puremahts qestion

Let y = f(x), then we have:
dy/dx = 1968ex + y
dy/dx - y = 1968ex
e-x(dy/dx) - ye-x = 1968
d(ye-x)/dx = 1968
ye-x = 1968x + C, where C is an arbitrary constant.
y = 1968xex + Cex
From the given:
dy/dx = 1968ex + y
d2y/dx2 = 1968ex + dy/dx
d2y/dx2 = 1968ex + 1968ex + y
d2y/dx2 = 3936ex + 1968xex + Cex
At x = 0, d2y/dx2 = 1, so
3936 + C = 1
C = -3935
Therefore,
y = f(x) = 1968xex - 3935ex

f'(x)=(1968e^x)+f(x) and f''(0)=1, find f(x)

f''(x)=(1968e^x)+f'(x) and f''(0)=1

So 1=1968+f'(0)

f'(0)=-1967

f(0)=-1967-1968=-3935

Let y=f(x)

Then dy/dx-y=1968e^x

d(e^(-x)y)/dx=1968

e^(-x)y=1968x+C

y=1968xe^x+Ce^x

Sub x=0 C=-3935

So f(x)=1968xe^x-3935e^x