Find a particular solution that satasfies the D.E f '(x)= 2x and its initial conditions f '(1)= 2, f(0)= -1

f " (x)  = 2x   f ' (1)   = 2, f (0)  = -1

f "(x)   = 2x        integrate to find the first derivative

f ' (x) =  x^2  + C₁      applying the initial condition to find C₁ ,  we have that

f '( 1)  = 2

2  =    (1)^2 + C₁

2  = 1 + C₁

1  = C₁

So....  f'(x) =  x^2 + 1

Integrate this again to find y(x)

f(x)  = (1/3)x^2 + x  + C₂

And applying  the  second initial condition to find C₂, we have that

f(0) = -1

-1 = (1/3)(0)^3 + 0 + C₂

C₂  = -1

So.....   f(x)  =  (1/3)x^3 +  x   - 1