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For the given differential equation, $$ y^{\prime \prime}-2 y^{\prime}-3 y=2 e^{-t} \cos t+t^{2}+t e^{3 t} $$

Short Answer

Expert verified
Question: Find the general solution of the following second-order linear inhomogeneous differential equation: $$ y'' - 2y' - 3y = 2e^{-t}\cos t + t^2 + te^{3t}. $$ Answer: The general solution of the given differential equation is: $$ y(t) = C_1 e^{3t} + C_2 e^{-t} -\frac{1}{5}t e^{3t} + \frac{9}{10}e^{-t} \cos t + \frac{1}{10}t^{2} + t, $$ where \(C_1\) and \(C_2\) are arbitrary constants.

Step by step solution

01

Solve the homogeneous equation

We first consider the associated homogeneous equation: $$ y^{\prime \prime} - 2 y^{\prime} - 3 y = 0. $$ This is a linear second-order differential equation with constant coefficients. We can solve it by assuming a solution of the form \(y_h = e^{rt}\) and substituting this into the equation. This yields a quadratic equation for \(r\): $$ r^{2} e^{rt} - 2 r e^{rt} - 3 e^{rt} = 0. $$ Since \(e^{rt} \neq 0\), we can divide both sides by this term and obtain the quadratic equation for \(r\): $$ r^{2} - 2 r - 3 = 0. $$ Now we solve for \(r\) by factoring this equation: $$ (r - 3)(r + 1) = 0. $$ This gives us the roots \(r_1 = 3\) and \(r_2 = -1\). Therefore, the homogeneous solution is given by the complementary function: $$ y_h(t) = C_1 e^{3t} + C_2 e^{-t}, $$ where \(C_1\) and \(C_2\) are arbitrary constants.
02

Find the particular integral

Now we turn our attention to the inhomogeneous equation. We need to find a particular integral \(y_p(t)\) such that: $$ y^{\prime \prime}-2 y^{\prime}-3 y=2 e^{-t} \cos t+t^{2}+t e^{3 t}. $$ To do this, we will assume the particular solution has the following form: $$ y_p(t) = Ate^{3t} + Be^{-t} \cos t + Ct^{2} + Dt. $$ The first term is due to the presence of \(te^{3t}\) in the inhomogeneous part. The second term comes from the \(e^{-t}\cos t\) in the inhomogeneous part, with an additional \(B e^{-t}\) term to avoid duplication with the complementary function. The third term is due to \(t^2\), and the fourth term is a polynomial necessary to cancel the non-oscillatory exponential term. Now we calculate the first and second derivatives of \(y_p(t)\), substitute them into the inhomogeneous equation, and find the coefficients \(A, B, C, D\) by matching terms on both sides.
03

Calculate derivatives and substitute

First, find the first derivative of \(y_p(t)\): $$ y_p^{\prime}(t) = 3Ate^{3t} + Ae^{3t} - Be^{-t} \cos t + Be^{-t} \sin t + 2Ct + D. $$ Next, find the second derivative of \(y_p(t)\): $$ y_p^{\prime \prime}(t) = 9Ate^{3t} + 6Ae^{3t} + Ae^{3t} + 2Be^{-t} \cos t - 2Be^{-t} \sin t + 2C. $$ Now, substitute \(y_p(t)\), \(y_p^{\prime}(t)\), and \(y_p^{\prime \prime}(t)\) into the inhomogeneous equation and simplify.
04

Equate coefficients and solve

After substitution and simplification, we equate coefficients of like terms on both sides and solve the following system of linear equations for the coefficients: $$ \begin{cases} 9A + 2B = 0, \\ 6A - 2B = 2, \\ A + 2C = 0, \\ D = 1. \end{cases} $$ Solving this system, we find that \(A = -\frac{1}{5}\), \(B = \frac{9}{10}\), \(C = \frac{1}{10}\), and \(D = 1\). Therefore, the particular integral is: $$ y_p(t) = -\frac{1}{5}t e^{3t} + \frac{9}{10}e^{-t} \cos t + \frac{1}{10}t^{2} + t. $$
05

Combine the complementary function and particular integral

Finally, the general solution to the given differential equation is the sum of the complementary function and the particular integral: $$ y(t) = y_h(t) + y_p(t) = C_1 e^{3t} + C_2 e^{-t} -\frac{1}{5}t e^{3t} + \frac{9}{10}e^{-t} \cos t + \frac{1}{10}t^{2} + t. $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Homogeneous Equations
In the realm of differential equations, a homogeneous equation is one where the terms are all dependent on the function and its derivatives. Specifically, it means the equation is set to zero, which implies any term involving the independent variable alone is absent.

Consider the equation \(y^{\prime \prime} - 2y^{\prime} - 3y = 0\). This is our homogeneous equation. The significance of a homogeneous equation lies in its ability to be solved independently by assuming a solution form, often using exponentials like \(e^{rt}\).

  • We find solutions by determining characteristic roots \(r\) using a simple process where the equation transforms into a polynomial.
  • Here, \(r^2 - 2r - 3 = 0\) is factored into \((r-3)(r+1) = 0\), yielding roots \(r_1 = 3\) and \(r_2 = -1\).

The roots indicate the form of the general solution, made by simple linear combinations of these exponential solutions. In this example, the complementary function, also known as the homogeneous solution, is given by \(y_h(t) = C_1 e^{3t} + C_2 e^{-t}\), where \(C_1\) and \(C_2\) are constants defining the family of solutions.
Particular Integral
Once the homogeneous part of the equation is tackled, the next step involves finding a solution that satisfies the full non-homogeneous equation. This calls for finding a particular integral, also denoted \(y_p(t)\), which shifts the problem from the perfect world of homogeneous solutions to the real dynamics of an equation with non-zero right-hand side.

The particular integral takes into account the specific parts of the non-homogeneous equation, such as \(2 e^{-t} \cos t+t^{2}+t e^{3 t}\). Different components like polynomials, exponential, or trigonometric functions in the non-homogeneous term directly suggest the form of \(y_p(t)\).

  • The first term \(te^{3t}\) implies a solution component \(At e^{3t}\).
  • \(e^{-t} \cos t\) suggests including \(Be^{-t} \cos t + Ce^{-t} \sin t\).
  • For the polynomial \(t^2\), the analogous solution is \(Dt^2 + Et\).

This comprehensive approach assures no overlap with any form already covered by the homogeneous solution, thus honoring the principle of superposition.
Linear Second-Order Differential Equations
Linear second-order differential equations involve the second derivative of the unknown function and are paramount in fields like physics and engineering for their application in modeling natural phenomena. Such equations are linear because every term is a linear function of the unknown and its derivatives.

The general form of a linear second-order differential equation looks like \(a y^{\prime \prime} + b y^{\prime} + c y = g(t)\), where \(a\), \(b\), and \(c\) are constants, and \(g(t)\) represents a non-homogeneous term if it exists (or zero if homogeneous).

When you have constants as coefficients like in our equation \(y^{\prime \prime}-2 y^{\prime}-3 y=2 e^{-t} \cos t+t^{2}+t e^{3 t}\), the process of finding the solution involves:
  • First, solving the homogeneous equation to find the complementary function, \(y_h(t)\).
  • Then, discovering a particular solution, \(y_p(t)\), addressing the non-homogeneous elements.

By combining \(y_h(t)\) and \(y_p(t)\), we obtain the general solution \(y(t)\). This illustrates the superposition principle where the total solution is a sum of the homogeneous and particular solutions, reflecting both natural responses and external forces.

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Most popular questions from this chapter

In each exercise, you are given the general solution of $$ y^{(4)}+a_{3} y^{\prime \prime \prime}+a_{2} y^{\prime \prime}+a_{1} y^{\prime}+a_{0} y=0, $$ where \(a_{3}, a_{2}, a_{1}\), and \(a_{0}\) are real constants. Use the general solution to determine the constants \(a_{3}, a_{2}, a_{1}\), and \(a_{0}\). [Hint: Construct the characteristic equation from the given general solution.] $$ y(t)=c_{1} e^{-t} \sin t+c_{2} e^{-t} \cos t+c_{3} e^{t} \sin t+c_{4} e^{t} \cos t $$

For each differential equation, (a) Find the complementary solution. (b) Find a particular solution. (c) Formulate the general solution. $$ y^{\prime \prime \prime}+y=t^{3} $$

The general solution of the nonhomogeneous differential equation \(y^{\prime \prime}+\alpha y^{\prime}+\beta y=g(t)\) is given, where \(c_{1}\) and \(c_{2}\) are arbitrary constants. Determine the constants \(\alpha\) and \(\beta\) and the function \(g(t)\). $$y(t)=c_{1} \sin 2 t+c_{2} \cos 2 t-1+\sin t$$

Consider the \(n\)th order homogeneous linear differential equation $$ y^{(n)}+a_{n-1} y^{(n-1)}+\cdots+a_{3} y^{\prime \prime \prime}+a_{2} y^{\prime \prime}+a_{1} y^{\prime}+a_{0} y=0, $$ where \(a_{0}, a_{1}, a_{2}, \ldots, a_{n-1}\) are real constants. In each exercise, several functions belonging to a fundamental set of solutions for this equation are given. (a) What is the smallest value \(n\) for which the given functions can belong to such a fundamental set? (b) What is the fundamental set? $$ y_{1}(t)=t, \quad y_{2}(t)=e^{t}, \quad y_{3}(t)=\cos t $$

Concavity of the Solution Curve In the discussion of direction fields in Section 1.3, you saw how the differential equation defines the slope of the solution curve at a point in the ty-plane. In particular, given the initial value problem \(y^{\prime}=f(t, y), y\left(t_{0}\right)=\) \(y_{0}\), the slope of the solution curve at initial condition point \(\left(t_{0}, y_{0}\right)\) is \(y^{\prime}\left(t_{0}\right)=f\left(t_{0}, y_{0}\right)\). In like manner, a second order equation provides direct information about the concavity of the solution curve. Given the initial value problem \(y^{\prime \prime}=f\left(t, y, y^{\prime}\right), y\left(t_{0}\right)=\) \(y_{0}, y^{\prime}\left(t_{0}\right)=y_{0}^{\prime}\), it follows that the concavity of the solution curve at the initial condition point \(\left(t_{0}, y_{0}\right)\) is \(y^{\prime \prime}\left(t_{0}\right)=f\left(t_{0}, y_{0}, y_{0}^{\prime}\right)\). (What is the slope of the solution curve at that point?) Consider the four graphs shown. Each graph displays a portion of the solution of one of the four initial value problems given. Match each graph with the appropriate initial value problem. (a) \(y^{\prime \prime}+y=2-\sin t, \quad y(0)=1, \quad y^{\prime}(0)=-1\) (b) \(y^{\prime \prime}+y=-2 t, \quad y(0)=1, \quad y^{\prime}(0)=-1\) (c) \(y^{\prime \prime}-y=t^{2}, \quad y(0)=1, \quad y^{\prime}(0)=1\) (d) \(y^{\prime \prime}-y=-2 \cos t, \quad y(0)=1, \quad y^{\prime}(0)=1\)

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