Problem 16
Question
In Exercises \(11-16,\) solve the initial value problem. Confirm your answer by checking that it conforms to the slope field of the differential equation. $$\frac{d y}{d x}=2 x \sqrt{x+2}\( and \)y=0\( when \)x=-1$$
Step-by-Step Solution
Verified Answer
The solution to the initial value differential equation is \( y = \frac{4}{5}(x+2)^{5/2} - \frac{4}{5} \).
1Step 1: Integrate the differential equation
The given equation can be written as \( dy = 2x\sqrt{x+2}dx \). Let's integrate both sides of the equation with respect to \(x\). Doing so will yield the general solution of the differential equation: \( y = \frac{4}{5}(x+2)^{5/2} + C \), where C is the constant of integration.
2Step 2: Apply the initial condition
The initial condition given is \( y = 0 \) when \( x = -1 \). Substituting these values into the general solution, we get: \( 0 = \frac{4}{5}(x+2)^{5/2} + C. \) Solving for \( C \) yields \( C = -4/5 \).
3Step 3: Write down the final specific solution
Substitute the found constant \( C = -4/5 \) back into the general solution, the specific solution conforming to the initial condition becomes: \( y = \frac{4}{5}(x+2)^{5/2} - \frac{4}{5} \).
Key Concepts
Differential EquationIntegral CalculusSlope FieldConstant of Integration
Differential Equation
A differential equation is a type of equation that relates a function to its derivatives. In simple terms, it shows how a certain quantity changes in relation to another. For example, in our exercise, the differential equation given is \( \frac{d y}{d x} = 2x \sqrt{x+2} \). This equation describes how the function \( y \) changes with respect to \( x \).
In solving an initial value problem, like the one presented, the goal is to find such a function \( y(x) \) that satisfies both the differential equation and a given starting condition. This initial condition is crucial because it allows us to find a unique solution out of the many possible solutions that might satisfy the differential equation.
In solving an initial value problem, like the one presented, the goal is to find such a function \( y(x) \) that satisfies both the differential equation and a given starting condition. This initial condition is crucial because it allows us to find a unique solution out of the many possible solutions that might satisfy the differential equation.
Integral Calculus
Integral calculus is a branch of mathematics focused on integration, which is essentially the reverse process of differentiation. In the context of solving a differential equation, integration is used to find the original function from its derivative.
In our exercise, we integrated the equation \( dy = 2x\sqrt{x+2}dx \). This integration resulted in the general solution \( y = \frac{4}{5}(x+2)^{5/2} + C \). Here, \( C \) is an arbitrary constant, known as the constant of integration. By performing this integration, we successfully transformed the derivative back into a function.
In our exercise, we integrated the equation \( dy = 2x\sqrt{x+2}dx \). This integration resulted in the general solution \( y = \frac{4}{5}(x+2)^{5/2} + C \). Here, \( C \) is an arbitrary constant, known as the constant of integration. By performing this integration, we successfully transformed the derivative back into a function.
Slope Field
A slope field is a graphical representation of a differential equation. It consists of small line segments or slope markers that represent the slope of the solution curve at various points in the plane. Slope fields are valuable because they enable us to visualize solutions to a differential equation without actually solving it explicitly.
In the context of the given initial value problem, checking against the slope field means ensuring that the found solution \( y = \frac{4}{5}(x+2)^{5/2} - \frac{4}{5} \) aligns with the visual representation provided by the slope field. This confirms both the accuracy of the analytical solution and its conformity with the geometric intuition offered by the slope field.
In the context of the given initial value problem, checking against the slope field means ensuring that the found solution \( y = \frac{4}{5}(x+2)^{5/2} - \frac{4}{5} \) aligns with the visual representation provided by the slope field. This confirms both the accuracy of the analytical solution and its conformity with the geometric intuition offered by the slope field.
Constant of Integration
The constant of integration, denoted as \( C \), appears during the process of integrating a differential equation and represents an infinite number of potential solutions. When integrating, there's a family of functions that can satisfy the equation due to this constant.
To find the specific solution to a problem, such as our initial value problem, we need additional information like the initial condition. In our example, the initial condition \( y = 0 \) when \( x = -1 \) was used to solve for \( C \), resulting in \( C = -4/5 \).
This makes sure that the selected solution not only complies with the differential equation but also passes through the specific point, thereby making it a unique solution.
To find the specific solution to a problem, such as our initial value problem, we need additional information like the initial condition. In our example, the initial condition \( y = 0 \) when \( x = -1 \) was used to solve for \( C \), resulting in \( C = -4/5 \).
This makes sure that the selected solution not only complies with the differential equation but also passes through the specific point, thereby making it a unique solution.
Other exercises in this chapter
Problem 15
In Exercises \(11-20,\) solve the initial value problem explicitly. $$\frac{d y}{d x}=-\frac{1}{x^{2}}-\frac{3}{x^{4}}+12\( and \)y=3\( when \)x=1$$
View solution Problem 16
In Exercises \(13-16,\) verify that \(\int f(u) d u \neq \int f(u) d x\) $$f(u)=\sin u\( and \)u=4 x$$
View solution Problem 16
In Exercises \(15-18,\) solve the differential equation. $$\frac{d u}{d x}=\frac{2}{x^{2}-1}$$
View solution Problem 16
In Exercises \(11-20,\) solve the initial value problem explicitly. $$\frac{d y}{d x}=5 \sec ^{2} x-\frac{3}{2} \sqrt{x}\( and \)y=7\( when \)x=0$$
View solution