solve the initial value problem.\n\frac{d^{2}y}{dx^{2}} = 6 - 36x, y(0)=5, and y(0)=6\ny(x)=

Explanation:

Step1: Integrate the second - derivative

Integrate $\frac{d^{2}y}{dx^{2}}=6 - 36x$ with respect to $x$ to get the first - derivative. $y'(x)=\int(6 - 36x)dx=6x-18x^{2}+C_1$

Step2: Use the initial condition for $y'$

Given $y'(0) = 5$, substitute $x = 0$ and $y'(0)=5$ into $y'(x)$: $y'(0)=6(0)-18(0)^{2}+C_1=5$, so $C_1 = 5$. Then $y'(x)=6x-18x^{2}+5$

Step3: Integrate the first - derivative

Integrate $y'(x)=6x-18x^{2}+5$ with respect to $x$ to get $y(x)$. $y(x)=\int(6x-18x^{2}+5)dx = 3x^{2}-6x^{3}+5x+C_2$

Step4: Use the initial condition for $y$

Given $y(0)=6$, substitute $x = 0$ and $y(0) = 6$ into $y(x)$: $y(0)=3(0)^{2}-6(0)^{3}+5(0)+C_2=6$, so $C_2 = 6$

Answer:

$y(x)=3x^{2}-6x^{3}+5x + 6$