find the function f that satisfies the following differential equations and initial conditions. f(x)=4x…

Explanation:

Step1: Integrate $F'''(x)$

Integrate $F'''(x)=4x$ with respect to $x$. Using the power - rule $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $F''(x)=\int4x dx=4\times\frac{x^{2}}{2}+C_1 = 2x^{2}+C_1$.

Step2: Find $C_1$

Substitute $x = 0$ and $F''(0)=0$ into $F''(x)$. Then $F''(0)=2\times0^{2}+C_1=0$, so $C_1 = 0$. Thus, $F''(x)=2x^{2}$.

Step3: Integrate $F''(x)$

Integrate $F''(x)=2x^{2}$ with respect to $x$. Using the power - rule, $F'(x)=\int2x^{2}dx=2\times\frac{x^{3}}{3}+C_2=\frac{2}{3}x^{3}+C_2$.

Step4: Find $C_2$

Substitute $x = 0$ and $F'(0)=6$ into $F'(x)$. Then $F'(0)=\frac{2}{3}\times0^{3}+C_2=6$, so $C_2 = 6$. Thus, $F'(x)=\frac{2}{3}x^{3}+6$.

Step5: Integrate $F'(x)$

Integrate $F'(x)=\frac{2}{3}x^{3}+6$ with respect to $x$. Using the power - rule, $F(x)=\int(\frac{2}{3}x^{3}+6)dx=\frac{2}{3}\times\frac{x^{4}}{4}+6x + C_3=\frac{1}{6}x^{4}+6x + C_3$.

Step6: Find $C_3$

Substitute $x = 0$ and $F(0)=1$ into $F(x)$. Then $F(0)=\frac{1}{6}\times0^{4}+6\times0 + C_3=1$, so $C_3 = 1$.

Answer:

$\frac{1}{6}x^{4}+6x + 1$