Find the particular solution of the differential equation that satisfies the initial condition

Find the particular solution of the differential equation that satisfies the initial condition

#14) f '' (x) = sin(x) + e^(2x)
f(0) = 1/4 ; f ' (0) = 1/2

#15) dx/dy = e^(x+y) x(1) = 0

Also I have one other question:

Use integration to find a general solution of the ifferential equation.
dy/dx - x cos x^2


Can somebody answer ASAP!?!?!?? PLEASEEEE

14) f '(x) = -cosx + e^(2x)/2 + c
Substitute 0 for x and 1/2 for f '(x) and solve for c.
1/2 = -cos0 + e^(2(0))/2 + c
1/2 = -1 + 1/2 + c
c = 1
f '(x) = -cosx + e^(2x)/2 + 1
f(x) = -sinx + e^(2x)/4 + x + d
1/4 = -sin0 + e^(2(0))/4 + 0 + d
1/4 = 1/4 + d
d = 0
f(x) = -sinx + e^(2x)/4 + x

15) dx/dy = e^(x + y)
dx/dy = e^x(e^y)
e^-x dx = e^y dy
∫ e^-x dx = ∫ e^y dy
-e^-x = e^y + c
-e^-1 = e^0 + c
-1/e = 1 + c
-1/e - e = c
-e^-x = e^y - 1/e - e

For #14 when you integrate f ''(x) it gives you f ' (x). Using f '(0) =1/2, solve for C. Integrate this and repeat to get f(x)