Question

For the following problems, find the general solution to the differential equation.\(y^{\prime}=\sin x e^{\cos x}\)

Short answer

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

Step-by-step solution

Identify the Type of Differential Equation

This differential equation is a first-order differential equation because it involves the first derivative of the dependent variable y with respect to x. It fits the form \( y' = f(x) \), which means it is separable in this instance.

Separate the Variables

To solve the equation, we need to separate the variables. The differential equation is given as \( y' = \sin x e^{\cos x} \). We rewrite this as \( \frac{dy}{dx} = \sin x e^{\cos x} \). Then, we can separate the variables by multiplying both sides by \( dx \) to get \( dy = \sin x e^{\cos x} \, dx \).

Integrate Both Sides

Now, we integrate both sides to find the general solution. Integrate the left side with respect to \( y \) and the right side with respect to \( x \):\[\int dy = \int \sin x e^{\cos x} \, dx\]The left integral simply becomes \( y \). The integral on the right can be recognized by a substitution method. Let \( u = \cos x \), which implies \( du = -\sin x \, dx \). This substitution changes the integral into \(-\int e^u \, du \), which results in \(-e^u + C = -e^{\cos x} + C \), where \( C \) is the constant of integration.

Construct the General Solution

Since the integral of the left side is \( y \) and the integral of the right side is \(-e^{\cos x} + C\), the general solution to the differential equation is:\[y = -e^{\cos x} + C\]

Key concepts

Separable Differential Equations

When dealing with differential equations, a crucial type you may encounter is the separable differential equation. These equations can be expressed in the form \( y' = f(x)g(y) \), where the derivatives are separated by moving all terms involving \( y \) to one side and all terms involving \( x \) to the other.
This separation simplifies the solving process significantly, as it allows each side to be integrated independently. In the original exercise, we had \( y' = \sin x \cdot e^{\cos x} \), fitting nicely into a separable form. By separating the variables, we rearranged it into \( dy = \sin x \cdot e^{\cos x} \ dx \), a crucial step towards finding the solution.
Structuring a differential equation into a separable form makes it feasible to integrate both sides directly, breaking it into easier sub-problems to tackle.

Integration by Substitution

The integration by substitution technique is a powerful tool used when tackling integrals of slightly complex functions. It's especially handy when dealing with products of functions or composite functions, as seen in our differential equation.
In substitution, you rewrite the integral using a new variable to simplify the integration process. For example, in our problem, after separating the variables, we performed the substitution \( u = \cos x \). This gives \( du = -\sin x \, dx \), cleverly turning the complex-looking integral \( \int \sin x \, e^{\cos x} \, dx \) into a simpler \( -\int e^u \, du \).
By converting the integral, we can now focus on integrating \( e^u \) easily, yielding \( -e^u + C \), and then revert back to the original variable, converting to \( -e^{\cos x} + C \). This technique allows us to solve for the integral precisely, unraveling the complexity one layer at a time.

Constant of Integration

Whenever we perform integration, especially indefinite integration, a significant component of our solution is the constant of integration, denoted as \( C \). This constant is vital because it accounts for any constant term that could have existed before differentiation.
Since integration is the reverse process of differentiation, determining the exact values is complex without additional information. Thus, we introduce \( C \) to represent all the possible vertical shifts of our function. In our exercise, after integrating both sides, we ended up with \( y = -e^{\cos x} + C \).
The presence of \( C \) illustrates that there are infinitely many solutions to the differential equation, each differing by a constant. In practical applications, further conditions, known as initial conditions or boundary conditions, are usually provided to pinpoint the specific solution by resolving the exact value of \( C \).
Understanding the role of the constant of integration is essential in applying these concepts meaningfully to solve real-world problems.