Question

Find the particular solution \(y=f(x)\) that satisfies the differential equation and initial condition. $$ f^{\prime}(x)=4 x ; \quad f(0)=6 $$

Short answer

The particular solution for the given initial value problem is \(f(x) = 2x^2 + 6\).

Step-by-step solution

Integration of the Differential Equation

To obtain the general solution of the differential equation, perform indeterminate integration on both sides: \(f'(x) = 4x\), hence \(f(x) = \int 4x dx\).

Use of Integral Calculation and Inclusion of an Arbitrary Constant

Applying standard integral rules, the result is: \(f(x)=2x^2+C\), where \(C\) is an arbitrary constant.

Apply the initial condition

The given initial condition \(f(0) = 6\) allows us to solve for the constant \(C\). Substituting \(f(0) = 6\) into the general solution \(f(x)=2x^2+C\) gives \(6 = 2*0^2+C \rightarrow C = 6\).

Writing the Particular Solution

Substitute the found constant \(C = 6\) into the general solution, which gives the particular solution \(f(x) = 2x^2 + 6\).

Key concepts

Integration

Integration is a fundamental process in calculus that is essentially the reverse of differentiation. In the context of differential equations, it is used to find a function from its derivative. In our exercise, the derivative given is \( f'(x) = 4x \). To retrieve the original function, we need to perform what is called an indefinite integration. This involves finding the antiderivative of the given function. The integral of \( 4x \) can be calculated using the power rule for integration:

• Power Rule states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)
• So, \( \int 4x \, dx = 2x^2 + C \)

This result represents a family of functions, all differing by a constant \( C \). Therefore, our integrated function becomes \( f(x) = 2x^2 + C \). This equation includes the same derivative, \( 4x \), when you differentiate it back, confirming the integration was performed correctly.

Initial Condition

An initial condition is a value that specifies a particular solution from a family of solutions provided by the general form of the integrated function. In this exercise, we have an initial condition \( f(0) = 6 \). Initial conditions are essential for determining the specific function that fits the context of a given problem, as different constants can represent different scenarios.
By substituting 0 for \( x \) and 6 for \( f(x) \) in the general solution \( f(x) = 2x^2 + C \), we can solve for \( C \):

• Substitute: \( 6 = 2(0)^2 + C \)
• This simplifies to \( 6 = C \)

Therefore, the arbitrary constant \( C \) is determined to be 6 using the initial condition.

Particular Solution

The process of finding a particular solution involves using the specific value of \( C \) obtained from the initial condition in the general solution. A particular solution is a single, unique function that satisfies both the differential equation and the specified initial condition. For our exercise, by substituting the constant \( C = 6 \) back into the general form, \( f(x) = 2x^2 + C \), we arrive at:

• The particular solution: \( f(x) = 2x^2 + 6 \)

This solution uniquely describes the behavior of the function \( f(x) \) in accordance with the initial conditions. It represents the specific case of the family of curves given by the general solution where \( C \) equals 6, fitting the precise requirements of the problem.