Question

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

$f(x)=6x$, $f(0)=5$

$f(x)=$

Explanation:

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

Integrate \( f'(x)=6x \) with respect to \( x \). Using the power rule \( \int x^n dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we have \( f(x)=\int6x dx\).
Since \( \int6x dx=6\int x dx\), and \( \int x dx=\frac{x^{2}}{2}+C\), then \( f(x)=6\times\frac{x^{2}}{2}+C = 3x^{2}+C\).

Step2: Determine the constant \( C \)

Use the initial condition \( f(0) = 5\). Substitute \( x = 0\) and \( f(0)=5\) into \( f(x)=3x^{2}+C\).
We get \( f(0)=3\times0^{2}+C\). So \( 5 = 0 + C\), which means \( C = 5\).

Answer:

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