Problem 70
Question
Express the solutions of the initial value problems in terms of integrals. $$\frac{d y}{d x}=\sqrt{1+x^{2}}, \quad y(1)=-2$$
Step-by-Step Solution
Verified Answer
The solution is \( y(x) = \int_{1}^{x} \sqrt{1 + t^2} \, dt - 2 \).
1Step 1: Rewrite the Differential Equation
The given differential equation is \( \frac{d y}{d x} = \sqrt{1 + x^2} \). This can be rewritten in differential form as \( dy = \sqrt{1 + x^2} \, dx \).
2Step 2: Integrate Both Sides
To find \( y(x) \), integrate both sides of the equation. \[ y(x) = \int \sqrt{1 + x^2} \, dx + C \] Where \( C \) is the constant of integration.
3Step 3: Determine the Constant of Integration
Use the initial condition \( y(1) = -2 \) to determine \( C \). Plug \( x = 1 \) and \( y = -2 \) into the expression:\[ -2 = \int \sqrt{1 + 1^2} \, dx + C \]First, evaluate the definite integral from 1 to some variable \( t \) and then solve for \( C \).
4Step 4: Solve the Expression
Since the integral does not have a common elementary form, express it as a definite integral:\[ y(x) = \int_{1}^{x} \sqrt{1 + t^2} \, dt - 2 \] This expresses \( y(x) \) in terms of a definite integral and satisfies \( y(1) = -2 \).
Key Concepts
Differential EquationsIntegrationInitial Conditions
Differential Equations
Differential equations are mathematical equations that involve an unknown function and its derivatives. They are essential in describing various phenomena like physics, biology, and engineering problems, where changes are observed over time or space.
The core principle is to find the unknown function that satisfies the given relationship. In this exercise, we have the differential equation:
Its task is to find the function \( y(x) \) that matches this condition. Differential equations can either be ordinary (ODEs) or partial differential equations (PDEs), depending on whether they involve derivatives with respect to a single variable or multiple variables. In our case, it's an ordinary differential equation (ODE) because it involves derivatives with respect to just \( x \). Understanding differential equations is crucial in comprehending how systems evolve over time or change in connection with other variables.
The core principle is to find the unknown function that satisfies the given relationship. In this exercise, we have the differential equation:
- \( \frac{d y}{d x} = \sqrt{1 + x^2} \)
Its task is to find the function \( y(x) \) that matches this condition. Differential equations can either be ordinary (ODEs) or partial differential equations (PDEs), depending on whether they involve derivatives with respect to a single variable or multiple variables. In our case, it's an ordinary differential equation (ODE) because it involves derivatives with respect to just \( x \). Understanding differential equations is crucial in comprehending how systems evolve over time or change in connection with other variables.
Integration
Integration is the process of finding the integral of a function, which is essentially the opposite operation of differentiation. It helps in determining the total accumulation of quantities and is widely used for finding areas under curves or solving differential equations.
In our exercise, the function \( \sqrt{1 + x^2} \) needs to be integrated to find \( y(x) \). When the differential form is given as:
This particular integral is more complex as it doesn’t have a straightforward elementary form.
Instead, it's solved as a definite integral from 1 to \( x \) to satisfy the initial condition, making it essential for solving the differential equation accurately.
In our exercise, the function \( \sqrt{1 + x^2} \) needs to be integrated to find \( y(x) \). When the differential form is given as:
- \( dy = \sqrt{1 + x^2} \, dx \)
- \( y(x) = \int \sqrt{1 + x^2} \, dx + C \)
This particular integral is more complex as it doesn’t have a straightforward elementary form.
Instead, it's solved as a definite integral from 1 to \( x \) to satisfy the initial condition, making it essential for solving the differential equation accurately.
Initial Conditions
Initial conditions are necessary for determining the specific solution of a differential equation. They provide the extra information required to solve for any arbitrary constants introduced during integration.
These conditions specify values of the unknown function and/or its derivatives at specific points. In our given problem,
Using this information, we can solve for the constant of integration \( C \) in the equation. By substituting \( x = 1 \) and \( y = -2 \) into the integrated function, we find \( C \), which customizes the general solution to match the initial problem conditions.
For the function's expression in terms of the integral, the solution becomes:
These conditions specify values of the unknown function and/or its derivatives at specific points. In our given problem,
- The initial condition is \( y(1) = -2 \)
Using this information, we can solve for the constant of integration \( C \) in the equation. By substituting \( x = 1 \) and \( y = -2 \) into the integrated function, we find \( C \), which customizes the general solution to match the initial problem conditions.
For the function's expression in terms of the integral, the solution becomes:
- \( y(x) = \int_{1}^{x} \sqrt{1 + t^2} \, dt - 2 \)
Other exercises in this chapter
Problem 69
Find the areas of the regions enclosed by the lines and curves in Exercises \(63-72\). $$y=x^{4}-4 x^{2}+4 \text { and } y=x^{2}$$
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Evaluate the integrals $$\int \frac{\sin \sqrt{\theta}}{\sqrt{\theta \cos ^{3} \sqrt{\theta}}} d \theta$$
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Archimedes \((287-212\) B.C.), inventor, military engineer, physicist, and the greatest mathematician of classical times in the Western world, discovered that t
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