Question

In Exercises \(29-32,\) solve the differential equation. $$\frac{d y}{d x}=x^{2} \ln x$$

Short answer

The solution to the differential equation \(\frac{d y}{d x} = x^{2}\ln(x)\) is \(y = \frac{1}{3}x^{3}\ln x - \frac{1}{9}x^{3}+c\).

Step-by-step solution

Rearrange the equation

Separating variables, we will need to arrange the function in such a way that on one side of the equation there are only terms with \(y\) and \(dy\), and on the other side terms with \(x\) and \(dx\). This allows to write the equation in the form \(f(y) dy = g(x) dx\).\nSo, in our case we can write it as: \(dy = x^{2}\ln(x)dx \) which is already separated.

Integrate both sides

Now, integrate both sides of the equation with respect to their respective variables.\nPerforming the integration of each side gives: \(\int dy = \int x^{2}\ln(x) dx\).

Calculate the integral of the right-hand side

Calculate the integral of the right-hand side, \(\int x^{2}\ln(x) dx\), we can use parts of integrations formula, \( \int \! u \, dv = u \, v - \int \! v \, du \).\nLet's \begin{align*} u = \ln x, dv = x^{2} dx \ du = \frac{1}{x} dx, v = \frac{1}{3}x^{3} \end{align*} Then, substitute into the integration by parts formula: \(\int x^{2} \ln x \, dx = \ln x \cdot \frac{1}{3}x^{3} - \int\frac{1}{3}x^{3} \cdot \frac{1}{x}dx = \frac{1}{3}x^{3} \ln x - \frac{1}{3}\int x^{2} dx = \frac{1}{3}x^{3} \ln x - \frac{1}{9}x^{3}\)

Calculate the integral of the left-hand side

On the left-hand side, the integration of dy over y is quite straightforward and results in y. So, \(\int dy = y\)

Combine results and write down the solution

Finally, combining these integrals gives the solution to the differential equation: \( y = \frac{1}{3}x^{3}\ln x - \frac{1}{9}x^{3}+c\), where \( c \) is the constant of integration, it could be found if initial conditions were given.

Key concepts

Integration by Parts

Imagine taking on a challenging puzzle where you break it down into smaller, more manageable pieces. That's what happens in integration by parts; it's a strategy to integrate products of functions that may not be straightforward to integrate as a whole. It follows a specific formula: ewline \[ \int u dv = uv - \int v du \] ewlineThink of it as a mathematical dance where we choose pairs to separate into parts, namely functions \(u\) and \(dv\), based on the assumption that their derivatives or antiderivatives (\(du\) and \(v\) respectively) will be simpler to work with. For instance, when faced with \( \int x^{2} \ln x \, dx \), we can cleverly pick \( u = \ln x\) and \(dv = x^{2} dx\) to make our lives easier.ewlineOnce we've calculated \( du \) and \( v \), we plug these into our formula. We're essentially breaking down the integral into parts, one of which involves a new integral that's often more manageable. The beauty of this method lies in its iterative nature; sometimes, we may need to apply integration by parts multiple times or even use it in combination with other techniques to solve an integral.

Separable Differential Equations

The beauty of separable differential equations rests in their simplicity—these equations are the 'untie the knot' puzzles of calculus. A differential equation is separable if you can manipulate it to isolate all the \(y\)'s and \(dy\)'s on one side of the equation and all the \(x\)'s and \(dx\)'s on the other, like segregating apples and oranges into their respective boxes. This allows you to treat each side of the equation independently.ewlineIn the world of differential equations, simplifying the equation to the form \( f(y) dy = g(x) dx \) means that you can address each side individually—integrate \( f(y) \) with respect to \( y \) and \( g(x) \) with respect to \( x \). It turns the complex relationship between \(x\) and \(y\) into a strikeout game where you neatly pair up each \( dy \) with a \( y \) and each \( dx \) with an \( x \), leading to a solution that describes the curves that these differential equations embody.

Integrating Factors

Confronted with a linear differential equation that seems tangled up? Integrating factors are the lubricant that eases the gears into place, transforming the equation into something that's easier to integrate. Picture an unruly bush that we tame with a growth serum to make it fit neatly in its space. Similarly, an integrating factor is a function we multiply through the differential equation to coax it into a form where the left side becomes the perfect differential of some product of functions.ewlineThis sophisticated tactic involves finding a function, often denoted as \( \mu(x) \) or \( \mu(y) \), that, when multiplied with the original equation, enables us to express the equation as a derivative of a product. It's a bit like assembly instructions for flat-pack furniture—once you figure out the correct configuration, everything falls into place smoothly. In essence, the integrating factor method is a powerful tool that can make the seemingly complicated job of solving certain differential equations as neat as closing a zipper.

Initial Value Problem

An initial value problem (IVP) in the realm of differential equations is much like being given a starting point in a maze; it guides you towards the unique path that leads to the exit. In mathematical terms, an IVP pairs a differential equation with a condition that specifies the value of the unknown function at a particular point, commonly referred to as the initial condition.ewlineSuch problems stipulate not only the rule for how a quantity changes (the differential equation) but also where to begin (the initial condition), such as \( y(x_0) = y_0 \). It's like setting the stage for a play with both a script and a starting scene. Solving an IVP allows us to pin down the exact solution from a family of potential candidates, just as lock coordinates help locate a treasure in a vast sea—the initial condition is the key that unlocks the specific function we seek.