'(0) = U(0) = 2

Hi David,

If U'(X) - U(X) = 0 then U'(X) = U(X) and you can either recognize immediately that

U(X) = A e^X

or you can proceed as follows

U'(X) = U(X) so
 ^U'(X)/_U(X) = 1
You should then recognize that the left side is the derivative of ln[U(X)] with respect to X so that
ln[U(X)] = X + C, where C is a constant. Applying the exponential function to both sides gives
U(X) = e^X+C = e^C e^X
Write A = e^C and again you get
U(X) = A e^X

The initial condition is that U(0) = 2 so

U(0) = A e⁰ = A 1 = 2 and hence A = 2
Hence the solution is U(X) = 2 e^x

For the second problem let V(X) = U'(X) and you have

V'(X) - V(X) = 0; V(0) = 2

This is exactly the first problem. Hence you know V(X) so you can solve

V(X) = U'(X); U(0) = 2

To find U(X).

Harley