Question

Solving initial value problems Solve the following initial value problems. $$y^{\prime}(x)=3 x^{2}-3 x^{-4}, y(1)=0$$

Short answer

Question: Solve the initial value problem represented by the differential equation \(y'(x)=3x^2-3x^{-4}\) and the initial condition \(y(1)=0\). Answer: The solution to the given initial value problem is \(y(x) = x^3 + x^{-3} - 2\).

Step-by-step solution

Identify the differential equation and initial condition

The given initial value problem is represented by the differential equation: $$y'(x)=3x^2-3x^{-4}$$ and the initial condition: $$y(1)=0$$

Integrate the right-hand side of the equation

To find the function y(x), we need to integrate the right-hand side of the given differential equation with respect to x: $$\int y'(x) dx = \int (3x^2-3x^{-4}) dx$$

Calculate the integral

Now that we have the integral, we can find its antiderivative: $$\int (3x^2-3x^{-4}) dx = \frac{3x^3}{3} - \frac{3x^{-3}}{-3} + C = x^3 + x^{-3} + C$$ So, our function y(x) is given by: $$y(x) = x^3 + x^{-3} + C$$

Apply the initial condition to find the integration constant

We can now use the initial condition, y(1)=0, to find the integration constant C: $$0 = 1^3 + 1^{-3} + C$$ Solving for C, we find that C = -2.

Write the final solution

Now, we can write our final solution by substituting the value of C back into the function y(x): $$y(x) = x^3 + x^{-3} - 2$$ Thus, the solution to the given initial value problem is \(y(x) = x^3 + x^{-3} - 2\).

Key concepts

Differential Equation

A differential equation is a mathematical equation that relates a function with its derivatives. These equations are crucial because they describe various phenomena in physics, engineering, and other sciences. In simple terms, if you have ever seen an equation with derivatives in it, you're looking at a differential equation.
For example, the given problem starts with the differential equation \(y'(x) = 3x^2 - 3x^{-4}\).
This equation tells us about the rate of change of a function \(y\) with respect to \(x\).
Solving a differential equation usually means finding a function or a set of functions that satisfy the equation when you substitute them in place of the unknowns.

• In our exercise, we seek a function \(y(x)\) whose derivative looks like \(3x^2 - 3x^{-4}\).

Understanding differential equations helps you comprehend how changes occur and predict future behavior of dynamic systems.

Integration

Integration is a fundamental concept in calculus. It is the process of finding the integral of a function, which can be considered the reverse of taking a derivative. When you integrate a function, you are, in essence, determining the accumulated sum of areas under a curve or the total accumulation of quantities.
In our exercise, to solve the differential equation \(y'(x) = 3x^2 - 3x^{-4}\), we perform integration on the right-hand side:

• \(\int (3x^2 - 3x^{-4}) \, dx\)

This process will help us find \(y(x)\), the function that gives us the specific rate of change indicated by the differential equation.
Integration is a crucial step in finding solutions to many equations and gives insight into the original function before it was differentiated.

Antiderivative

When solving a differential equation, finding an antiderivative is essential. The antiderivative, or indefinite integral, is a function whose derivative is the original function you started with.
In our problem, the antiderivative of the function \(3x^2 - 3x^{-4}\) is computed as:

• \(\int (3x^2 - 3x^{-4}) \, dx = x^3 + x^{-3} + C\)

The \(C\) represents the constant of integration, a necessary component because an antiderivative is not unique—there are infinitely many solutions that differ by a constant.
By getting the antiderivative, you essentially reverse the process of differentiation, allowing you to find a broader family of functions that could solve the differential equation.

Initial Condition

The initial condition is a vital part of solving an initial value problem. It provides specific information about the function at a particular point, allowing you to find the exact solution that fits your problem.
In the given exercise, the initial condition is \(y(1)=0\).
This initial condition specifies the value of \(y(x)\) when \(x = 1\). By applying this condition, we can determine the constant \(C\) from the antiderivative \(x^3 + x^{-3} + C\).

• Here, we substitute and solve: \(0 = 1^3 + 1^{-3} + C\), finding \(C = -2\).

This step is crucial because without it, we would only have a general solution containing an unknown constant, instead of the particular solution that uniquely satisfies the given differential equation and its initial condition.