Question

For the following problems, use a software program or your calculator to generate the directional fields. Solve explicitly and draw solution curves for several initial conditions. Are there some critical initial conditions that change the behavior of the solution?[T] $y^{\prime }=y\ln (x)$

Short answer

Initial conditions around $x=1$ can change solution behavior from decay to growth.

Step-by-step solution

Set Up the Differential Equation

We start with the given differential equation $y^{\prime }=y\ln (x)$. This is a first-order differential equation where the rate of change of $y$ depends on both $y$ and a function of $x$.

Separate Variables

Separate the variables to solve the equation. Rewriting the equation gives $y^{\prime }=y\ln (x)$. Dividing by $y$ and multiplying by $dx$, we have $\frac{dy}{y}=\ln (x)dx$.

Integrate Both Sides

Integrate both sides of the separated equation, $\int \frac{dy}{y}=\int \ln (x)dx$. The left side integrates to $\ln |y|+C_1$; use integration by parts for the right side to get $x\ln (x)-x+C_2$.

Solve for $y$

Exponentiate both sides to solve for $y$. Starting from $\ln |y|=x\ln (x)-x+C$, we have $|y|=e^{(x\ln (x)-x+C)}$. Choosing positive $y$, note this can be rewritten as $y=A{e}^{x\ln (x)-x}$, where $A=e^C$.

Analyze Initial Conditions

Examining initial conditions, we determine different values of $A$. If initial conditions make $A>1$, solutions grow more than if $0<A<1$. If an initial condition sets $A=0$, the solution is trivially $y=0$.

Use Software to Generate Directional Fields

Use a software calculator to plot the directional fields for the given differential equation. The graphical representation helps visualize how solutions behave for different initial condition values.

Conclusion and Behavior Analysis

Critically, when $x<1$, $\ln (x)<0$ causing decay, and when $x>1$, $\ln (x)>0$ causing growth. Initial conditions around $x=1$ change the behavior from decay to growth.

Key concepts

Directional Fields

Directional fields, often called slope fields, are helpful graphical representations used in differential equations. They provide a visual method to see how the solutions of a differential equation behave over a region. By plotting tiny line segments at grid points, each with a slope given by the differential equation, you can visualize potential solution curves. This is because each line segment's slope corresponds to the derivative values at those points.

In the problem, a directional field for the equation $y^{\prime }=y\ln (x)$ helps us understand visually how different solutions behave based on various initial conditions. If you use software or a calculator to generate such a field, you'll get a graph where all these small lines show the gradient or direction of the potential solution curves. By drawing a solution curve, you can connect these line segments, effectively mapping out the solution for specific starting conditions the equation models.

These fields are particularly powerful because they can illustrate stability and changes, known as critical points, helping you predict whether solutions will spiral outward, converge to a point, or exhibit any other unique behaviors. This way, you can gain insights into the qualitative behavior of solutions even without precise calculations.

Separation of Variables

Separation of variables is a crucial technique for solving some differential equations. The goal is to rearrange the equation so that each variable and its derivative are on opposite sides. For our differential equation $y^{\prime }=y\ln (x)$, we separate variables by manipulating the terms to isolate $y$ on one side and $x$ on the other.

First, divide both sides by $y$ to have $\frac{dy}{y}$. Then, multiply both sides by $dx$, giving $\frac{dy}{y}=\ln (x)dx$. This makes both sides ready for integration because they are expressed with their respective differentials.

Using this method effectively converts the problem into one integrating each variable independently. This way, we can solve more complex differential equations by breaking them down into manageable parts. Remember that after integration, you might need to apply techniques like integration by parts for more complex integrals, which is especially useful if $\ln (x)$ is part of the integration on the $x$ side.

Initial Conditions

Initial conditions are specific pieces of information giving the value of the function (or its derivatives) at a particular point. In differential equations, they determine the specific solution curve out of many possible ones generated by the general solution. In simpler terms, they help pinpoint exactly which curve among many the solution should follow.

For our equation $y^{\prime }=y\ln (x)$, different initial conditions will give us different constants after integration. For example, choosing an initial condition like $y(x_0)=y_0$ will determine a particular value for the constant $A$ in the general solution $y=A{e}^{x\ln (x)-x}$.

By experimenting with different initial conditions, we see how small changes can significantly affect the behavior of solutions, especially near critical points like $x=1$. These variations can show us decay or growth patterns and even identify non-standard behaviors such as flat solutions with $y=0$. Understanding these concepts is essential when modeling real-life systems where initial states often define future behavior.

Integration by Parts

Integration by parts is a powerful technique used to solve integrals of products of functions. When one of the functions easily differentiates, and the other easily integrates, this method can simplify the process. The integration by parts formula is based on the product rule for differentiation and is given by:$$\int udv=uv-\int vdu$$

In our differential equation $y^{\prime }=y\ln (x)$, we find the need to integrate $\ln (x)$ over $x$. Since $\ln (x)$ doesn't integrate directly, integration by parts becomes essential. Here, we set $u=\ln (x)$ and $dv=dx$, leading to $du=\frac{1}{x}dx$ and $v=x$.

Proceeding with integration by parts, we calculate:

• Integrate $v\cdot du$ $=x\cdot \frac{1}{x}=1$
• This results in the integral $\int \ln (x)dx=x\ln (x)-x+C$

This integration helps us find the antiderivative needed for $\int \ln (x)dx$, thus allowing us to solve for $y$ and continue our journey to unraveling the particular solutions of the differential equation. Mastery of integration by parts enriches your mathematical toolkit, especially for more complex integrations involving tricky functions.