Question

solve the initial value problem.

\\( \frac { d ^ { 3 } y } { d x ^ { 3 } } = 18, y ^ { \prime \prime } ( 0 ) = 4, y ^ { \prime } ( 0 ) = 5 \\), and \\( y ( 0 ) = 5 \\)

\\( y ( x ) = \\)

Explanation:

Step1: Integrate the third - derivative

Integrate \(\frac{d^{3}y}{dx^{3}} = 18\) with respect to \(x\).
Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), \(\int18dx=18x + C_1\). So \(y''(x)=18x + C_1\).
Substitute \(x = 0\) and \(y''(0)=4\) into \(y''(x)\): \(4=18\times0 + C_1\), so \(C_1 = 4\). Then \(y''(x)=18x + 4\).

Step2: Integrate the second - derivative

Integrate \(y''(x)=18x + 4\) with respect to \(x\).
\(\int(18x + 4)dx=18\times\frac{x^{2}}{2}+4x+C_2=9x^{2}+4x + C_2\). So \(y'(x)=9x^{2}+4x + C_2\).
Substitute \(x = 0\) and \(y'(0)=5\) into \(y'(x)\): \(5=9\times0^{2}+4\times0 + C_2\), so \(C_2 = 5\). Then \(y'(x)=9x^{2}+4x + 5\).

Step3: Integrate the first - derivative

Integrate \(y'(x)=9x^{2}+4x + 5\) with respect to \(x\).
\(\int(9x^{2}+4x + 5)dx=9\times\frac{x^{3}}{3}+4\times\frac{x^{2}}{2}+5x+C_3=3x^{3}+2x^{2}+5x + C_3\).
Substitute \(x = 0\) and \(y(0)=5\) into \(y(x)\): \(5=3\times0^{3}+2\times0^{2}+5\times0 + C_3\), so \(C_3 = 5\).

Answer:

\(y(x)=3x^{3}+2x^{2}+5x + 5\)