Question

\(y\left(\frac{d y}{d x}\right)^{2}+2 x \frac{d y}{d x}-y=0 ; \quad y(0)=\sqrt{5}\)

Short answer

Answer: \(y(x) = \int \tanh(2\ln|-2x| + 2C) dx + K\)

Step-by-step solution

1. Rewrite the equation to make it easier to separate variables

It is helpful to rewrite the given equation in a more familiar form. Let \(v = \frac{dy}{dx}\) be the first derivative of y(x) with respect to x. Replace all the \(\frac{dy}{dx}\) terms with v. The equation will now look like this: \(yv^2 + 2xv - y = 0\).

2. Separate the variables

To separate the variables, we need to isolate y and its derivatives on one side of the equation and the independent variable x on the other side. We can start by factoring y out of the equation: \(y(v^2 - 1) = -2xv\). Now, divide both sides by \((v^2 - 1)\) to isolate y and its derivatives: \(y = -\frac{2xv}{v^2 - 1}\). Next, since \(v = \frac{dy}{dx}\), we have \(\frac{dy}{-2xv} = \frac{dx}{v^2 - 1}\). Now the variables are separated.

3. Integrate both sides

Integrate both sides of the equation with respect to their respective variables: \(\int \frac{-v\,d\!y}{2x} = \int \frac{d\!x}{v^2 - 1}\). Use substitution to simplify the left side of the equation: Let \(u = -2x\), then \(du = -2dx\). The left side of the equation becomes: \(\int \frac{dy}{u}\), which when integrated, yields \(\ln|u| = \ln|-2x| + k_1\). Similarly, integrate the right side of the equation using the arctanh function: \(\int \frac{dx}{v^2 - 1} = \frac{1}{2} \operatorname{arctanh}(v) + k_2\). Here, \(k_1\) and \(k_2\) are constants of integration. Finally, equate both side of the equation: \(\ln|-2x| + k_1 = \frac{1}{2} \operatorname{arctanh}(v) + k_2\).

4. Apply the initial condition y(0) = √5

Apply the initial condition \(y(0) = \sqrt{5}\) to the equation: \(\ln|-2(0)| + k_1 = \frac{1}{2} \operatorname{arctanh}(v(0)) + k_2\). Adding both constants of integration gives a single constant, so we can simplify the equation: \(\frac{1}{2} \operatorname{arctanh}(v) - \ln|-2x| = C\).

5. Solve for y(x)

Now, we can rewrite the equation in terms of y(x) and solve for the dependent variable. Recall that \(v = \frac{dy}{dx}\). So we can write: \(\frac{1}{2} \operatorname{arctanh}\left(\frac{dy}{dx}\right) - \ln|-2x| = C\). To find y(x), solve for \(\frac{dy}{dx}\) and then integrate both sides with respect to x: \(\frac{1}{2} \operatorname{arctanh}(v) = \ln|-2x| + C \implies v = \tanh(2\ln|-2x| + 2C)\). Finally, solve the first-order differential equation: \(y(x) = \int \tanh(2\ln|-2x| + 2C) dx + K\). Here, K is a constant of integration. This equation presents y(x) that satisfies the given first-order differential equation and the initial condition.

Key concepts

Integrating Factors

When solving linear first-order differential equations, integrating factors are a useful tool. They transform a non-exact differential equation into an exact equation, making it solvable through integration. An integrating factor is generally a function of the independent variable, say \(\mu(x)\), that we multiply through the entire differential equation to achieve exactness. To find the integrating factor, there's usually a formula derived from the coefficient of \(y\) in the standard form of the equation: \(\frac{dy}{dx} + P(x)y = Q(x)\).

For our exercise at hand, the method of integrating factors wasn't directly applied, as the equation given is not a linear first-order differential equation. However, understanding integrating factors can be highly beneficial. It prepares you for complex scenarios where separation of variables is not possible, and an integrating factor becomes the way forward.

Separation of Variables

Separation of variables is a technique used to solve a particular type of differential equation where the variables can be moved onto opposite sides of the equation. The goal is to rewrite the equation so that each variable and its differential are on the same side. This process results in an equation that can be integrated directly to find the solution.

In our exercise, separation of variables was key in progressing towards the solution. After the substitution \(v = \frac{dy}{dx}\), the variables were separated by dividing by \(v^2 - 1\) and rearranging, ultimately leading to two integrable functions. Successful separation allowed the integration of both sides of the equation, which pushed us closer to solving for \(y(x)\). It is essential when using this method to ensure you’re working with an equation where separation of variables is allowed, as it is not applicable to all differential equations.

Initial Value Problem

An initial value problem (IVP) involves a differential equation along with a specific value, called the 'initial condition' that the solution must satisfy at a particular point. This problem is crucial as it helps to determine the specific solution out of the potentially infinite family of solutions that only a differential equation provides.

In this scenario, we were given the initial condition \(y(0) = \sqrt{5}\). This condition is used after integrating to find the constant of integration. IVPs are fundamental for realistic modeling of physical processes because they anchor mathematical predictions to real-world measurements. Solving an IVP provides us with a unique solution corresponding to the physical context of the problem at hand.