Question

Find the solution of the following initial value problems. $$g^{\prime }(x)=7x(x^6-\frac{1}{7});g(1)=2$$

Short answer

Question: Find the particular solution of the given initial value problem: $g^{\prime}(x)=7x(x^{6}-\frac{1}{7})$, with the initial condition $g(1) = 2$. Answer: The particular solution of the given initial value problem is $$g(x)=\frac{7x^8}{8}-x^2+\frac{9}{8}.$$

Step-by-step solution

Find the general solution of the given differential equation

In order to solve this initial value problem, we first need to find the general solution, (i.e., the solution without using the initial condition). To do that, we shall integrate the given expression for g'(x). We have: $$g^{\prime }(x)=7x(x^6-\frac{1}{7})$$ By integrating this with respect to x, we can obtain the general solution: $$g(x)=\int 7x(x^6-\frac{1}{7})dx$$

Integrate the expression

Let's integrate the given expression term by term: $$g(x)=7\int x(x^6-\frac{1}{7})dx=7(\int x^{7}dx-\int \frac{1}{7}xdx)$$ Now, we integrate the two terms separately: $$g(x)=7[\frac{x^8}{8}-\frac{1}{7}\frac{x^2}{2}]+C$$ Simplify the equation: $$g(x)=\frac{7x^8}{8}-x^2+C$$ Here, C is the constant of integration.

Use the initial condition to find the constant of integration

We are provided with the initial condition, g(1) = 2. We will plug the given value of g(x) into the general solution equation and solve for the constant of integration C. $$2=\frac{7(1{)}^8}{8}-(1{)}^2+C$$ Simplify and solve for C: $$C=2-\frac{7}{8}+1$$ $$C=\frac{9}{8}$$

Write the particular solution

We have found the constant of integration that satisfies the initial condition. Now, the particular solution can be obtained by substituting C back into the general solution: $$g(x)=\frac{7x^8}{8}-x^2+\frac{9}{8}$$ This is the solution of the given initial value problem.

Key concepts

Differential Equation

A differential equation is a mathematical equation that relates some function with its derivatives. In simpler terms, it describes a relationship between a variable-dependent function and its rates of change. When we deal with a differential equation like in our exercise, we aim to find a function that satisfies the equation. Differential equations are indispensable in physics, engineering, economics, and other fields, as they model a wide range of phenomena. The basic form of a first-order differential equation is $f^{\prime }(x)=g(x)$. The exercise presents a differential equation in the form $g^{\prime }(x)=7x(x^6-\frac{1}{7})$, where $g^{\prime }(x)$ denotes the derivative of $g$ with respect to $x$.

The objective is to find $g(x)$ given this derivative and an initial condition. The process includes steps that involve integration to move from the derivative to the original function. The differential equation in this exercise is particularly straightforward because it is already given in an explicit form for the derivative, making the integration process more direct.

Integration Techniques

Term-by-Term Integration

Integrating a function is the inverse process of differentiating it. In the context of our exercise, the crucial detail is integrating term by term. To integrate $g^{\prime }(x)=7x(x^6-\frac{1}{7})$, we multiply out the terms and then apply the integral to each term separately. This basic technique is useful when dealing with polynomials as we can integrate each term independently and then add the results to get the general solution. Techniques such as substitution or integration by parts might be required for more complex functions, but for polynomials, the power rule suffices. In our exercise, we see the power rule applied to both $x^7$ and $x$ after simplifying the original expression. Bullet points for clear steps for integration technique include:

• Expanding the equation if necessary.
• Integrating each term independently using proper rules.
• Applying algebra to simplify the integral result.

Constant of Integration

In calculus, every indefinite integral, or antiderivative, includes a constant of integration. This is because integrating a function gives us a family of functions that differ only by a constant. When we do not have any additional information, this constant remains arbitrary and is usually denoted as $C$. Our exercise demonstrates when we integrate $g^{\prime }(x)$, we add $C$ to the result to represent all possible antiderivatives of the function. Integral constants play a crucial role because they help ensure that all particular solutions of a differential equation can be represented. These particular solutions can be found by applying initial conditions or boundary conditions that determine the precise value of the constant. Without the constant of integration, we would lose an entire family of possible solutions and the general solution of the differential equation would be incomplete.

Particular Solution

Moving from the general to the particular solution of a differential equation includes applying additional information, typically in the form of initial values or boundary conditions. A 'particular solution' refers to a solution that satisfies both the differential equation and the initial conditions. It's specific because it's a single function out of the entire family of solutions provided by the general solution. In our exercise, we used the initial condition $g(1)=2$ to find the specific value of our constant of integration. By substituting $x=1$ and $g(x)=2$ into our general solution, we determined that $C=\frac{9}{8}$. The final equation $g(x)=\frac{7x^8}{8}-x^2+\frac{9}{8}$ is the particular solution as it is the only function that satisfies both the differential equation and the given initial condition.