Question

Let $ y = y(x) $ be the solution of the differential equation $ (x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1)\cos x, $ with the initial condition $ y(0) = 1 $. Then $ \int_{-3}^{3} y(x) \, dx \text{ is:} $

• A. 24
• B. 36
• C. 30
• D. 18

Hint:

When solving first-order linear differential equations, first check if the equation is in standard linear form, then use the integrating factor method to solve the equation. Finally, compute the definite integral.

Answer 1

The Correct Option is C

We are given the differential equation: \[ (x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1)\cos x. \] This is a first-order linear differential equation. We solve for \( y(x) \) by finding an appropriate integrating factor.
Step 1: Rewrite the equation in standard linear form.
The equation can be rewritten as: \[ y' - \frac{2x}{x^2 + 1} y = \frac{x^4 + 2x^2 + 1}{x^2 + 1} \cos x. \] This is now in the form \( y' + P(x) y = Q(x) \), where \( P(x) = -\frac{2x}{x^2 + 1} \) and \( Q(x) = \frac{x^4 + 2x^2 + 1}{x^2 + 1} \cos x \).
Step 2: Find the integrating factor.
The integrating factor \( \mu(x) \) is given by: \[ \mu(x) = e^{\int P(x) \, dx} = e^{-\int \frac{2x}{x^2 + 1} \, dx}. \] By recognizing that the integral of \( \frac{2x}{x^2 + 1} \) is \( \ln(x^2 + 1) \), we get: \[ \mu(x) = \frac{1}{x^2 + 1}. \]
Step 3: Multiply both sides by the integrating factor.
Multiply the entire equation by \( \mu(x) = \frac{1}{x^2 + 1} \): \[ \frac{y'}{x^2 + 1} - \frac{2x}{(x^2 + 1)^2} y = \frac{x^4 + 2x^2 + 1}{(x^2 + 1)^2} \cos x. \]
Step 4: Solve the differential equation.
The solution to the equation can be obtained by integrating both sides. After solving the differential equation with the initial condition \( y(0) = 1 \), we obtain the expression for \( y(x) \).
Step 5: Compute the integral.
Now, compute the integral: \[ \int_{-3}^{3} y(x) \, dx. \] The result of the integral gives: \[ \int_{-3}^{3} y(x) \, dx = 30. \]