Question

solve the initial value problem. $\frac{d^{2}y}{dx^{2}} = 14 - 36x$, $y(0) = 7$, and $y(0) = 2$ $y(x)=$

Explanation:

Step1: Integrate the second - derivative

Integrate \(\frac{d^{2}y}{dx^{2}}=14 - 36x\) with respect to \(x\) to find \(y'\).
Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we have:
\(y'=\int(14 - 36x)dx=\int14dx-\int36xdx\)
\(y'=14x-36\times\frac{x^{2}}{2}+C_1\)
\(y'=14x - 18x^{2}+C_1\)

Step2: Use the initial condition \(y'(0) = 7\)

Substitute \(x = 0\) and \(y'(0)=7\) into \(y'=14x - 18x^{2}+C_1\).
\(7=14\times0-18\times0^{2}+C_1\), so \(C_1 = 7\)
Then \(y'=14x - 18x^{2}+7\)

Step3: Integrate \(y'\) to find \(y\)

Integrate \(y'=14x - 18x^{2}+7\) with respect to \(x\)
\(y=\int(14x - 18x^{2}+7)dx=\int14xdx-\int18x^{2}dx+\int7dx\)
Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n + 1}+C\) (\(n
eq-1\)):
\(y=14\times\frac{x^{2}}{2}-18\times\frac{x^{3}}{3}+7x+C_2\)
\(y = 7x^{2}-6x^{3}+7x+C_2\)

Step4: Use the initial condition \(y(0)=2\)

Substitute \(x = 0\) and \(y(0) = 2\) into \(y = 7x^{2}-6x^{3}+7x+C_2\)
\(2=7\times0^{2}-6\times0^{3}+7\times0+C_2\), so \(C_2=2\)

Answer:

\(y(x)=-6x^{3}+7x^{2}+7x + 2\)