Problem 31 - Calculus Single Variable

Question

Find the solution of the given initial value problem. $$ x^{2} y^{\prime}(x)=y(x) \sin (1 / x) \quad y(2 / \pi)=3 $$

Step-by-Step Solution

Answer

The solution is \( y(x) = 3 e^{\cos(1/x)} \).

Step 1: Recognize the Differential Equation Type

The problem is an initial value problem involving a differential equation. The given equation is \(x^{2} y^{\prime}(x)=y(x) \sin(1/x)\). Recognize that this is a separable differential equation, meaning it can be separated into functions of \(y\) and \(x\) on opposite sides of the equation.

Step 2: Separate Variables

To separate the variables, rewrite the equation as \(\frac{dy}{y} = \frac{\sin(1/x)}{x^2} dx\). This rearrangement allows both sides to be integrated with respect to their own variables.

Step 3: Integrate Both Sides

Integrate both sides of the equation independently. The left side becomes \( \ln |y|\) and the right side requires integration by parts or a table of integrals to integrate \(\frac{\sin(1/x)}{x^2}\). This step might involve substitutions or advanced techniques.

Step 4: Solve the Resulting Integral

The integral \(\int \frac{\sin(1/x)}{x^2} dx\) can be solved using the substitution \(u = 1/x\), which gives \(du = -1/x^2 dx\). Thus the integral transforms to \(-\int \sin(u) du\), resulting in \(\cos(u)+C \) or \(\cos(1/x)+C\).

Step 5: Exponentiate to Solve for y

After integration, exponentiate to solve for \(y\): \(|y| = e^{\cos(1/x) + C}\). This implies \(y = \pm e^{C} e^{\cos(1/x)}\). To simplify, write \(y = A e^{\cos(1/x)}\) where \(A\) is a constant.

Step 6: Apply Initial Condition

Use the initial condition \(y(2/\pi)=3\) to solve for the constant \(A\). Substitute \(x = 2/\pi\) and \(y = 3\) into \(y = A e^{\cos(1/x)}\): \(3 = A e^{\cos(\pi/2)}\). Since \(\cos(\pi/2) = 0\), \( e^{0}=1\), we find that \(A = 3\).

Step 7: Write Final Solution

The solution to the initial value problem is \( y(x) = 3 e^{\cos(1/x)} \). This equation satisfies both the differential equation and the initial condition.

Key Concepts

Separable Differential Equations

Differential equations can often appear complex, but certain types have characteristics that make them easier to solve. One of these types is the separable differential equation. A differential equation is considered separable when it can be rearranged to have all terms involving the variable \(y\) on one side and all terms involving \(x\) on the other side. This is done to make integration possible on both sides independently. In the given exercise, the equation \(x^{2} y^{\prime}(x)=y(x) \sin(1/x)\) was successfully rewritten as \(\frac{dy}{y} = \frac{\sin(1/x)}{x^2} dx\). This separation allows integration and ultimately aids in finding the solution. Recognizing whether an equation is separable is a key skill in solving differential equations efficiently.

Integration by Substitution

When solving integrals, direct integration is not always possible. Sometimes, substitution becomes necessary to simplify the integral into a more manageable form. In our case, integrating \(\frac{\sin(1/x)}{x^2} dx\) directly would be challenging. Instead, by using the substitution \(u = 1/x\), the process is simplified. With \(du = -1/x^2 dx\), the integral changes to \(-\int \sin(u) du\). This shows how substitution allowed the transformation of a complex integral into one that’s more familiar and thus easier to solve. Integration by substitution is a versatile and powerful technique, frequently used in calculus to tackle otherwise difficult integrals.

Exponential Functions

Exponential functions are a common element in mathematical solutions, particularly with differential equations. After integrating, the solution \(|y| = e^{\cos(1/x) + C}\) was found. This involved an exponential function because the natural logarithm \(\ln|y|\) had been integrated, leading to an exponentiation step to solve for \(y\). An exponential function like \(e^{x}\) grows rapidly and has significant applications in modeling growth processes. In this case, \(y = A e^{\cos(1/x)}\) represents a solution that traces back to the problem's original exponential structure. Understanding the properties of exponential functions helps in predicting the behavior of the solution they form.

Initial Conditions in Differential Equations

Initial conditions are crucial when solving differential equations as they determine specific solutions from a family of possible solutions. In the initial value problem we solved, the condition \(y(2/\pi) = 3\) was key in finding the constant \(A\) in the solution \(y = A e^{\cos(1/x)}\). By substituting the initial condition into the solution, we determined \(A = 3\). Without initial conditions, the solution of a differential equation remains general, depicting an entire set of solutions rather than a single, specific function. Initial conditions anchor the solution to one particular function, effectively tailoring the equation to meet specific scenarios or datasets. Understanding how to apply these conditions is crucial for accurately solving initial value problems.