Question

Find the general solution of each of the following differential equations. \(y^{\prime} \sqrt{x^{2}+1}+x y=x\)

Short answer

The general solution is \( y = 1 - C_1 e^{-\arctan(x)} \).

Step-by-step solution

- Write the Differential Equation in Standard Form

The given differential equation is \( y' \sqrt{x^{2}+1} + xy = x \). First, let's rewrite it in a more standard form for easier manipulation: \[ y' \sqrt{x^{2}+1} = x - xy \].

- Separate the Variables

To separate the variables, divide both sides by \sqrt{x^{2}+1} : \[ y' = \frac{x - xy}{\sqrt{x^{2}+1}} \].Next, divide both sides by \left( 1 - y \right) : \[ \frac{y'}{1 - y} = \frac{x}{\sqrt{x^{2}+1}} \].

- Integrate Both Sides

Integrate both sides to find the general solution. Starting with the left side, we have: \[ \int \frac{dy}{1 - y} \].For the right side, integrate:\[ \int \frac{x}{\sqrt{x^{2}+1}} dx \].Notice \( \frac{1}{\sqrt{x^{2}+1}} \) is the derivative of \( \arctan(x) \). Therefore, \[ \int \frac{dy}{1 - y} = -\ln|1 - y| \]and \[ \int \frac{x}{\sqrt{x^{2}+1}} dx = \arctan(x) \].

- Combine and Simplify

Combine the integrated results: \[ -\ln|1 - y| = \arctan(x) + C \],where \( C \) is the integration constant.Exponentiate both sides to solve for \( y \) : \[ |1 - y| = e^{-\arctan(x) + C} \].Let \( e^{C} = C_1 \), then the general solution is \[ |1 - y| = C_1 e^{-\arctan(x)} \].

- Solve for y

Solve for \( y \) explicitly: \[ y = 1 - C_1 e^{-\arctan(x)} \],where \( C_1 \) is a new arbitrary constant. This is the general solution.

Key concepts

general solution

When solving a differential equation, the goal is often to find a general solution. The general solution is a formula that gives all possible solutions to the differential equation. In our example, the differential equation given is: \( y' \sqrt{x^{2}+1} + xy = x \). After some manipulation and integration, we derived the general solution: \( y = 1 - C_1 e^{-\tan^{-1}(x)} \). This general solution includes an arbitrary constant \( C_1 \), which means that there are infinitely many solutions corresponding to different values of \( C_1 \).
The general solution must satisfy the original differential equation for any value of the arbitrary constant. To verify, you can differentiate the solution and substitute it back into the original equation. If all terms cancel out correctly, you have the correct solution. Remember, the purpose of the general solution is to represent all possible behaviors of the differential equation.

separation of variables

Separation of variables is a powerful technique used in solving differential equations, like the one given as \( y' \sqrt{x^{2}+1} + xy = x \). The idea is to manipulate the equation so that each variable appears on a different side of the equation. This allows us to integrate each side separately.
In our example, we start with: \[ y' \sqrt{x^{2}+1} = x - xy \]. We then divide by \( \sqrt{x^{2}+1} \) to get: \[ y' = \frac{x - xy}{\sqrt{x^{2}+1}} \]. Next, we divide by \( 1 - y \), yielding: \[ \frac{y'}{1 - y} = \frac{x}{\sqrt{x^{2}+1}} \].
This separation allows us to integrate each side independently. The left side depends only on \( y \), and the right side depends only on \( x \). This is the core concept of separating variables and it greatly simplifies finding the general solution.

integration

Integration is a fundamental process in calculus used to find the antiderivative or the area under a curve. When solving a differential equation through separation of variables, integration is the step following the separation.
After separating variables in our example to \( \frac{y'}{1 - y} = \frac{x}{\sqrt{x^{2}+1}} \), we integrate both sides. For the left side, we compute: \[ \int \frac{dy}{1 - y} = - \ln|1 - y| \]. For the right side, recognize that \( \frac{1}{\sqrt{x^{2}+1}} \) is the derivative of \( \tan^{-1}(x) \), giving us: \[ \int \frac{x}{\sqrt{x^{2}+1}} \, dx = \tan^{-1}(x) \].
This integration step is crucial because it transforms the derived expressions into their respective antiderivatives, aiding us in determining the general solution to the differential equation.

standard form of differential equations

The standard form of a differential equation provides a consistent structure, making it easier to apply solution techniques like separation of variables or recognizing patterns. In our example, the given differential equation is: \( y' \sqrt{x^{2}+1} + xy = x \).
To convert it to a more standard form, we rearrange terms: \( y' \sqrt{x^{2}+1} = x - xy \). Dividing by \( \sqrt{x^{2}+1} \): \[ y' = \frac{x - xy}{\sqrt{x^{2}+1}} \]. Then, divide by \( 1 - y \): \[ \frac{y'}{1 - y} = \frac{x}{\sqrt{x^{2}+1}} \].
Writing differential equations in standard form not only helps in simplifying the problem but also makes it easier to identify suitable methods for finding the solution. This form ensures that variables are properly separated or the equation fits a recognizable type, making the problem-solving process more straightforward.