Find the solution of the following initial value problem.

[tex]\[ f^{\prime}(x) = 6x - 5; \quad f(0) = 9 \][/tex]

The solution to the initial value problem is [tex]\( f(x) = \square \)[/tex]



Answer :

Let's solve the initial value problem step by step.

1. Determine the indefinite integral of [tex]\( f'(x) \)[/tex]:

We start with the derivative of the function, [tex]\( f'(x) = 6x - 5 \)[/tex]. To find [tex]\( f(x) \)[/tex], we need to integrate [tex]\( f'(x) \)[/tex] with respect to [tex]\( x \)[/tex].

[tex]\[ f(x) = \int (6x - 5) \, dx \][/tex]

2. Compute the integral:

To integrate [tex]\( 6x - 5 \)[/tex]:

[tex]\[ \int (6x - 5) \, dx = \int 6x \, dx - \int 5 \, dx \][/tex]

Now handle each term separately:

[tex]\[ \int 6x \, dx = 6 \int x \, dx = 6 \left( \frac{x^2}{2} \right) = 3x^2 \][/tex]

[tex]\[ \int 5 \, dx = 5x \][/tex]

Putting it all together:

[tex]\[ f(x) = 3x^2 - 5x + C \][/tex]

Here, [tex]\( C \)[/tex] is the constant of integration.

3. Use the initial condition to determine [tex]\( C \)[/tex]:

We are given the initial condition [tex]\( f(0) = 9 \)[/tex].

Substitute [tex]\( x = 0 \)[/tex] and [tex]\( f(0) = 9 \)[/tex] into the equation [tex]\( f(x) = 3x^2 - 5x + C \)[/tex]:

[tex]\[ f(0) = 3(0)^2 - 5(0) + C = 9 \][/tex]

Simplifying, we have:

[tex]\[ C = 9 \][/tex]

4. Write the final solution:

Substitute [tex]\( C \)[/tex] back into the function:

[tex]\[ f(x) = 3x^2 - 5x + 9 \][/tex]

Therefore, the solution to the initial value problem is:

[tex]\[ f(x) = \boxed{3x^2 - 5x + 9} \][/tex]