Chapter 3: Problem 14
Find the solution of the given initial value problem. $$ y^{\prime \prime}+4 y=t^{2}+3 e^{\prime}, \quad y(0)=0, \quad y^{\prime}(0)=2 $$
Short Answer
Expert verified
Question: Find the solution of the initial value problem: \(y^{\prime \prime} + 4y = t^2 + 3e^t, \ y(0) = 0, y^{\prime}(0) = 2\).
Answer: The solution of the given initial value problem is: \(y(t) = -\frac{3}{5}\cos(2t) + \frac{1}{2}\sin(2t) + \frac{1}{4}t^2 + \frac{3}{5}e^t\).
Step by step solution
01
Solve the homogeneous equation
First, consider the homogeneous equation:
$$
y^{\prime \prime} + 4y = 0
$$
The corresponding characteristic equation is:
$$
r^2 + 4 = 0
$$
Solving for \(r\), we get complex roots \(r = 0 \pm 2i\), where \(i\) is an imaginary unit. The general solution for the homogeneous equation is:
$$
y_h(t) = A\cos(2t) + B\sin(2t)
$$
02
Find the particular solution for the nonhomogeneous equation
The given equation is:
$$
y^{\prime \prime} + 4y = t^2 + 3e^t
$$
To find the particular solution, we consider two cases for the right side of the equation:
- Case 1: \(g(t) = t^2\)
Assume a particular solution of the form \(y_{p1}(t) = At^2 + Bt + C\) where A, B, C are constants to be determined.
Differentiating \(y_{p1}\) twice with respect to t:
$$
y_{p1}^{\prime} = 2At + B, \quad y_{p1}^{\prime\prime} = 2A
$$
Substituting these values into the nonhomogeneous equation:
$$
2A + 4(At^2 + Bt + C) = t^2 \Rightarrow \\
2A + 4At^2 + 4Bt + 4C = t^2
$$
Using the method of undetermined coefficients, we find that:
$$
A=\frac{1}{4}, \quad B=0, \quad C=0
$$
Thus, \(y_{p1}(t) = \frac{1}{4}t^2\).
- Case 2: \(g(t) = 3e^t\)
Assume a particular solution of the form \(y_{p2}(t) = De^t\), where D is a constant to be determined.
Differentiating \(y_{p2}\) twice with respect to t:
$$
y_{p2}^{\prime} = De^t, \quad y_{p2}^{\prime\prime} = De^t
$$
Substituting these values into the nonhomogeneous equation:
$$
De^t + 4De^t = 3e^t \Rightarrow \\
(1+4)D = 3
$$
We find that \(D = \frac{3}{5}\) and \(y_{p2}(t) = \frac{3}{5}e^t\).
The total particular solution is the sum of the two cases: \(y_p(t) = y_{p1}(t) + y_{p2}(t)\), which gives us:
$$
y_p(t) = \frac{1}{4}t^2 + \frac{3}{5}e^t
$$
03
Combine the solutions and apply the initial conditions
The general solution is the sum of homogeneous and particular solutions:
$$
y(t) = y_h(t) + y_p(t) = A\cos(2t) + B\sin(2t) + \frac{1}{4}t^2 + \frac{3}{5}e^t
$$
Now, we apply the initial conditions. For \(y(0) = 0\), we have:
$$
0 = A\cos(0) + B\sin(0) + \frac{1}{4}(0)^2 + \frac{3}{5}e^{0} \Rightarrow \\
0 = A + 0 + 0 + \frac{3}{5} \Rightarrow \\
A = -\frac{3}{5}
$$
For \(y^{\prime}(0) = 2\), we first find the derivative of \(y(t)\):
$$
y^{\prime}(t) = -2A\sin(2t) + 2B\cos(2t) + \frac{1}{2}t + \frac{3}{5}e^t
$$
Now, we apply the condition:
$$
2 = -2A\sin(0) + 2B\cos(0) + \frac{1}{2}(0) + \frac{3}{5}e^{0} \Rightarrow \\
2 = 0 + 2B + 0 + \frac{3}{5} \Rightarrow \\
B = \frac{1}{2}
$$
04
Write the final solution
By substituting the values of A and B in the general solution, we get the final solution to the given initial value problem:
$$
y(t) = -\frac{3}{5}\cos(2t) + \frac{1}{2}\sin(2t) + \frac{1}{4}t^2 + \frac{3}{5}e^t
$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are a type of mathematical equation involving functions and their derivatives. They describe the relationship between a function and its rates of change, essentially representing physical phenomena where change is a fundamental aspect, such as motion, growth, decay, and other time-dependent processes.
In solving a differential equation, we're essentially looking for an unknown function that satisfies the equation. For example, the initial value problem from the exercise:
\[\begin{equation}y^{\' \' }+4 y=t^{2}+3 e^{t}, \quad y(0)=0, \quad y^{\' }(0)=2 \end{equation}\]
involves finding a function y(t) that satisfies the second-order differential equation and the given initial conditions. It's important to understand that the 'prime' notation (\left(y^{\' }\right)) refers to differentiation with respect to the independent variable, commonly time (t).
Differential equations can be categorized broadly into ordinary differential equations (ODEs), involving functions of a single variable, and partial differential equations (PDEs), involving functions of multiple variables. The given problem is an ODE, as it only involves the function y(t) and its derivatives with respect to a single variable t.
In solving a differential equation, we're essentially looking for an unknown function that satisfies the equation. For example, the initial value problem from the exercise:
\[\begin{equation}y^{\' \' }+4 y=t^{2}+3 e^{t}, \quad y(0)=0, \quad y^{\' }(0)=2 \end{equation}\]
involves finding a function y(t) that satisfies the second-order differential equation and the given initial conditions. It's important to understand that the 'prime' notation (\left(y^{\' }\right)) refers to differentiation with respect to the independent variable, commonly time (t).
Differential equations can be categorized broadly into ordinary differential equations (ODEs), involving functions of a single variable, and partial differential equations (PDEs), involving functions of multiple variables. The given problem is an ODE, as it only involves the function y(t) and its derivatives with respect to a single variable t.
Homogeneous Equation
When solving differential equations, distinguishing between homogeneous and nonhomogeneous equations is crucial. A homogeneous linear differential equation has the form:\[\begin{equation}L(y) = 0\end{equation}\]
where L is a linear differential operator on the function y. These equations are characterized by the fact that all terms involve the function y or its derivatives, and there is no separate function of the independent variable (e.g., t) standing alone.
The exercise presents the homogeneous part of the problem as:\[\begin{equation}y^{\' \' } + 4y = 0\end{equation}\]
With no additional functions of t on the right side, it fits the definition of a homogeneous equation. The general solution to a homogeneous linear differential equation is composed of a linear combination of functions, often involving sines and cosines when complex roots are present, as demonstrated in the exercise with the solution:\[\begin{equation}y_h(t) = A\cos(2t) + B\sin(2t)\end{equation}\]
The constants A and B are determined later using the initial conditions. Homogeneous equations are vital because their solutions form the complementary part of the general solution to a nonhomogeneous equation.
where L is a linear differential operator on the function y. These equations are characterized by the fact that all terms involve the function y or its derivatives, and there is no separate function of the independent variable (e.g., t) standing alone.
The exercise presents the homogeneous part of the problem as:\[\begin{equation}y^{\' \' } + 4y = 0\end{equation}\]
With no additional functions of t on the right side, it fits the definition of a homogeneous equation. The general solution to a homogeneous linear differential equation is composed of a linear combination of functions, often involving sines and cosines when complex roots are present, as demonstrated in the exercise with the solution:\[\begin{equation}y_h(t) = A\cos(2t) + B\sin(2t)\end{equation}\]
The constants A and B are determined later using the initial conditions. Homogeneous equations are vital because their solutions form the complementary part of the general solution to a nonhomogeneous equation.
Particular Solution
A particular solution to a nonhomogeneous differential equation is a specific solution that satisfies the nonhomogeneous equation across its entire domain. Nonhomogeneous differential equations can be written in the form:\[\begin{equation}L(y) = g(t)\end{equation}\]
where g(t) is a non-zero function which makes the equation nonhomogeneous. For the nonhomogeneous part of the exercise:\[\begin{equation}y^{\' \' }+4 y=t^{2}+3 e^{t}\end{equation}\]
the particular solution, denoted by y_p(t), is found by forecasting the form this solution might take based on the nonhomogeneous term g(t). This often involves guessing a solution that resembles g(t) but with undetermined coefficients which are later calculated.
In our exercise, we determined two parts for y_p(t) derived from g(t) = t^2 and g(t) = 3e^t, resulting in:\[\begin{equation}y_p(t) = y_{p1}(t) + y_{p2}(t) = \frac{1}{4}t^2 + \frac{3}{5}e^t\end{equation}\]
Once a particular solution is found, it's added to the general solution of the corresponding homogeneous equation to form the general solution of the nonhomogeneous equation. The final result then incorporates initial conditions to produce a unique solution that satisfies all conditions of the problem.
where g(t) is a non-zero function which makes the equation nonhomogeneous. For the nonhomogeneous part of the exercise:\[\begin{equation}y^{\' \' }+4 y=t^{2}+3 e^{t}\end{equation}\]
the particular solution, denoted by y_p(t), is found by forecasting the form this solution might take based on the nonhomogeneous term g(t). This often involves guessing a solution that resembles g(t) but with undetermined coefficients which are later calculated.
In our exercise, we determined two parts for y_p(t) derived from g(t) = t^2 and g(t) = 3e^t, resulting in:\[\begin{equation}y_p(t) = y_{p1}(t) + y_{p2}(t) = \frac{1}{4}t^2 + \frac{3}{5}e^t\end{equation}\]
Once a particular solution is found, it's added to the general solution of the corresponding homogeneous equation to form the general solution of the nonhomogeneous equation. The final result then incorporates initial conditions to produce a unique solution that satisfies all conditions of the problem.
