Find the function that satisfies the condition given for the initial value problem

f '(x) = 3 - 2x, y(0) = 5

and

f "(x) - 3x² - 2x, f(0) = 4, f '(1) = 3

thanks

f'(x) = 3 − 2x

f(x) = ∫ (3 − 2x) dx
f(x) = 3x − x² + C
f(0) = 5 -----> C = 5

f(x) = −x² + 3x + 5

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f''(x) = 3x² − 2x

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

f'(x) = x³ − x² + 3

f(x) = ∫ (x³ − x² + 3) dx
f(x) = 1/4 x⁴ − 1/3 x³ + 3x + C
f(0) = 4 -----> C = 4

f(x) = 1/4 x⁴ − 1/3 x³ + 3x + 4

Mαthmφm