differential equation

Question

solve f'(x)=-4x^2+3x-1 given that f(1)=0

Answer 1

f'(x) = -4x² + 3x - 1

 

f(x) = ∫f'(x)dx = ∫(-4x² + 3x - 1)dx = (-4/3)x³ + (3/2)x² - x + C

 

f(1) = 0 ⇒ -4/3 + 3/2 - 1 + C = 0

 

C = -(-8 + 9 - 6)/6 = 5/6

 

f(x) = (-4/3)x³ + (3/2)x² - x + 5/6