Chapter 4: Problem 15
Solve (showing details): $$\begin{aligned}y_{1}^{\prime} &=4 y_{2}+3 e^{3 t} \\\y_{2}^{\prime} &=2 y_{2}-15 e^{-3 x} \\\y_{1}(0) &=2, y_{2}(0)=2\end{aligned}$$
Short Answer
Expert verified
\( y_2 = -15e^{-x} + 17e^{-2x} \) and \( y_1 = 60e^{-x} - 34e^{-2x} + e^{3t} - 25 \).
Step by step solution
01
Identify the System
We are given a system of first-order differential equations with initial conditions:1. \( y_1' = 4y_2 + 3e^{3t} \)2. \( y_2' = 2y_2 - 15e^{-3x} \)Initial conditions:- \( y_1(0) = 2 \)- \( y_2(0) = 2 \)
02
Solve Equation for \( y_2 \)
The equation for \( y_2 \) is \( y_2' = 2y_2 - 15e^{-3x} \). This is a linear non-homogeneous differential equation. The general solution is obtained using the method of integrating factors. Let \( \,I(t) = e^{\int 2 \,dt} = e^{2t} \, \).Multiply both sides by the integrating factor:\[ e^{2x}y_2' - 2e^{2x}y_2 = -15e^{-x} \, e^{2x} \]Simplifying gives:\[ \frac{d}{dx}(e^{2x}y_2) = -15e^{-x+2x} = -15e^{x} \]Integrate both sides:\[ e^{2x}y_2 = -15\int e^{x} \, dx = -15(e^{x}) + C \]\[ e^{2x}y_2 = -15e^{x} + C \]Solving for \( y_2 \):\[ y_2 = -15e^{-x} + Ce^{-2x} \]Apply the initial condition \( y_2(0) = 2 \):\[ 2 = -15e^{0} + C \times e^{0} \]\[ 2 = -15 + C \]\[ C = 17 \]Thus, \( y_2 = -15e^{-x} + 17e^{-2x} \).
03
Solve Equation for \( y_1 \)
Now substitute \( y_2 = -15e^{-x} + 17e^{-2x} \) into the equation for \( y_1 \):\[ y_1' = 4(-15e^{-x} + 17e^{-2x}) + 3e^{3t} \]This simplifies to:\[ y_1' = -60e^{-x} + 68e^{-2x} + 3e^{3t} \]The solution is given by integrating:\[ y_1 = \int (-60e^{-x} + 68e^{-2x} + 3e^{3t}) \, dt \]Integrating term by term: \[ y_1 = -60\int e^{-x} \, dt + 68\int e^{-2x} \, dt + 3\int e^{3t} \, dt \]\[ y_1 = -60(-e^{-x}) + 68(-\frac{1}{2}e^{-2x}) + 3\frac{1}{3}e^{3t} + C \, \] \[ y_1 = 60e^{-x} - 34e^{-2x} + e^{3t} + C \]Apply the initial condition \( y_1(0) = 2 \):\[ 2 = 60e^{0} - 34e^{0} + e^{0} + C \]\[ 2 = 60 - 34 + 1 + C \]\[ 2 = 27 + C \]\[ C = -25 \]Thus, \( y_1 = 60e^{-x} - 34e^{-2x} + e^{3t} - 25 \).
04
Conclusion of the System Solutions
The solutions for the system of differential equations are:- \( y_2 = -15e^{-x} + 17e^{-2x} \)- \( y_1 = 60e^{-x} - 34e^{-2x} + e^{3t} - 25 \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-order differential equations
First-order differential equations are mathematical expressions involving derivatives of a function with respect to one variable. They are termed 'first-order' because they involve the first derivative of the function. In this exercise, we looked at two equations with derivatives indicating rates of change: the rate at which \(y_1\) changes with respect to \(t\) and the rate at which \(y_2\) changes with respect to \(x\). These equations were presented:
- \( y_1' = 4y_2 + 3e^{3t} \)
- \( y_2' = 2y_2 - 15e^{-3x} \)
Integrating factor method
The integrating factor method is a powerful technique for solving first-order linear non-homogeneous differential equations. The idea is to multiply the entire differential equation by a carefully chosen function, called the integrating factor, which simplifies the equation into something more straightforward to solve.
For the equation \( y_2' = 2y_2 - 15e^{-3x} \), we selected \( e^{2x} \) as the integrating factor. This factor is computed from the coefficient of \( y_2 \) in the equation. By multiplying through by the integrating factor, we can convert the left-hand side into the derivative of a product:
For the equation \( y_2' = 2y_2 - 15e^{-3x} \), we selected \( e^{2x} \) as the integrating factor. This factor is computed from the coefficient of \( y_2 \) in the equation. By multiplying through by the integrating factor, we can convert the left-hand side into the derivative of a product:
- \( e^{2x}y_2' - 2e^{2x}y_2 = -15e^{-x+2x} \)
- Which simplifies to \( \frac{d}{dx}(e^{2x}y_2) = -15e^{x} \)
Linear non-homogeneous differential equation
Linear non-homogeneous differential equations contain terms that are not dependent on the function or its derivatives alone; they also have a `non-homogeneous' term or input. These equations often arise in practical contexts where an external force or input is affecting the system.
In our work, \( y_2' = 2y_2 - 15e^{-3x} \) is a great example of such an equation. The term \(-15e^{-3x}\) represents an external input affecting the variable \( y_2 \). Solving these equations requires addressing both the homogeneous part (which would result if the non-homogeneous term were zero) and the particular part, which is due to the non-homogeneous term.
The general solution combines both elements:
In our work, \( y_2' = 2y_2 - 15e^{-3x} \) is a great example of such an equation. The term \(-15e^{-3x}\) represents an external input affecting the variable \( y_2 \). Solving these equations requires addressing both the homogeneous part (which would result if the non-homogeneous term were zero) and the particular part, which is due to the non-homogeneous term.
The general solution combines both elements:
- A homogeneous solution: \( Ce^{-2x} \)
- A particular solution: \(-15e^{-x} \)
Initial conditions
Initial conditions are specific values provided for the function at a certain point, crucial for identifying the constants within the general solution of differential equations. They "anchor" the function at particular points, ensuring a unique solution appropriate to the context described.
In the given problem, initial conditions are:
\( y_2 = -15e^{-x} + 17e^{-2x} \).
Likewise for \( y_1 \), substituting its condition solved for another constant, leading to a specific solution fitting all conditions faithfully. Initial conditions ensure the solutions are meaningful and precisely adapted to the problem described.
In the given problem, initial conditions are:
- \( y_1(0) = 2 \)
- \( y_2(0) = 2 \)
\( y_2 = -15e^{-x} + 17e^{-2x} \).
Likewise for \( y_1 \), substituting its condition solved for another constant, leading to a specific solution fitting all conditions faithfully. Initial conditions ensure the solutions are meaningful and precisely adapted to the problem described.