Question

Solving a First-Order Linear Differential Equation In Exercises \(51-58\) , solve the first-order differential equation by any appropriate method. $$ y \cos x-\cos x+\frac{d y}{d x}=0 $$

Short answer

The solution to the first-order differential equation \(y \cos x - \cos x + \frac{dy}{dx} = 0\) is \(y=1-e^{C-\sin x}\).

Step-by-step solution

Rearrange the given equation

The given differential equation is \(y \cos x - \cos x + \frac{dy}{dx} = 0\). It can be rearranged to isolate the derivative term on one side of the equation: \(\frac{dy}{dx} = \cos x(1-y)\).

Separate Variables and Integrate

Now we can separate variables for which we get \(\frac{dy}{1-y} = \cos x dx\). We integrate both sides of the equation: \(\int \frac{dy}{1-y} = \int \cos x dx\) and solve both integrals, yielding: \(-\ln|1 - y| = \sin x + C\).

Rewrite the solution

With the logarithmic expression on one side, we can exponentiate both sides to cancel the logarithm, finally rewriting to solve for \(y\). The final form of the solution is \(y=1-e^{C-\sin x}\).

Key concepts

Linear Differential Equations

Linear differential equations are equations involving a function and its derivatives. These equations are called 'linear' because every term of the function and its derivatives is either multiplied by a constant or depends linearly on the function. In the given exercise, the differential equation is:

• \( y \cos x - \cos x + \frac{dy}{dx} = 0 \)

This equation is of the first order because it involves the first derivative \( \frac{dy}{dx} \). For such equations, our main goal is to find a function \( y = f(x) \) that satisfies the equality. Linear equations are easier to handle because they follow predictable patterns that allow us to apply specific solving techniques.

Separation of Variables

The separation of variables is a powerful method used to solve differential equations, especially those like first-order types. This approach involves rearranging the equation so that all terms involving the dependent variable \( y \) appear on one side and all terms involving the independent variable \( x \) appear on the other.

• In our problem, we separated the equation as follows: \[ \frac{dy}{1-y} = \cos x \cdot dx \]

By getting to this form, we can independently integrate both sides with respect to their variables, which will eventually lead us toward the solution. Proper separation is often crucial—it simplifies the task of solving differential equations significantly.

Integration Techniques

Integration techniques are essential in solving differential equations. Once we have separated the variables, the next step is to integrate both sides. For our problem:

• We integrate the left side: \[ \int \frac{dy}{1-y} \] Using the integral formula for a natural logarithm, this becomes \(-\ln|1-y|\).
• For the right side: \[ \int \cos x \, dx \] The integration leads to \(\sin x\).

Integration requires a solid grasp of fundamental rules and theorems, such as recognizing standard forms and applying substitutions when necessary. Integrating correctly enables us to find an antiderivative, which helps to reconstruct the original function or solve the equation.

Exponential Functions

Exponential functions play a critical role in differential equations, particularly when dealing with integrals and logarithms. In the solution, we encounter a logarithmic form \(-\ln|1-y| = \sin x + C\). To isolate \( y \), we need to "undo" the logarithm by exponentiating both sides:

• This step transforms the equation into: \[ |1-y| = e^{-\sin x - C} \]

By doing so, we rewrite the solution in terms of \( y \):

• \( y = 1 - e^{C-\sin x} \)

This form reflects the influence of the exponential function, which frequently appears in the solutions to linear differential equations, particularly when integrating factors involving logarithms. Understanding exponentials and their inverses is key for manipulating and solving differential equations effectively.