Problem 13

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. $$\frac{d u}{d x}=x \sec ^{2} x$ and $u=1$ when $x=0$$

Step-by-Step Solution

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Answer

The solution to the initial value problem with differential equation $ du/dx = x \sec ^{2} x $ and initial value u(0) = 1 is $ u = x \tan x -\ln |\sec x| + 1 $

1Step 1: Apply the formula for integration by parts

The formula for integration by parts is $ \int udv = uv - \int vdu $. We choose $ u=x $ and $ dv=\sec^2x dx $. This gives $ du=dx $ and $ v=\tan x $. So, the integral of $ x \sec ^{2} x dx $ becomes $ \int udv = uv - \int vdu = x \tan x - \int tan x dx $.

2Step 2: Simplify the integral

The integral $ \int \tan x dx = \ln | \sec x | $, giving $ \int x \sec ^{2} x dx = x \tan x - \ln |\sec x| + C $ where C is the constant of integration.

3Step 3: Solve for u from du/dx equation

From the differential equation $ du/dx = x \sec ^{2} x $ and step 2, we get $ u = x \tan x -\ln |\sec x| + C $

4Step 4: Apply the initial condition

We are given that u=1 when x=0. So, substituting these values into our equation from Step 3 gives $ 1 = 0-\ln|\sec(0)| + C$. This simplifies to $ 1 = 0 - \ln|1| + C = C $.

5Step 5: Confirm the final answer

Substitute the constant $ C $ found in Step 4 into the equation from Step 3 to get the final answer. So, the solution to the differential equation is $ u = x \tan x -\ln |\sec x| + 1 $

Key Concepts

Integration by PartsDifferential EquationsSlope FieldsConstant of Integration

Integration by Parts

Integration by Parts is a powerful technique used in calculus to solve certain types of integrals. It applies when you have a product of functions, usually denoted as $ u(x) $ and $ dv $. The formula is given as $ \int u \, dv = uv - \int v \, du $.

In our exercise, $ u = x $ and $ dv = \sec^2 x \, dx $. This means we differentiate $ u $ to find $ du = dx $ and integrate $ dv $ to get $ v = \tan x $.

Choose functions: Identify which part of the integrand to differentiate and which to integrate.
Differentiate $ u $ to get $ du $.
Integrate $ dv $ to get $ v $.
Substitute into the formula to simplify the integral.

In summary, Integration by Parts breaks down a complex integral into simpler parts that are easier to solve.

Differential Equations

Differential Equations are equations that involve derivatives of a function. They can describe a variety of phenomena, such as motion and growth processes. In our example, the differential equation is $ \frac{du}{dx} = x \sec^2 x $.

Understand the form: It usually involves the derivative of the function, in this case, $ u $.
To solve, integrate both sides of the equation.
The goal is to find the original function $ u(x) $.

The process often requires integration techniques, such as Integration by Parts, to isolate the dependent variable $ u $. Solving differential equations often involves finding an unknown function, not just a number.

Slope Fields

Slope Fields, also known as direction fields, visually represent differential equations. They help us understand the behavior of solutions even before solving the differential equation.

Each point on the field has a tiny line with a slope that matches the derivative at that point.
The graphical representation shows the general direction or trend of solutions.
It's a useful tool for verifying solutions.

In the context of our problem, the solution we found should align with these slope directions, confirming the accuracy of our integral solution.

Constant of Integration

The Constant of Integration is a fundamental part of solving indefinite integrals. It represents an arbitrary constant added to an antiderivative.

When integrating, the result is a family of functions that differ by a constant. In our solution, after integrating, we add $ C $, the constant of integration. This is crucial because it accounts for any initial conditions given in the problem.

When solving differential equations, use initial conditions to solve for $ C $.
This process personalizes the general solution to meet specific conditions, like $ u(0) = 1 $.

The constant of integration ensures that we find the precise solution that meets the initial value problem's requirements.

Problem 13

Other exercises in this chapter

Problem 12

In Exercises $11-20,$ solve the initial value problem explicitly. $$\frac{d y}{d x}=2 e^{x}-\cos x$ and $y=3$ when $x=0$$

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Problem 13

In Exercises $13-16,$ verify that $\int f(u) d u \neq \int f(u) d x$ $$f(u)=\sqrt{u}$ and $u=x^{2}(x>0)$$

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Problem 13

In Exercises $5-14,$ evaluate the integral. $$\int \frac{8 x-7}{2 x^{2}-x-3} d x$$

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Problem 13

In Exercises $11-14$ , find the solution of the differential equation $d y / d t=k y, k$ a constant, that satisfies the given conditions. $$y(0)=50, \quad y

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