Problem 5 (a) Verify that the given functi... [FREE SOLUTION]

Chapter 3: Problem 5

(a) Verify that the given function, $y_{P}(t)$, is a particular solution of the differential equation. (b) Determine the complementary solution, $y_{C}(t)$. (c) Form the general solution and impose the initial conditions to obtain the unique solution of the initial value problem. $$y^{\prime \prime}+y^{\prime}=2 t, \quad y(1)=1, \quad y^{\prime}(1)=-2, \quad y_{p}(t)=t^{2}-2 t$$

Short Answer

Expert verified

Question: Verify that $y_P(t) = t^2 - 2t$ is a particular solution of the given differential equation $y'' + y' = 2t$, and solve the initial value problem with initial conditions $y(1) = 1$ and $y'(1) = -2$. Answer: The unique solution for the initial value problem is $y(t) = e^{-t} + (2 - e^{-1}) + t^2 - 2t$.

Step by step solution

Calculate derivatives of $y_P(t)$

Find the first and second derivatives of the given function $y_P(t) = t^2 - 2t$. 1. The first derivative: $\displaystyle\frac{d}{dt}(t^{2}-2 t)=2 t-2$, denoted as $y_P'(t) = 2t - 2$. 2. The second derivative: $\displaystyle\frac{d^{2}}{dt^2}(2t-2)=2$, denoted as $y_P''(t) = 2$.

Verify that $y_P(t)$ is a particular solution

Substitute $y_P(t)$ and its derivatives into the given differential equation ($y''+ y' = 2t$): $y_P''(t) + y_P'(t) = 2 + (2t - 2) = 2t$. Thus, the given function $y_P(t) = t^2 - 2t$ is indeed a particular solution to the given differential equation.

Find the Homogeneous Differential Equation

Set the right side of the given differential equation to $0$ to find the homogeneous differential equation: $y'' + y' = 0$

Find the Characteristic Equation and its Roots

For the homogeneous differential equation $y'' + y' = 0$, the characteristic equation is $r^2 + r = 0$. Factoring the equation yields $r(r+1) = 0$. Therefore, the roots are $r = -1$ and $r = 0$

Determine the Complementary Solution $y_C(t)$

Using the roots of the characteristic equation, the complementary solution can be given by $y_C(t) = Ae^{-t} + B$.

Form the General Solution

The general solution of the nonhomogeneous differential equation is the sum of the complementary solution and the particular solution: $y(t) = y_C(t) + y_P(t) = Ae^{-t} + B + t^2 - 2t$.

Impose the Initial Condition $y(1) = 1$

Now, using the initial condition $y(1) = 1$, we substitute $t = 1$ into the general solution: $1 = Ae^{-1} + B + (1 - 2)$ Solve for $A$ and $B$: $1 = Ae^{-1} + B - 1$ $Ae^{-1} + B = 2$ (Equation 1)

Impose the Initial Condition $y'(1) = -2$

First, find the derivative of the general solution: $y'(t) = -Ae^{-t} - 2t + 2$ Now, using the initial condition $y'(1) = -2$, we substitute $t = 1$ into the derived function: $-2 = -Ae^{-1} - 2(1) + 2$ Solve for $A$: $-Ae^{-1} = -1 \Rightarrow A = e$

Solve for $B$ and Form the Unique Solution

Substitute $A = e$ into Equation 1, and solve for $B$: $e^{-1} + B = 2\Rightarrow B = 2 - e^{-1}$ Since we now have the values of $A$ and $B$, we substitute them into the general solution to form the unique solution for the initial value problem: $y(t) = e^{-t} + (2 - e^{-1}) + t^2 - 2t$

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Key Concepts

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

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation that comes with specific initial conditions that help to determine a unique solution. In the given exercise, the differential equation provided is $y'' + y' = 2t$, and the initial conditions are $y(1) = 1$ and $y'(1) = -2$. This information ensures that when you find solutions, they are not just any possible solutions but ones tailored to start from these particular "initial" values.Initial conditions are essential because, without them, you might end up with many possible solutions to a differential equation. By applying these conditions, we narrow it down to just one unique solution that satisfies both the equation and the starting (initial) values. This process is like setting starting rules for a scenario in which only one specific outcome is possible given those rules.

Particular Solution

In differential equations, a particular solution is one that satisfies the nonhomogeneous differential equation given in the problem. This solution does not satisfy the associated homogeneous equation but rather incorporates the effects of any external "forcing" terms, like the function on the right side of the equation.For example, in our exercise, the given function $y_P(t) = t^2 - 2t$ is tested to be a particular solution by substituting it into the nonhomogeneous differential equation $y'' + y' = 2t$. By calculating its derivatives and ensuring that it satisfies the equation, we verify it's a particular solution. This specific form accounts for the "2t" term in the differential equation, meaning it accounts for whatever external factors or "inputs" are affecting the system described by the equation.

Complementary Solution

The complementary solution $y_C(t)$ is a critical part of solving a differential equation. It is the solution to the associated homogeneous equation, which has the right-hand side set to zero $y'' + y' = 0$. Solving for $y_C(t)$ involves finding the characteristic equation from the homogeneous form.By factoring the characteristic equation $r^2 + r = 0$ into roots, we find $r = -1$ and $r = 0$. These roots help us form the complementary solution, $y_C(t) = Ae^{-t} + B$. The complementary solution represents the natural behavior of the system without any external forces and offers one of the components to the overall general solution of the nonhomogeneous equation. It essentially shows how the system would react if it was left to "itself" without outside influences.

Homogeneous Differential Equation

A homogeneous differential equation is one where all terms depend solely on the function and its derivatives, set equal to zero. In our exercise, the homogeneous form of the differential equation is $y'' + y' = 0$. The right-hand side zero indicates it's free from external "forcing" influences.The goal is to solve this equation separately to find the complementary solution. By determining the characteristic equation, $r^2 + r = 0$, derived from the homogeneous form, it's possible to find the roots that will help form the solution. The homogeneous equation uncovers the natural dynamics of the system without outside influence, giving us insight into how the system behaves in its "default" state, providing solutions based only on internal factors.

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Short Answer

Step by step solution

Calculate derivatives of \(y_P(t)\)

Verify that \(y_P(t)\) is a particular solution

Find the Homogeneous Differential Equation

Find the Characteristic Equation and its Roots

Determine the Complementary Solution \(y_C(t)\)

Form the General Solution

Impose the Initial Condition \(y(1) = 1\)

Impose the Initial Condition \(y'(1) = -2\)

Solve for \(B\) and Form the Unique Solution

Key Concepts

Initial Value Problem

Particular Solution

Complementary Solution

Homogeneous Differential Equation

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