Problem 27 - Differential Equations and Linear Algebra

Question

Solve the given initial-value problem. $$\frac{d y}{d x}=\frac{y-\sqrt{x^{2}+y^{2}}}{x}, \quad y(3)=4$$

Step-by-Step Solution

Answer

The final solution of the given initial-value problem is: \[y(x) = \frac{1}{\frac{1}{4} + \ln 3 - \ln|x|}\]

Step 1: Identify the ODE type and method of solution

The given ODE is a first-order ODE. It has the form \(\frac{dy}{dx} = \frac{y - \sqrt{x^2 + y^2}}{x}\), and the problem also provides an initial condition y(3) = 4. The ODE is separable since we can rewrite it in the form where the function y(x) and x are separated on each side of the equation. We will solve this using the method of separation of variables.

Step 2: Separate variables

To separate variables, we need to move all y terms on one side of the equation and all x terms on the other side. To do this, we can write the given ODE as follows: \[\frac{dy}{y - \sqrt{x^2 + y^2}} = \frac{dx}{x}\]

Step 3: Integrate both sides

Now that we have separated the variables, we can integrate both sides of the equation with respect to their respective variables: \[\int \frac{dy}{y - \sqrt{x^2 + y^2}} = \int \frac{dx}{x}\] The left-hand side is a bit complicated, but we can use the substitution method to simplify it. Let \(u = \sqrt{x^2 + y^2}\), then \(u^2 = x^2 + y^2\), and \(2u\frac{du}{dy} = 2y\). So, \(\frac{du}{dy} = \frac{y}{u}\). Now we rewrite the left-hand side integral in terms of u: \[\int \frac{y}{y^2-u^2} dy = \int \frac{u}{u^2 - (u^2 - y^2)} du\] Now the integral on the left side is simpler: \[\int \frac{u}{y^2} du = \int \frac{dx}{x}\] Now, we can integrate: \[\int \frac{u}{y^2} du = \ln|x| + C_1\] And on the left side: \[\int \frac{u}{y^2} du = -\frac{1}{y} + C_2\]

Step 4: Solve for the constant

Now that we have integrated both sides, we can solve for the constant using the initial condition y(3) = 4. Plugging the initial condition into our integrated equation, we get: \[-\frac{1}{4} = \ln 3 + C_1\] Solving for the constant, we get \(C_1 = -\frac{1}{4} - \ln 3\).

Step 5: Write the final solution

Now, we write the final solution in terms of y and x: \[-\frac{1}{y} = \ln|x| - \frac{1}{4} - \ln 3\] Now, we want to isolate y: \[-\frac{1}{y} +\frac{1}{4} +\ln 3 = \ln|x|\] \[\frac{1}{y} = \frac{1}{4} + \ln 3 - \ln|x|\] Now, take the reciprocal of both sides of the equation to find the solution: \[y(x) = \frac{1}{\frac{1}{4} + \ln 3 - \ln|x|}\] This is the final solution of the initial-value problem.

Key Concepts

First-Order ODE

A first-order ordinary differential equation (ODE) is an equation involving a function and its first derivative. It is expressed in the general form \( \frac{dy}{dx} = f(x, y) \), where \( y \) is the unknown function of the independent variable \( x \) and \( f(x, y) \) represents the relationship between \( y \) and \( x \) that we are aiming to find. In our example, \( \frac{dy}{dx} = \frac{y - \sqrt{x^2 + y^2}}{x} \) is a first-order ODE because it contains only the first derivative of the function \( y(x) \).

The presence of an initial condition, such as \( y(3) = 4 \) in our case, turns the differential equation into an initial-value problem. This condition anchors the solution to a specific point and ensures that the resulting function passes through \( (3, 4) \) on the \( xy \) plane, providing a unique solution to the equation.

Separation of Variables

Separation of variables is a technique used to solve some ODEs in which the equation can be manipulated to isolate all terms involving \( y \) on one side and those involving \( x \) on the other. The steps typically involve algebraic manipulation to achieve the form \( g(y)dy = h(x)dx \) where \( g(y) \) and \( h(x) \) are functions of \( y \) and \( x \) respectively.

In the given problem, we have \( \frac{dy}{y - \sqrt{x^2 + y^2}} = \frac{dx}{x} \). This form allows us to integrate both sides respectively to \( y \) and \( x \) to solve for the function \( y(x) \). The integral sign represents the sum of an infinite number of infinitesimally small parts, and in this context, it traces out the curve of the function we are seeking to determine.

Method of Substitution

The method of substitution is used to simplify integrals that are difficult or impossible to solve in their original form. By introducing a new variable \( u \) that is a function of \( x \) or \( y \) (or both), we can transform the integral into a simpler one. For example, we can let \( u = \sqrt{x^2 + y^2} \) and find a relationship for \( du \) in terms of \( dy \) to make the integral easier to evaluate.

In our equation, after the substitution \( u = \sqrt{x^2 + y^2} \) is made, we obtain a new differential equation that is easier to integrate because it involves the new variable \( u \) in place of the more complex original terms. This technique turns a challenging integral into a tractable one, paving the way toward finding the solution to the differential equation.

Integration of Differential Equations

After manipulating a differential equation into a form that separates the variables or through the method of substitution, the next step is to integrate both sides with respect to their variables. This process involves finding the antiderivative or integral of each side, which undoes the differentiation and brings us closer to the solution \( y(x) \).

In the initial-value problem provided, once we separated the variables and made an appropriate substitution, we ended up with an integral that we could evaluate. After integrating both sides of the equation, constants of integration (denoted by \( C_1 \) and \( C_2 \) in general) appear. These constants are determined using the initial conditions given in the problem, like \( y(3) = 4 \) in our case. By evaluating the integral at this point, we can solve for the constants and find the specific solution that satisfies both the differential equation and the initial condition.