Questions: Find the integrating factor μ(x) for the following ODE 2 y′+(x sin x) y=x^2

Solution Steps

To solve the given first-order linear ordinary differential equation (ODE), we need to find the integrating factor \(\mu(x)\). The standard form of a first-order linear ODE is \(y' + P(x)y = Q(x)\). We can rewrite the given ODE in this form and then determine the integrating factor \(\mu(x)\) using the formula \(\mu(x) = e^{\int P(x) \, dx}\).

1. Rewrite the given ODE in the standard form \(y' + P(x)y = Q(x)\).
2. Identify \(P(x)\) from the standard form.
3. Compute the integrating factor \(\mu(x) = e^{\int P(x) \, dx}\).

The given ordinary differential equation (ODE) is

\[ 2y' + (x \sin x)y = x^2. \]

We can rewrite it in standard form as

\[ y' + \frac{x \sin x}{2} y = \frac{x^2}{2}. \]

Here, we identify \(P(x) = \frac{x \sin x}{2}\) and \(Q(x) = \frac{x^2}{2}\).

The integrating factor \(\mu(x)\) is calculated using the formula

\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{x \sin x}{2} \, dx}. \]

After performing the integration, we find that

\[ \mu(x) = e^{-\frac{x \cos x}{2} + \frac{\sin x}{2}}. \]

Final Answer