Question

Show that the equation \(\frac{d y}{d x}=\frac{y}{x}\) subject to the initial condition \(\mathrm{y}(0)=0\) has an infinite number of solutions of the form \(\mathrm{y}=\mathrm{Cx}\). The same equation subject to the initial condition \(\mathrm{y}(0)=\mathrm{a} \neq 0\) has no solution.

Short answer

Question: Show that the first-order ordinary differential equation (ODE) \(\frac{dy}{dx} = \frac{y}{x}\) subject to y(0) = 0 has an infinite number of solutions of the form y = Cx. Additionally, show that if the initial condition is y(0) = a ≠ 0, there is no solution to the ODE. Answer: For the initial condition y(0) = 0, the ODE has an infinite number of solutions of the form y = Cx, where C is any suitable constant. However, when the initial condition is y(0) = a ≠ 0, there is no solution for the given ODE.

Step-by-step solution

Solve the given ODE

The equation \(\frac{dy}{dx} = \frac{y}{x}\) already has variables separable. So, rewrite it as \(\frac{dy}{y} = \frac{dx}{x}\). Integrate both sides of this equation with respect to the respective variable: $$\int \frac{1}{y} dy = \int \frac{1}{x} dx.$$ Now, we have \(\ln(|y|) = \ln(|x|) + C\), where C is the constant of integration. To remove the natural logarithm, we can exponentiate both sides: $$|y| = \pm e^C |x|.$$ Alternatively, we can denoted the constant \(\pm e^C\) as \(k\). Thus, the general solution for this equation is: $$y = kx.$$

Apply initial condition y(0) = 0

Now, let's substitute the initial condition y(0) = 0 into the general solution: $$0 = k(0).$$ As this equation holds true for any value of k, the given ODE with the initial condition y(0) = 0 has an infinite number of solutions of the form y = Cx, where C is any suitable constant.

Address initial condition y(0) = a ≠ 0

Let's try substituting the initial condition y(0) = a ≠ 0 into the general solution: $$a = k(0).$$ The equation evaluates to a = 0, which contradicts the initial condition y(0) = a ≠ 0. Hence, there's no solution for the given ODE with the initial condition y(0) = a ≠ 0.

Key concepts

Initial Condition Problem

Understanding an initial condition problem is central to solving many differential equations found in physics, engineering, and other disciplines. This type of problem involves not just finding the general solution to a differential equation, but also determining a specific solution that adheres to given initial conditions. In other words, it starts with a differential equation that describes a physical system and narrows down the infinite number of possible solutions to the one that meets certain criteria at the outset—typically the value of the function (or its derivatives) at a specific point.

For instance, consider the equation \(\frac{d y}{d x}=\frac{y}{x}\) with different initial conditions. If we set \(y(0)=0\), we are essentially looking for a curve that passes through the origin (0,0) on an x-y graph. This problem presents us with an infinite number of straight line solutions through the origin, each represented by \(y=Cx\), where \(C\) is a constant that can vary. In contrast, if the initial condition is \(y(0)=a eq 0\), we are looking for a solution that passes through the point (0, a), but such a solution doesn't exist for this particular equation because it would contradict the behavior dictated by the equation itself. The solution to an initial condition problem not only satisfies the differential equation but also fulfills the specified initial conditions.

Separable Differential Equations

When we come across separable differential equations, they pose a relatively straightforward opportunity for finding solutions. These are a type of ordinary differential equations (ODEs) in which the variables can be separated on opposite sides of the equation. After separation, each side of the equation depends on only one variable, making them easier to integrate and solve.

In our example, \(\frac{d y}{d x} = \frac{y}{x}\) can be manipulated into a separable form by dividing both sides by \(y\) and multiplying both sides by \(dx\), effectively isolating \(dy\) on one side and \(dx\) on the other. This leads to \(\frac{1}{y} dy = \frac{1}{x} dx\), setting the stage for direct integration. What's key here is recognizing that separability allows us to treat the derivatives as ratios and integrate both sides as regular functions. The step-by-step solution demonstrates how, after integration, we obtain the natural logarithm of the absolute value of both \(y\) and \(x\), plus a constant. This method hinges on the ability to separate variables and often results in finding a general solution, which shows all possible equations that satisfy the differential equation.

Integration of Functions

Integration is one of the fundamental operations in calculus, alongside differentiation. While differentiation involves finding the rate at which a quantity changes, integration is essentially the reverse process—it sums up small quantities over an interval to find a total amount.

When we integrate a function, we are finding the area under the curve of that function on a graph. This process is crucial when working with differential equations as it allows us to move from a derivative (or a rate of change) back to the original function that describes the cumulative effect of that change. In the problem provided, after separating the variables, we integrate \(\frac{1}{y} dy\) and \(\frac{1}{x} dx\), which results in the natural logarithms on both sides of the equation. It's important to note that integrating involves adding a constant of integration because indefinite integration can result in an infinite number of antiderivatives.

In solving differential equations, a key step is often integrating functions after variables have been separated, as shown in the step-by-step solution. This fundamental process allows us to reconstruct the function from its rate of change, enabling us to propose a general solution to the differential equation.