V 29 10ekt V 39 10ekt V 39 39ekt V 39 10ekt Find the

V = 29 - 10e^-kt

V = 39 - 10e^-kt

V = 39 - 39e^-kt

V = 39 - 10e^kt

Find the particular solution determined by the given condition.

f\'(x) = 3x² - 2x; f(0) = 18

f(x) = 3x³ + 2x² + 18

f(x) = 3x³ + x² + 18

f(x) = x³ - x² + 18

f(x) = x³ + 2x² + 18

Solution

1)differential equation not visible in first question

2)given f \'(x) = 3x² - 2x

integrate with respect to x ,xⁿ dx =(1/(n+1))xⁿ⁺¹+C

f(x) = 3(1/3)x³ - 2(1/2)x²+C

f(x) = x³ - x²+C

given f(0)=18

18 = 0³ - 0²+C

=>C=18

f(x) = x³ - x²+18 is the particular solution