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$$

Short answer

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

Step-by-step solution

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 \( \int udv = uv - \int vdu = x \tan x - \int tan x dx \).

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.

Solve for u from du/dx equation

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

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 \( 1 = 0 - \ln|1| + C = C \).

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 \( u = x \tan x -\ln |\sec x| + 1 \)