Problem 11 - Single Variable Calculus

Question

Solve the differential equation. $$ x y^{\prime}-y=2 x(\ln x)^{2} $$

Step-by-Step Solution

Answer

The short answer is: \( y(x) = x \int 2x^{-1} (\ln x)^2 \, dx + Cx \) where C is the integration constant. Unfortunately, we cannot find a closed-form expression for the integral in this case, so this is the final form of the analytical solution.

Step 1: Rewrite the differential equation

First, let's rewrite the given differential equation in terms of y'(x), so that we can identify if it is separable. Dividing both sides by x: $$ y'(x)-\frac{y}{x} = 2(\ln x)^2 $$

Step 2: Check for separability

Now, let's check if the differential equation is separable. To do this, we seek to represent the equation in the form of: $$ g(y) \cdot y'(x) = f(x) $$ Unfortunately, y'(x) cannot be factored out from the left-hand side. Thus, the differential equation is not separable.

Step 3: Apply an integrating factor

Since the differential equation is not separable, we should look for an integrating factor. The given equation is a first-order linear differential equation of the form: $$ y'(x) + P(x)y(x) = Q(x) $$ where \(P(x) = -\frac{1}{x}\) and \(Q(x) = 2 (\ln x)^2\). The integrating factor is given by: $$ I(x) = e^{\int P(x) \, dx}. $$

Step 4: Compute the integrating factor

Now, compute the integrating factor: $$ I(x) = e^{\int -\frac{1}{x} \, dx} = e^{-\ln x} = x^{-1}. $$

Step 5: Multiply the differential equation by the integrating factor

Now, multiply the given differential equation by the integrating factor: $$ x^{-1} \left( y'(x) - \frac{y}{x} \right) = 2x^{-1}(\ln x)^2 $$ This new equation will be in the form: $$ \left( y(x) \cdot I(x) \right)' = 2x^{-1}(\ln x)^2. $$

Step 6: Integrate both sides of the equation

Now, integrate both sides of the equation with respect to x: $$ \int \left( y(x) \cdot x^{-1} \right)' \, dx = \int 2x^{-1}(\ln x)^2 \, dx $$ The left side integral is straightforward: $$ y(x) \cdot x^{-1} = \int 2x^{-1}(\ln x)^2 \, dx + C, $$ where C is the integration constant.

Step 7: Solve for y(x)

Now, we will multiply both sides by x to isolate y(x): $$ y(x) = x \int 2x^{-1}(\ln x)^2 \, dx + Cx. $$ At this stage, we have derived the general formula for the solution. However, we cannot find a closed-form expression for the integral in this case. So, this is the final form of the analytical solution.

Key Concepts

Integrating Factor

The integrating factor is a useful technique for solving certain types of differential equations, specifically the first-order linear differential equations. Imagine you have an equation that isn’t easily separable. Here, the integrating factor becomes your best friend. It helps turn a difficult-to-solve differential equation into one that is more manageable.

For the given problem, the differential equation is identified as a first-order linear equation in the form:

• \( y' + P(x)y = Q(x) \)
• Where \(P(x) = -\frac{1}{x}\) and \(Q(x) = 2(\ln x)^2\).

The integrating factor \(I(x)\) for such equations is determined by:

• Calculating \(I(x) = e^{\int P(x) \, dx}\).
• For this exercise, it becomes \(e^{-\ln x} = x^{-1}\).

Once you have \(I(x)\), multiply through the original equation by this factor, transforming it into a simpler form. This clever trick is like unlocking a puzzle! Feel how satisfying it is to see the equation get easier to handle.

Separability

Separability is another method used to solve differential equations, but it only works for equations that can be rewritten in a way that allows terms involving \(y\) to be separated from terms involving \(x\). A separable differential equation has the form:

• \(g(y) \, y' = f(x)\)

For this problem, the attempt to separate reveals that it is not possible to isolate \(y'\) as such. This suggests that other methods, such as using an integrating factor, are necessary.

The equation \(y'(x) - \frac{y}{x} = 2(\ln x)^2\) cannot be easily split into separate functions like \(g(y)\) and \(f(x)\). Therefore, recognizing that the equation isn't separable helps guide us to pick the right tool, which is the integrating factor method. Being able to differentiate between separable and non-separable forms can save a lot of time in solving these puzzles.

First-order Linear Differential Equation

A first-order linear differential equation is a foundational type of equation in mathematics. Understanding what makes an equation linear and first-order helps greatly in knowing the types of tools you can use to solve it

Such an equation typically looks like:

• \(y' + P(x)y = Q(x)\)

The key features are:

• \(y'\) is the first derivative of \(y\) - this indicates it's first-order
• The function \(P(x)\) and \(Q(x)\) can be any function of \(x\) but do not involve higher-order derivatives or powers of \(y\).

In our exercise, with the equation transformed into \(y'(x) - \frac{y}{x} = 2(\ln x)^2\), we identify \(P(x)\) as \(-\frac{1}{x}\) and \(Q(x)\) as \(2(\ln x)^2\). Recognizing this form is the crucial step to applying the integrating factor method, like finding a hidden map to solving the puzzle that's in front of you.

Knowing about first-order linear differential equations not only helps solve them but also paves the way to understanding more complex differential systems.

Integration Constant

In solving differential equations, an integration constant \(C\) comes into play during the integration process. It arises because an indefinite integral can represent a whole family of functions, differing only by a constant term.

While integrating both sides of our transformed equation, which is

• \(\int (y(x) \cdot x^{-1})' \, dx = \int 2x^{-1}(\ln x)^2 \, dx\)

the constant \(C\) reflects possible shifts in the resulting solution. This constant becomes vital, especially when you need particular solutions that satisfy initial conditions or constraints on the problem.

Without \(C\), you might end up with an incomplete picture of the solution. In practical scenarios, initial conditions or additional information will help you solve for \(C\), tailoring the general solution into something specific and useful.

Think of the integration constant as the element that provides flexibility, allowing you to fit the general solution correctly to particular cases. Therefore, never forget \(C\) – it holds the key to finalizing your solution adventure!