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Magazine Chapter 8: Problem 1
\(y^{\prime \prime}+11 y^{\prime}+24 y=0, y(0)=-1, y^{\prime}(0)=0\)
Short Answer
Expert verified
The solution is \( y(t) = -\frac{8}{5} e^{-3t} + \frac{3}{5} e^{-8t} \).
Step by step solution
01
- Write the characteristic equation
Given the differential equation is a second-order linear homogeneous differential equation: \[ y'' + 11y' + 24y = 0 \]First, write the corresponding characteristic equation: \[ r^2 + 11r + 24 = 0 \].
02
- Solve the characteristic equation
Solve the characteristic equation: \[ r^2 + 11r + 24 = 0 \]Factorize it to find the roots: \[ (r + 3)(r + 8) = 0 \]Thus, the roots are: \[ r_1 = -3 \] and \[ r_2 = -8 \].
03
- Write the general solution
Using the roots from the characteristic equation, write the general solution: \[ y(t) = C_1 e^{-3t} + C_2 e^{-8t} \].
04
- Apply initial conditions
Use the initial conditions to find the constants. The initial conditions are y(0) = -1 and y'(0) = 0.First, apply y(0) = -1:\[ y(0) = C_1 e^{0} + C_2 e^{0} = -1 \]This simplifies to: \[ C_1 + C_2 = -1 \].
05
- Find the derivative and apply initial conditions
Now, find the first derivative of the general solution: \[ y'(t) = -3C_1 e^{-3t} - 8C_2 e^{-8t} \].Apply the initial condition y'(0) = 0:\[ y'(0) = -3C_1 e^{0} - 8C_2 e^{0} = 0 \]This simplifies to: \[ -3C_1 - 8C_2 = 0 \].
06
- Solve the system of equations
Solve the system of equations formed from the initial conditions:\[ C_1 + C_2 = -1 \]\[ -3C_1 - 8C_2 = 0 \]From the second equation: \[ -3C_1 = 8C_2 \]\[ C_1 = -\frac{8}{3}C_2 \]Substitute \( C_1 = -\frac{8}{3}C_2 \) into the first equation:\[ -\frac{8}{3}C_2 + C_2 = -1 \]\[ -\frac{5}{3}C_2 = -1 \]\[ C_2 = \frac{3}{5} \].Now, substitute \( C_2 \) back into \( C_1 + C_2 = -1 \):\[ C_1 + \frac{3}{5} = -1 \]\[ C_1 = -1 - \frac{3}{5} = -\frac{8}{5} \].
07
- Write the particular solution
Use the values of \( C_1 \) and \( C_2 \) to write the particular solution:\[ y(t) = -\frac{8}{5} e^{-3t} + \frac{3}{5} e^{-8t} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
characteristic equation
To solve a second-order linear homogeneous differential equation, the first step is to create the characteristic equation. Given the equation: \[ y'' + 11y' + 24y = 0 \]we replace the derivatives with powers of \(r\), resulting in the characteristic equation: \[ r^2 + 11r + 24 = 0 \].This characteristic equation helps us find the roots, which are essential for constructing the general solution.
initial conditions
Initial conditions provide specific values for the solution and its derivative at a particular point, usually at \(t = 0\). In this problem, we are given \( y(0) = -1 \) and \( y'(0) = 0 \).
- The first condition, \( y(0) = -1 \), tells us the value of the function at \( t = 0 \).
- The second condition, \( y'(0) = 0 \), tells us the value of the first derivative at \( t = 0 \).
general solution
The general solution of a differential equation is based on the roots of the characteristic equation. For our problem, the characteristic equation is:\[ r^2 + 11r + 24 = 0 \].The roots are found through factorization:\[ (r + 3)(r + 8) = 0 \],leading to roots \( r_1 = -3 \) and \( r_2 = -8 \).Thus, the general solution is:\[ y(t) = C_1 e^{-3t} + C_2 e^{-8t} \],where \( C_1 \) and \( C_2 \) are constants determined by the initial conditions.
roots
The roots of the characteristic equation are crucial in finding the general solution. In our case, the characteristic equation is:\[ r^2 + 11r + 24 = 0 \].By factorizing, we get the roots:\[ (r + 3)(r + 8) = 0 \],which simplifies to roots:
- \( r_1 = -3 \)
- \( r_2 = -8 \)
particular solution
The particular solution of the differential equation is found by applying the initial conditions to the general solution. Given our initial conditions \( y(0) = -1 \) and \( y'(0) = 0 \), we substitute into:\[ y(t) = C_1 e^{-3t} + C_2 e^{-8t} \].First, we apply \( y(0) = -1 \):\[ C_1 + C_2 = -1 \].Next, we find the first derivative:\[ y'(t) = -3C_1 e^{-3t} - 8C_2 e^{-8t} \],and apply \( y'(0) = 0 \):\[ -3C_1 - 8C_2 = 0 \].Solving this system of equations gives us the constants:
- \( C_1 = -\frac{8}{5} \)
- \( C_2 = \frac{3}{5} \)
