Question

Find the solution of the given initial value problem. Plot the graph of the solution and describe how the solution behaves as $x\to 0$. $2x^2{y}^{\prime \prime }+x{y}^{\prime }-3y=0,y(1)=1,y^{\prime }(1)=4$

Short answer

2) What form does the general solution of the Cauchy-Euler equation take? 3) What are the values of m1 and m2 (the two values of m)? 4) What are the values of C1 and C2 for the initial value problem? 5) What is the complete solution for the initial value problem? 6) Describe the behavior of the solution as x approaches 0. Answers: 1) The given equation is a Cauchy-Euler equation. 2) The general solution of the Cauchy-Euler equation takes the form $y(x)=C_1{x}^{m_1}+C_2{x}^{m_2}$. 3) The values of m1 and m2 are $m_1=1$ and $m_2=-\frac{3}{2}$. 4) The values of C1 and C2 for the initial value problem are $C_1=1$ and $C_2=-2$. 5) The complete solution for the initial value problem is $y(x)=1x^1-2x^{-\frac{3}{2}}$. 6) As x approaches 0, the solution behaves as a vertical line going to negative infinity ($y\to -\mathrm{\infty }$).

Step-by-step solution

Recognize the given equation as a Cauchy–Euler equation

The given equation is in the form of a Cauchy–Euler equation $2x^2{y}^{″}+x{y}^{\prime }-3y=0$, where $a=1$ and $b=-3$.

Assume the solution in the form of y(x) = x^m

Let's assume the solution in the form $y(x)=x^m$. We find the first and second derivatives: $y^{\prime }(x)=m{x}^{m-1}$ $y^{″}(x)=m(m-1)x^{m-2}$

Substitute this assumed solution into the given equation

Now, we substitute $y(x)$, $y^{\prime }(x)$, and $y^{″}(x)$ into the given equation: $2x^2(m(m-1)x^{m-2})+x(m{x}^{m-1})-3(x^m)=0$

Solve for m

Simplify the equation: $m(m-1)2x^m+m{x}^m-3x^m=0$ $x^m(2m^2-2m+m-3)=0$ This equation must hold for all $x>0$. Therefore, we should have: $2m^2-m-3=0$ Solving this quadratic equation, we get two values for m: $m_1=1$ and $m_2=-\frac{3}{2}$.

Write the general solution for the Cauchy–Euler equation

Using the values of $m_1$ and $m_2$, the general solution for the Cauchy–Euler equation is: $y(x)=C_1{x}^1+C_2{x}^{-\frac{3}{2}}$

Use the initial conditions to find the values of C1 and C2

We are given two initial conditions: $y(1)=1$ and $y^{\prime }(1)=4$ Use these initial conditions to find $C_1$ and $C_2$: 1) $y(1)=C_1(1^1)+C_2(1^{-\frac{3}{2}})=1$ 2) $y^{\prime }(x)=C_1(1)-\frac{3}{2}{C}_2{x}^{-\frac{5}{2}}$ and $y^{\prime }(1)=C_1-\frac{3}{2}{C}_2=4$ Solving these equations, we get $C_1=1$ and $C_2=-2$

Plot the graph of the solution and describe its behavior as x→0

The complete solution for the initial value problem is: $y(x)=1x^1-2x^{-\frac{3}{2}}$ Plotting the graph of the solution, we can see that it has an asymptote at $x=0$. Since the term $2x^{-\frac{3}{2}}$ dominates as $x$ approaches 0, the solution will behave as a vertical line going to negative infinity ($y\to -\mathrm{\infty }$) as $x\to 0$.

Key concepts

Differential Equations

Differential equations are powerful tools in mathematics for modeling dynamic systems where the rate of change of one variable is proportional to another variable. They can describe a plethora of phenomena in natural sciences, engineering, economics, and beyond. A differential equation relates a function to its derivatives, with the degree of the highest derivative involved indicating the order of the differential equation. The equation presented in the exercise, $2x^2{y}^{\prime \prime }+x{y}^{\prime }-3y=0$, is a second-order homogeneous linear differential equation and is specifically known as a Cauchy-Euler equation because it contains terms like $x^2{y}^{\prime \prime }$ where the power of $x$ matches the order of the derivative it multiplies.

Understanding how to solve such equations requires familiarity with specific techniques. In this case, recognizing the Cauchy-Euler form guides us to assume a solution of the form $y(x)=x^m$, where $m$ is a constant to be determined. The ability to correctly identify the type of differential equation at hand is crucial as it directs the solver to the appropriate method for finding a solution.

Initial Value Problem

An initial value problem (IVP) is a differential equation accompanied by specific conditions that determine the starting state of the system. These conditions, referred to as 'initial values,' are critical as they allow for a unique solution to be found. In the context of the given Cauchy-Euler equation, the IVP is defined not only by the equation itself but also by the initial values $y(1)=1$ and $y^{\prime }(1)=4$.

The process of solving an IVP involves integrating the differential equation and then applying the given initial conditions to solve for the constants in the general solution. This step is vital to ensure that the solution is tailored to the specific starting scenario described by the initial values. Finding the constants $C_1$ and $C_2$ as outlined in the problem ensures that the resulting function $y(x)$ precisely matches the system's behavior at $x=1$.

Theory of Ordinary Differential Equations

The theory of ordinary differential equations (ODEs) concerns itself with the properties and solutions of differential equations involving only one independent variable. In our exercise, we deal with an ODE of a function $y(x)$ with $x$ being the independent variable. A key feature of ODEs is the existence and uniqueness theorem, which provides conditions under which an ODE has a unique solution. This is closely related to IVPs, as the theorem guarantees a unique solution given sufficient and appropriate initial conditions.

In the study of ODEs, one often classifies them into linear and nonlinear ODEs, homogeneous and non-homogeneous ODEs, and further categorizes them based on their order. The Cauchy-Euler equation falls into the category of second-order linear homogeneous ODEs, which often allows for solutions to be expressed as a combination of power functions of the independent variable. The theory provides us with a systematic approach to finding solutions, using associated characteristic equations, and dictates how initial or boundary conditions shape the final solution.

Asymptotic Behavior

Asymptotic behavior in the context of differential equations refers to how solutions behave as the independent variable approaches a certain point or infinity. For the specific solution of the IVP in this exercise, $y(x)=x-2x^{-\frac{3}{2}}$, the term $2x^{-\frac{3}{2}}$ is dominant as $x$ approaches zero, predicting the solution will sharply decrease towards negative infinity, indicating the presence of an asymptote at $x=0$.

Understanding the asymptotic behavior of solutions is crucial as it provides insight into the long-term trends or potential singularities of the system being modeled. When considering physical phenomena or engineering problems, knowing the asymptotic behavior can help to predict system stability, determine safe operating conditions, or highlight levels of performance that the system may be unable to sustain.