Problem 31 - Differential Equations and Linear Algebra

Question

Determine which of the five types of differential equations we have studied the given equation falls into (see Table \(1.12 .1),\) and use an appropriate technique to find the general solution. $$y^{\prime}+y \sin x=\sin x$$

Step-by-Step Solution

Answer

The given differential equation \(y'+y\sin x = \sin x\) is a first-order linear differential equation. To solve it, we use the integrating factor method. The integrating factor is given as \(IF = e^{\int \sin x \, dx} = e^{-\cos x}\). Multiplying the equation by the integrating factor, integrating both sides, and using integration by parts, we get the general solution: $$y(x) = \cos x - \sin x + C e^{\cos x}$$

Step 1: Find the Integrating Factor (IF)

We need to find the integrating factor, which is given by the formula: $$IF = e^{\int P(x) \, dx}$$ In our case, \(P(x) = \sin x\). Let's find the integrating factor: $$IF = e^{\int \sin x \, dx} = e^{-\cos x}$$

Step 2: Multiply the Differential Equation by the Integrating Factor

Next, we multiply the given differential equation by the integrating factor: $$(e^{-\cos x})(y^{\prime}+y \sin x)=(e^{-\cos x})\sin x$$ This simplifies to: $$e^{-\cos x}y^{\prime}+e^{-\cos x}y \sin x=e^{-\cos x}\sin x$$

Step 3: Rearrange and Integrate Both Sides

We can now write the equation in the form \((e^{-\cos x}y)' = e^{-\cos x}\sin x\). Next, we need to integrate both sides with respect to \(x\): $$\int (e^{-\cos x}y)' \, dx = \int e^{-\cos x}\sin x \, dx$$ The left side can be integrated directly: $$e^{-\cos x}y = \int e^{-\cos x}\sin x \, dx + C_1$$ Now, we need to find the integral on the right side.

Step 4: Integrate the Right Side using Integration by Parts

We need to evaluate the integral \(\int e^{-\cos x}\sin x \, dx\). Let's use integration by parts with \(u = \sin x\) and \(dv = e^{-\cos x} \, dx\). Then, we have: $$du = \cos x \, dx$$ $$v = -e^{-\cos x}$$ By integration by parts formula, we get: $$\int e^{-\cos x}\sin x \, dx = -e^{-\cos x}\sin x + \int e^{-\cos x}\cos x \, dx$$ Let's use integration by parts again with \(u = \cos x\) and \(dv = e^{-\cos x} \, dx\). Then, we have: $$du = -\sin x \, dx$$ $$v = -e^{-\cos x}$$ By integration by parts formula, we get: $$\int e^{-\cos x}\sin x \, dx = -e^{-\cos x}\sin x + e^{-\cos x}\cos x - \int e^{-\cos x}(-\sin x) \, dx$$ Now, we can rewrite the integral on the right side as: $$\int e^{-\cos x}\sin x \, dx = -e^{-\cos x}\sin x + e^{-\cos x}\cos x + C_2 - \int e^{-\cos x}\sin x \, dx$$

Step 5: Solve for y(x)

Now we can substitute the integral back into the equation from step 3: $$e^{-\cos x}y = -e^{-\cos x}\sin x + e^{-\cos x}\cos x + C_2 - \int e^{-\cos x}\sin x \, dx + C_1$$ $$e^{-\cos x}y = -e^{-\cos x}\sin x + e^{-\cos x}\cos x + C_3$$ Finally, we can isolate \(y(x)\) by multiplying both sides by \(e^{\cos x}\): $$y(x) = \cos x - \sin x + C$$ Here, \(C = C_3 e^{\cos x}\). So, the general solution to the given differential equation is: $$y(x) = \cos x - \sin x + C e^{\cos x}$$

Key Concepts

Integrating Factor

The integrating factor is a crucial concept when solving first-order linear differential equations. It's a function, often denoted by \( IF \), that we multiply with the entire differential equation to simplify it. This simplification allows us to combine terms into a single derivative, making integration straightforward.

The integrating factor is typically derived from the equation format \( y' + P(x)y = Q(x) \). The formula to find the integrating factor is:
\[ IF = e^{\int P(x) \, dx} \]
In this exercise, the differential equation given is \( y' + y \sin x = \sin x \). Here, \( P(x) \) is \( \sin x \). Computing the integrating factor, we find:
\[ IF = e^{-\cos x} \]
Multiplying through the original equation by this integrating factor transforms it, aiding in further operations like setting up for potential integration by parts.

Integration by Parts

Integration by parts is a powerful technique for simplifying integrals that are products of functions. It is based on the product rule for derivatives and is usefully expressed by the formula:
\[ \int u \, dv = uv - \int v \, du \]
It helps when we can't straightforwardly integrate a product of functions. This method is especially helpful for integrals involving exponential functions and trigonometric terms together, as in this problem.

In the context of the given differential equation, we used integration by parts to evaluate \( \int e^{-\cos x} \sin x \, dx \). We set \( u = \sin x \) and \( dv = e^{-\cos x} \, dx \) in the first instance, performing the integration by parts process to break down the complex integral into manageable pieces.

Repeating this process even enabled us to simplify further expressions, showcasing the versatility of the method when the right choices of \( u \) and \( dv \) are made.

General Solution

Finding the general solution to a differential equation involves unearthing a function, \( y(x) \), that satisfies the given differential equation for all its variables. After identifying an integrating factor and deploying integration by parts, we arrive at a solution that reflects all possible particular solutions.

From the steps outlined, after integrating and manipulating the differential equation, the final form we arrived at was:
\[ y(x) = \cos x - \sin x + C e^{\cos x} \]
Here, \( C \) is an arbitrary constant essential for representing the general solution. It adapts the equation for initial or boundary conditions which can specify a particular solution.

• \( \cos x \) and \( \sin x \) represent the particular solution that fits the non-homogeneous part of the equation, \( \sin x \).
• \( C e^{\cos x} \) incorporates the constant of integration, accounting for any initial conditions.

The general solution gives a more flexible view, allowing us to fit an infinite set of possible solutions within the given framework.