show that y=xsinx is a solution to y'' + y=2cosx


To show that \( y = x \sin x \) is a solution to the differential equation \( y'' + y = 2 \cos x \), we need to compute \( y' \) and \( y'' \), substitute them into the equation, and verify that both sides are equal.

Given \( y = x \sin x \):

### Step 1: Compute the first derivative \( y' \).
Using the product rule for differentiation:

\[
y' = \frac{d}{dx} (x \sin x) = \sin x + x \cos x
\]

### Step 2: Compute the second derivative \( y'' \).
Now differentiate \( y' = \sin x + x \cos x \) using the product rule again:

\[
y'' = \frac{d}{dx} (\sin x + x \cos x) = \cos x + (\cos x - x \sin x)
\]
\[
y'' = \cos x + \cos x - x \sin x = 2 \cos x - x \sin x
\]

### Step 3: Substitute \( y \) and \( y'' \) into the original equation.

The original equation is:

\[
y'' + y = 2 \cos x
\]

Substituting \( y = x \sin x \) and \( y'' = 2 \cos x - x \sin x \):

\[
(2 \cos x - x \sin x) + (x \sin x) = 2 \cos x
\]

Simplifying:

\[
2 \cos x - x \sin x + x \sin x = 2 \cos x
\]

The terms \( -x \sin x \) and \( +x \sin x \) cancel each other out:

\[
2 \cos x = 2 \cos x
\]

This is true, so the equation holds.

### Conclusion:
The function \( y = x \sin x \) satisfies the differential equation \( y'' + y = 2 \cos x \). Therefore, \( y = x \sin x \) is a solution.

Would you like any details or have questions? Here are some related questions:

1. How does the product rule apply in this context?
2. Can we solve this differential equation using other methods?
3. What are the general solutions for second-order differential equations?
4. How do you verify if a function is a solution to a differential equation?
5. How does differentiation affect trigonometric functions like \( \sin x \) and \( \cos x \)?

**Tip:** When solving differential equations, always verify by substituting back into the original equation.

show that y=xsinx is a solution to y'' + y=2cosx

Explore how to show that y=xsinx is a solution to the differential equation y'' + y = 2cosx. This detailed solution covers differentiation using the product rule and substitution into second-order linear differential equations. Ideal for undergraduate students studying Calculus II or Differential Equations.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation