If f‴(x)=e^x, f″(0)=3 and f′(0)=2, then f(x)= ? + C.

Question

If f‴(x)=e^x, f″(0)=3 and f′(0)=2,

then f(x)= ? + C.

Answer 1

If f‴(x)=e^x, f″(0)=3 and f′(0)=2 ...

∫f'''(x) - e^x dx = C₁ = f''(x) - e^x

Putting x = 0, we get C₁ = 3 - 1 = 2, so f''(x) = e^x +2

∫f''(x) - e^x - 2 dx = C₂ = f'(x) - e^x - 2x

Putting x= 0, we get C₂ = 2 -1 - 0 = 1, so f'(x) = e^x + 2x +1

∫f'(x) - e^x - 2x - 1 dx = C = f(x) - e^x - x^2 - x

So f(x) = e^x + x^2 + x + C

Answer 2

Integrating ,f''=e^x+c

F(0)=3--> 1+c=3 --> c=2

F''= e^x+2

Integrating again,

F'= e^x+2x+c

F'(0)=2 --> 1+2+c=2--> c=-1

Y'=e^x+2x-1

Y=e^x+x^2-x+c