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Magazine Chapter 7: Problem 18
$$\begin{array}{l}{y^{\prime \prime}-y^{\prime}-2 y=3 \delta(t-1)+e^{t}} \\\ {y(0)=0, \quad y^{\prime}(0)=3}\end{array}$$
Short Answer
Expert verified
After all the steps, the solution for given second order non-homogeneous differential equation is \(y(t) = -1/4 * e^{2t} + e^{-t} + 1/4 * e^{t} + 3H(t-1)\).
Step by step solution
01
Non-homogeneous term analysis
First, identify the non-homogeneous terms in the equation, which in this case are \(3 \delta(t-1)\) and \(e^{t}\). The approach to each is different. The delta function will cause a jump at \(t = 1\) and the exponential function contributes to the homogeneous solution.
02
Non-homogeneous solution for the Dirac function
To handle the Dirac function, solve the homogeneous equation disregarding the exponential function. Then, to account for the jump at \(t = 1\) caused by \(3 \delta(t-1)\), impose a unit jump in the derivative of \(y\) at \(t = 1\) leading to an output jump of \(3\). This leads to an additional term in \(y(t)\), given as \(3H(t-1)\) where \(H(t)\) is the Heaviside step function.
03
Homogeneous Solution and Superposition
To handle the \(e^{t}\) term, obtain the homogeneous solution of the given second order differential equation \(y''-y'-2y=0\). Following this, find the real roots of the quadratic equation given by the characteristic polynomial \(m^2-m-2=0\), which are \(m=2,m=-1\). So the homogeneous solution becomes \(y_h(t)=C1e^{2t}+C2e^{-t}\).
04
Particular Solution for the exponential function
To find a particular solution for the non-homogeneous equation with the exponential function, guess \(y_p(t)=Ae^{t}\), plug this into the differential equation, and solve for \(A\). This gives \(A=1/4\). Hence, the particular solution is \(y_p(t)=(1/4)e^{t}\).
05
Combine solutions
Combine the homogeneous solution \(y_h(t)\) and the particular solution \(y_p(t)\), and the discontinuous solution \(3H(t-1)\). Therefore the general solution becomes \(y(t) = C1e^{2t} + C2e^{-t} + (1/4)e^{t} + 3H(t-1)\)
06
Apply Initial Conditions
Apply the initial conditions provided in the problem: \(y(0)=0, \quad y^{\prime}(0)=3\). Solve the resulting system of equations to find the values of \(C1\) and \(C2\). After calculation, we get \(C1 = -1/4\) and \(C2 = 1\). Replacing these in the equation, the solution becomes \(y(t) = -1/4 * e^{2t} + e^{-t} + 1/4 * e^{t} + 3H(t-1)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Non-Homogeneous Solution
A non-homogeneous solution to a differential equation involves both the homogeneous and particular solutions. In differential equations, the non-homogeneous component is an external force or input affecting the system. For the given problem, this includes terms like the Dirac Delta function and the exponential function.
- The non-homogeneous part is specifically not zero, meaning there are outside influences or inputs.
- The differential equation has the non-homogeneous terms: 3未(t-1) and et.
Dirac Delta Function
The Dirac Delta function, often referred to as 未(t-c), represents an impulse or a "kick" at a specific point (t=c). Its key property is that it is zero everywhere except at t = c, where it becomes infinite, while maintaining an area of one under the curve.
- In our equation, 未(t-1) signifies an impulse at t = 1.
- This function shifts the solution from continuous to experiencing a sudden change at this impulse point.
Heaviside Step Function
A Heaviside step function, noted as H(t-c), is a simple function that "steps" up from zero to one at a particular point, c. It is used to model scenarios where an input suddenly turns on at a specific time.
- In our context, H(t-1) means the function jumps from 0 to 1 at time t=1.
- It reflects how the Dirac Delta function causes a change in the system's behavior at t=1.
Homogeneous Solution
Finding the homogeneous solution is the first step towards solving differential equations. The homogeneous equation is derived from the original equation by removing the non-homogeneous parts, setting them to zero.
- For the equation at hand, this would mean solving y'' - y' - 2y = 0.
- Such an equation usually translates into a characteristic equation. Here, solving m虏 - m - 2 = 0 provides roots that are essential for the homogeneous solution.
Initial Conditions
Initial conditions play a critical role in finding the unique solution for a differential equation. They specify the system's state at a starting point, often allowing us to solve for arbitrary constants present in the general solution.
- For our problem, the initial conditions are y(0) = 0 and y'(0) = 3.
- These conditions help in calculating specific values needed to make the general solution specific to a given situation.
