Question

find the particular solution of the differential equation that satisfies the initial condition(s).

$f(x)=4$, $f(2)=12$, $f(2)=17$

$f(x)=$

Explanation:

Step1: Integrate \(f''(x)\) to find \(f'(x)\)

Integrate \(f''(x)=4\) with respect to \(x\).
Using the power rule \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C\) (\(n = 0\) for the constant function \(y = 4\)), we have \(f'(x)=\int4dx=4x + C_1\).

Step2: Use the initial condition \(f'(2)=12\) to find \(C_1\)

Substitute \(x = 2\) and \(f'(2)=12\) into \(f'(x)=4x + C_1\).
\(12=4\times2+C_1\).
\(12 = 8+C_1\), so \(C_1=4\). Then \(f'(x)=4x + 4\).

Step3: Integrate \(f'(x)\) to find \(f(x)\)

Integrate \(f'(x)=4x + 4\) with respect to \(x\).
\(f(x)=\int(4x + 4)dx=\int4xdx+\int4dx\).
Using the power rule \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C\) (\(n = 1\) for \(y = 4x\) and \(n = 0\) for \(y = 4\)), we get \(f(x)=4\times\frac{x^{2}}{2}+4x+C_2=2x^{2}+4x + C_2\).

Step4: Use the initial condition \(f(2)=17\) to find \(C_2\)

Substitute \(x = 2\) and \(f(2)=17\) into \(f(x)=2x^{2}+4x + C_2\).
\(17=2\times2^{2}+4\times2+C_2\).
\(17=8 + 8+C_2\).
\(17=16+C_2\), so \(C_2 = 1\).

Answer:

\(f(x)=2x^{2}+4x + 1\)