find the function F that satisfies the following differential equations.

Question

f'''(x)=4x, f''(0)=0, f'(0)=1, f(0)=3

Answer 1

first we integrate 4x to recover f''

int of (4x)=2x^2+c

that means f''(x) +2x^2+c

initial condition f''(0)=0

then c=0 and f''(x) =2x^2

now we integrate 2x^2 to recover f'

int of (2x^2)= 2/3 x^3+c

same method with initial conditions f'(0)=1 we can find c

1=2/3 (0)^3+c ==> c=1

so f'(x) =2/3 x^3+1

to the last step we will recover f which is the target function 

int of (2/3 x^3+1)= 1/6 x^4+ x+c

again use initial conditions

f(0)=3

so 3= 1/6 (0)^4+(0)+c 

c=3

therefore f(x) =1/6 x^4+x+3

is the answer