Question

(a) Obtain the general solution of the differential equation. (b) Impose the initial conditions to obtain the unique solution of the initial value problem. (c) Describe the behavior of the solution as $t\to -\mathrm{\infty }$ and $t\to \mathrm{\infty }$. In each case, does $y(t)$ approach $-\mathrm{\infty },+\mathrm{\infty }$, or a finite limit? $$y^{\prime \prime }-4y^{\prime }+4y=0,y(-1)=2,y^{\prime }(-1)=1$$

Short answer

Short Answer: For the given second-order homogeneous linear differential equation, the unique solution is $y(t)=\frac{3}{2}{e}^{2t}+t{e}^{2t}$. As $t\to -\mathrm{\infty }$, the solution $y(t)$ approaches a finite limit (0), and as $t\to \mathrm{\infty }$, the solution $y(t)$ approaches positive infinity.

Step-by-step solution

Identify the characteristic equation from the given differential equation.

The given differential equation is: $$y^{\prime \prime }-4y^{\prime }+4y=0$$ Comparing this equation with the standard form of a homogeneous linear differential equation, we can form the characteristic equation as: $$m^2-4m+4=0$$

Solve the characteristic equation for m.

Now, we solve the characteristic equation: $$m^2-4m+4=(m-2{)}^2=0$$ It has a repeated root: $m_1=m_2=2$.

Write the general solution of the differential equation.

Since we have a repeated root, the general solution of the given differential equation can be written as: $$y(t)=C_1{e}^{2t}+C_{2}t{e}^{2t}$$ where $C_1$ and $C_2$ are arbitrary constants to be determined using the initial conditions.

Impose the initial conditions to find the unique solution.

We are given the initial conditions $y(-1)=2$ and $y^{\prime }(-1)=1$. Let's impose these conditions: $$y(-1)=C_1{e}^{-2}+C_2(-1)e^{-2}=2$$ To find $y^{\prime }$, we differentiate the general solution $y(t)$ with respect to $t$: $$y^{\prime }(t)=2C_1{e}^{2t}+C_2{e}^{2t}(2t+1)$$ Now, imposing the other initial condition: $$y^{\prime }(-1)=2C_1{e}^{-2}+C_2{e}^{-2}(2(-1)+1)=1$$ Solve the system of equations for $C_1$ and $C_2$: $$\{\begin{array}{l}{C}_1-C_2{e}^{-2}=2e^2\\ 2C_1-C_2{e}^{-2}(-1)=e^2\end{array}$$ Solving this system, we get $C_1=\frac{3}{2}{e}^2$ and $C_2=e^2$. Plug these values back into the general solution to get the unique solution: $$y(t)=\frac{3}{2}{e}^{2t}+t{e}^{2t}$$

Analyze the behavior of the solution for $t\to -\mathrm{\infty }$ and $t\to \mathrm{\infty }$.

We have the unique solution: $$y(t)=\frac{3}{2}{e}^{2t}+t{e}^{2t}$$ As $t\to -\mathrm{\infty }$: $$y(t)\to 0$$ This means that as $t$ approaches negative infinity, $y(t)$ approaches 0 (a finite limit). As $t\to \mathrm{\infty }$: $$y(t)\to \mathrm{\infty }$$ This means that as $t$ approaches positive infinity, $y(t)$ approaches positive infinity. In summary, as $t\to -\mathrm{\infty }$, the solution $y(t)$ approaches a finite limit (0), and as $t\to \mathrm{\infty }$, the solution $y(t)$ approaches positive infinity.

Key concepts

Characteristic Equation

Understanding the characteristic equation is pivotal when tackling differential equations, particularly linear homogeneous equations with constant coefficients.

When faced with a differential equation like $y^{″}-4y^{\prime }+4y=0$, our objective is to find a solution that satisfies this equation. This is where the characteristic equation comes into play. The characteristic equation is found by converting the differential equation into a polynomial one, substituting $y=e^{mt}$ to get $m^2-4m+4=0$. By solving this algebraic equation, we determine the values of $m$ that are essential in outlining the nature of the solution to the original equation.

The roots of the characteristic equation may be real and distinct, real and repeated, or complex. In the case of real and repeated roots, as seen in the exercise, the general solution will involve a linear combination of exponential terms, including both $e^{mt}$ and $t{e}^{mt}$ to accommodate the multiplicity of the roots.

Initial Value Problem

The initial value problem (IVP) serves as a launcher to obtain a unique solution from the plethora of general solutions offered by differential equations. An IVP consists of a differential equation coupled with initial conditions which specify the values of the function, and possibly its derivatives, at a particular point.

This process narrows down the infinite set of possible solutions to one that precisely aligns with the given conditions. In our exercise, with $y(-1)=2$ and $y^{\prime }(-1)=1$, these conditions guide us in finding the constants $C_1$ and $C_2$ from the general solution $y(t)=C_1{e}^{2t}+C_{2}t{e}^{2t}$. The solution to the IVP ultimately gives us a precise description of the behavior of the system modeled by the differential equation at any given time.

Behavior of Solutions

Analyzing the behavior of solutions over time, particularly as time approaches infinity, is a critical part of understanding differential equations. In practice, it informs us about the stability and long-term predictions of the system in question.

For instance, as $t\to -\mathrm{\infty }$, if the solution approaches a finite value, we have a stable system in that direction. Conversely, as $t\to \mathrm{\infty }$, if the solution increases without bound, we infer that the system exhibits an unstable behavior in that direction. In our example, the solution $y(t)$ trends towards zero as $t$ becomes negatively infinite, indicating a settling behavior, and trends toward positive infinity as $t$ increases, illustrating an unbounded growth.

Homogeneous Linear Differential Equation

Homogeneous linear differential equations are an important class of equations characterized by their linearity, homogeneity, and constant coefficients. These equations often emerge in mathematical models spanning physics, engineering, and economics, where they describe systems exhibiting consistent proportional change.

The general form is given by $a_n{y}^{(n)}+a_{n-1}{y}^{(n-1)}+...+a_1{y}^{\prime }+a_{0}y=0$, where each $a_i$ is a constant and the function's derivatives are raised to a power denoted by their superscripts. Homogeneity means that if $y(t)$ is a solution, so is $Cy(t)$, where $C$ is a constant. This feature leads to a solution space that is a vector space, often spanned by exponential functions whose exponents are dictated by the roots of the characteristic equation associated with the differential equation.