Question

Find the solution of the given initial value problem and plot its graph. How does the solution behave as \(t \rightarrow \infty ?\) $$ y^{\mathrm{iv}}-4 y^{\prime \prime \prime}+4 y^{\prime \prime}=0 ; \quad y(1)=-1, \quad y^{\prime}(1)=2, \quad y^{\prime \prime}(1)=0, \quad y^{\prime \prime \prime}(1)=0 $$

Short answer

In conclusion, given the initial value problem \(y^{(4)} - 4y'''+ 4y'' = 0,\) \(y(1) = -1, y'(1) = 2, y''(1) = 0,\) and \(y'''(1) = 0\), the particular solution is: $$y(t) = \frac{1}{5} - \frac{4}{5}t + \frac{3}{10} e^{2t} - \frac{1}{10} t e^{2t}$$ As \(t \rightarrow \infty\), the solution is dominated by the exponential terms, yielding: $$y(t) \sim \frac{3}{10} e^{2t} - \frac{1}{10} t e^{2t}$$

Step-by-step solution

Characteristic equation

Create the characteristic equation associated with the given differential equation: $$r^4 - 4r^3 + 4r^2 = 0$$

Solve the characteristic equation

Factorize the characteristic equation and find its roots: $$r^2(r^2 - 4r + 4) = r^2(r-2)^2 = 0$$ This leads to possible roots: \(r_1 = 0, r_2 = 0, r_3 = 2, r_4 = 2\)

Write the general solution

Based on the characteristics equation roots, we can form the general solution of the given ODE as: $$y(t) = C_1 + C_2t + C_3e^{2t} + C_4te^{2t}$$

Apply initial conditions

Now, apply the given initial conditions to find the constants \(C_1, C_2, C_3,\) and \(C_4:\) 1. \(y(1) = -1:\) Substitute \(t=1\) into the general solution and get $$-1 = C_1 + C_2 + C_3e^{2} + C_4e^{2}$$ 2. \(y'(1) = 2:\) Differentiate the general solution with respect to \(t\) and substitute \(t=1\): $$y'(t) = C_2 + 2C_3e^{2t} + (2C_4 + C_4)t e^{2t}$$ $$2 = C_2 + 2C_3e^{2} + 3C_4e^{2}$$ 3. \(y''(1) = 0:\) Differentiate \(y'(t)\) with respect to \(t\) and substitute \(t=1\): $$y''(t) = 4C_3e^{2t} + 8(2C_4 + C_4)e^{2t}$$ $$0 = 4C_3e^{2} + 24C_4e^{2}$$ 4. \(y'''(1) = 0:\) Differentiate \(y''(t)\) with respect to \(t\) and substitute \(t=1\): $$y'''(t) = 8C_3e^{2t} + 48C_4e^{2t}$$ $$0 = 8C_3e^{2} + 48C_4e^{2}$$ Solve the system of equations to find the constants, obtaining \(C_1 = \frac{1}{5}, C_2 = -\frac{4}{5}, C_3 = \frac{3}{10},\) and \(C_4 = -\frac{1}{10}.\) Hence, the particular solution is: $$y(t) = \frac{1}{5} - \frac{4}{5}t + \frac{3}{10} e^{2t} - \frac{1}{10} t e^{2t}$$

Analyze solution behavior at \(t \rightarrow \infty\)

To find the behavior of the solution as \(t \rightarrow \infty\), observe the terms in \(y(t):\) Considering the exponential terms \(e^{2t}\) and \(te^{2t}\), as \(t\) goes to infinity, these terms will dominate over the linear terms. Thus, the solution behavior when \(t \rightarrow \infty\) will depend primarily on the exponential terms: $$y(t) \sim \frac{3}{10} e^{2t} - \frac{1}{10} t e^{2t}$$

Key concepts

Characteristic Equation

The concept of a characteristic equation is fundamental when dealing with certain types of differential equations, specifically linear ones with constant coefficients. To derive the characteristic equation, one assumes that the solution to the differential equation can be written in the form of an exponential function, often \(e^{rt}\). By substituting this form into the differential equation, we convert the problem into an algebraic one in terms of the characteristic polynomial, usually denoted by \(r\).

The roots of this characteristic equation tell us a lot about the possible solutions to the original differential equation.

• Real roots suggest exponential growth or decay.
• Complex roots indicate oscillatory behavior.
• Repeated roots point towards polynomial terms multiplying the exponential solution.

This characterizes the solution set and helps in understanding how the system evolves over time.

Differential Equation

Differential equations are equations that encompass functions and their derivatives. These play a crucial role in modeling various physical phenomena, such as motion, heat, and waves. In mathematics, specifically in the provided problem, you see a higher-order linear differential equation with constant coefficients:
\( y^{\mathrm{iv}} - 4 y^{\prime \prime \prime} + 4 y^{\prime \prime} = 0 \)

This is a homogeneous differential equation because all terms involve the dependent variable \(y(t)\) and its derivatives.

• "Homogeneous" means there is no free-standing constant term.
• "Linear" indicates that each term is either the derivative of \(y\) or \(y\) itself, not a nonlinear function thereof.

Differential equations lay out the relationship within the rate of change (thanks to derivatives) of a quantity, making them foundational in describing dynamic systems. Working through their solutions involves finding the function \(y(t)\) that satisfies the equation for given initial conditions.

Exponential Growth

Exponential growth occurs when the rate of change of a variable is proportional to the current value of that variable, leading to rapid and increasing changes over time. In the context of the characteristic equation roots, particularly when real and positive, the exponential growth behavior is observed in the components of the solution.

In the solution \\( y(t) = \frac{1}{5} - \frac{4}{5}t + \frac{3}{10} e^{2t} - \frac{1}{10} t e^{2t} \)\

• Terms like \(e^{2t}\) and \(t e^{2t}\) exhibit exponential growth as \(t\) increases.
• Exponential components dominate other types of growth (such as linear or polynomial) as time \(t\) moves towards larger values.

This kind of behavior is critical in systems that replicate over time, such as populations, investments, or certain physical phenomena exhibiting exponential response.

Solution Behavior as t Approaches Infinity

The solution behavior as \(t \ ightarrow \infty\) provides insights into the long-term behavior of the system described by the differential equation. We need to consider the dominant terms that dictate this behavior.

In the context of the provided solution:\\( y(t) = \frac{1}{5} - \frac{4}{5}t + \frac{3}{10} e^{2t} - \frac{1}{10} t e^{2t} \)\

• The exponential terms \(e^{2t}\) and \(t e^{2t}\) grow significantly faster than any polynomial terms as \(t\) increases.
• The \(y(t)\) function primarily depends on these exponential parts for its asymptotic behavior.
• Even with negative coefficients, the terms with \(e^{2t}\) dominate, leading to exponential growth at large \(t\).

In practical terms, understanding solution behavior as \(t \ ightarrow \infty\) is essential in predicting how a solution behaves in extremes, which can be crucial in real-world applications like engineering systems and natural processes.