Question

Find the general solution of each of the following differential equations. \(y^{\prime} \cos x+y=\cos ^{2} x\)

Short answer

The general solution is \( y = 1 + C e^{ -\sin x} \).

Step-by-step solution

Rewrite the equation

Start by rewriting the given differential equation. The equation is given as: \( y' \cos x + y = \cos^2 x \).

Identify the integrating factor

To solve this linear first-order differential equation, identify the integrating factor. The standard form for a first-order linear differential equation is \( y' + P(x)y = Q(x) \). For our equation, \( P(x) = \cos x \). The integrating factor, \( \mu(x) \), is given by:\[ \mu(x) = e^{\int P(x) \, dx } = e^{\int \cos x \, dx} \]

Compute the integrating factor

Integrate \( \cos x \) with respect to \( x \):\[ \int \cos x \, dx = \sin x \]Thus, the integrating factor is:\[ \mu(x) = e^{\sin x} \]

Multiply through by the integrating factor

Multiply both sides of the original differential equation by the integrating factor \( e^{\sin x} \):\( e^{\sin x} y' \cos x + e^{\sin x} y = e^{\sin x} \cos^2 x \)

Simplify the left-hand side

Recognize that the left-hand side is the derivative of \( y e^{\sin x} \):\[ (y e^{\sin x})' = e^{\sin x} \cos^2 x \]

Integrate both sides

Integrate both sides with respect to \( x \):\[ \int (y e^{\sin x})' \, dx = \int e^{\sin x} \cos^2 x \, dx \]The left-hand side integrates to:\[ y e^{\sin x} \]The integral on the right-hand side can be simplified using the substitution \( u = \sin x \), \( du = \cos x \, dx \):\[ \int e^u \cos x \, dx = \int e^u \, du = e^u + C = e^{\sin x} + C \]

Solve for y

Finally, solve for \( y \):\[ y e^{\sin x} = e^{\sin x} + C \]Divide by \( e^{\sin x} \) to isolate \( y \):\[ y = 1 + C e^{ -\sin x} \]

Key concepts

first-order linear differential equation

A differential equation relates a function to its derivatives. A first-order linear differential equation is a specific type, where the highest order derivative is the first derivative. The general form is:
\( y' + P(x)y = Q(x) \).
Here, \(y'\) is the first derivative and \(P(x)\) and \(Q(x)\) are functions of \(x\).
In our problem, the given equation is: \( y' \, \cos x + y = \cos^2 x \). By dividing the entire equation by \( \cos x \), we can rewrite it in the standard form:
\( y' + \frac{1}{ \cos x } y = \cos x \). Notice how it fits the general form: \( y' + P(x) y = Q(x) \) with \( P(x) = \cos x \) and \( Q(x) = \cos^2 x \).

integrating factor

To solve first-order linear differential equations, we use the method of integrating factors. This technique makes it easier to integrate the equation and find a solution. The integrating factor, \( \mu(x) \), is calculated using:
\( \mu(x) = e^{ \int P(x) \, dx } \). For our problem, \( P(x) = \cos x \). Let's find the integrating factor step by step:
1. Compute the integral of \( P(x) \):
\( \int \cos x \, dx = \sin x \). 2. Exponentiate the result to find the integrating factor:
\( \mu(x) = e^{ \sin x } \). Now, we multiply the original differential equation by this integrating factor to simplify it and make the left-hand side a perfect derivative. This allows us to integrate both sides of the equation more easily.

solving differential equations

After finding the integrating factor \( \mu(x) = e^{ \sin x } \), we proceed to solve the differential equation. Multiplying both sides by the integrating factor gives us:
\( e^{ \sin x } y' \, \cos x + e^{ \sin x } y = e^{ \sin x } \cos^2 x \). We recognize that the left-hand side is the derivative of \( y e^{ \sin x } \):
\[ ( y e^{ \sin x })' = e^{ \sin x } \cos^2 x \]. To solve, integrate both sides with respect to \( x \):
\[ \int ( y e^{ \sin x })' \, dx = \int e^{ \sin x } \cos^2 x \, dx \]. Using the substitution \( u = \sin x \), \( du = \cos x \, dx \), the right-hand side integrates to:
\[ \int e^u \, du = e^u + C = e^{ \sin x } + C \]. Thus, we get:
\[ y e^{ \sin x } = e^{ \sin x } + C \]. Finally, solve for \( y \) by dividing both sides by \( e^{ \sin x } \):
\[ y = 1 + C e^{- \sin x } \]. This is the general solution of the differential equation.