Question

Solve the differential.f"(x)=3xGiven f'(0)=3 and f(1)=4

Answer 1

Given:

$f^{^{\prime}^{\prime}}(x)\text{ =3x}$

Solving the differential equation:

$\begin{gathered} \int f^{\prime}^{\prime}(x)\text{ dx= }\int3x\text{ dx} \\ f^{\prime}(x)\text{ = 3}\frac{x^2}{2}+c_1 \end{gathered}$

Applying the initial value f'(0) = 3

$f^{\prime}(x)\text{ = }\frac{3}{2}x^2\text{ + 3}$

Integrating further:

$\begin{gathered} \int f^{\prime}(x)dx\text{ = }\int\frac{3}{2}x^2\text{ dx + 3}\int dx \\ f(x)\text{ = }\frac{3}{2}\frac{x^3}{3}\text{ + 3x + c} \\ f(x)\text{ = }\frac{1}{2}x^3\text{ + 3x + c} \end{gathered}$

Applying the initial value f(1) =4

$\begin{gathered} 4\text{ = }\frac{1}{2}(1)^3\text{ + 3\lparen1\rparen + c} \\ 4\text{ = }\frac{1}{2}\text{ + 3 + c} \\ Solving\text{ for c} \\ c\text{ = 4-}\frac{1}{2}\text{ -3} \\ c\text{ = }\frac{8-1-6}{2} \\ c=\text{ }\frac{1}{2} \end{gathered}$

Hence, the solution is:

$f(x)\text{ = }\frac{1}{2}x^3\text{ + 3x + }\frac{1}{2}$