Questions: Find the particular solution of the differential equation dy/dx + y cos(x) = 2 cos(x) satisfying the initial condition y(0) = 4. Answer: y= Your answer should be a function of x.

Solution Steps

To solve the given first-order linear differential equation, we can use the method of integrating factors. First, identify the integrating factor, which is \( e^{\int \cos(x) \, dx} \). Multiply the entire differential equation by this integrating factor to make the left-hand side an exact derivative. Then, integrate both sides with respect to \( x \) and apply the initial condition to find the particular solution.

We start with the differential equation given by

\[ \frac{dy}{dx} + y \cos(x) = 2 \cos(x). \]

The integrating factor \( \mu(x) \) is calculated as

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

Multiplying the entire differential equation by the integrating factor gives us

\[ e^{\sin(x)} \frac{dy}{dx} + e^{\sin(x)} y \cos(x) = 2 e^{\sin(x)} \cos(x). \]

This can be rewritten as

\[ \frac{d}{dx}(e^{\sin(x)} y) = 2 e^{\sin(x)} \cos(x). \]

Integrating both sides results in

\[ e^{\sin(x)} y = 2 e^{\sin(x)} + C_1, \]

where \( C_1 \) is a constant of integration.

To find the particular solution, we apply the initial condition \( y(0) = 4 \):

\[ e^{\sin(0)} y(0) = 2 e^{\sin(0)} + C_1 \implies 4 = 2 + C_1 \implies C_1 = 2. \]

Substituting \( C_1 \) back into the equation gives us

\[ e^{\sin(x)} y = 2 e^{\sin(x)} + 2. \]

Dividing through by \( e^{\sin(x)} \) yields

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

Final Answer