Question

By integrating twice find the general solution of \(y^{\prime \prime}=12 x^{2}\)

Short answer

Answer: The general solution of the given second-order differential equation is \(y(x) = x^4 + C_1x + C_2\), where \(C_1\) and \(C_2\) are constants of integration.

Step-by-step solution

Find \(y'\) by integrating the given equation with respect to \(x\)

To find \(y'\), integrate the given equation with respect to \(x\): $$ y' = \int 12x^2 \, dx $$ You can use polynomial integration rules to solve this integral: $$ y' = 4x^3 + C_1 $$ Here, \(C_1\) is the constant of integration.

Integrate \(y'\) once more to find \(y\)

Now, we will find the function \(y(x)\) by integrating \(y'\) with respect to \(x\): $$ y(x) = \int (4x^3 + C_1) \, dx $$ Again, use polynomial integration rules to solve this integral: $$ y(x) = x^4 + C_1x + C_2 $$ Here, \(C_2\) is another constant of integration.

Write the general solution

The general solution of the given second-order differential equation is: $$ y(x) = x^4 + C_1x + C_2 $$ Where \(C_1\) and \(C_2\) are constants of integration.

Key concepts

Integral Calculus

Integral calculus is all about finding functions that describe accumulated quantities, such as area under curves. In mathematics, it complements differential calculus, as it focuses on the reverse process of differentiation.
If differentiation can be thought of as "breaking down" a function into parts, integration is like "reassembling" them back.
When dealing with differential equations—equations involving derivatives—integral calculus comes into play to recover the original function. For instance, in the problem we're tackling, we start with a second derivative and use integration to recover the first derivative and eventually the original function.
Integration involves two main operations:

• Indefinite Integration: Finding a general form of the antiderivative (no limits of integration).
• Definite Integration: Evaluating the antiderivative between set limits, thus providing a specific accumulated quantity.

In this exercise, we focus on indefinite integration, which introduces a constant of integration, reflecting the fact that the antiderivative is not unique without additional conditions (initial values). Each constant represents a family of curves parallel to each other.

Polynomial Integration

Polynomial integration is a straightforward process, especially when dealing with simple polynomials, like the ones in our example. Integration here involves applying the power rule in reverse.
This means if we have a term like \(n x^m\), the integral would be \(\frac{n}{(m+1)} x^{m+1}\), as long as \(m eq -1\). This formula helps increase the degree of the polynomial by one during the integration process.
Here’s how we apply this concept:

• The original differential equation \(y'' = 12x^2\) when integrated for the first time leads us to \(y' = \int 12x^2 \, dx = 4x^3 + C_1\).
• Integrate \(y'\) again: \(y = \int (4x^3 + C_1) \, dx\). This results in \(y = x^4 + C_1x + C_2\).

Notice how each step involves dealing with simple powers of \(x\), making polynomial integration a convenient tool in solving differential equations.

General Solution

In calculus, finding the general solution for a differential equation means determining the most comprehensive form of the solution that includes constants of integration.
These constants, which emerge from indefinite integration, account for the initial conditions that the specific problem might have. They demonstrate that the solution is a family of functions, not a singular form.
For the exercise presented, the constants \(C_1\) and \(C_2\) are such parameters. They allow our solution \(y(x) = x^4 + C_1x + C_2\) to fit a range of possible scenarios based on different initial conditions.
Without additional information about specific values of \(y\) or its derivatives at given points (like \(y(0)\)), we can only provide this general form.
Remember, successful use of the general solution requires interpreting these constants in relation to practical contexts or further constraints in problems.