Find the general solution of the given differential equation. 4y" + 20y' + 25y = 0 y(t) =
Find the general solution of the given differential equation. 4y" + 20y' + 25y = 0 y(t) =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
Find the general solution of the given differential equation.
\[ 4y'' + 20y' + 25y = 0 \]
\( y(t) = \) [Your Solution Here]
**Explanation:**
This is a second-order linear homogeneous differential equation with constant coefficients. To solve this type of equation, follow these steps:
1. **Characteristic Equation:**
- Assume a solution of the form \( y(t) = e^{rt} \).
- Substitute \( y(t) = e^{rt} \), \( y'(t) = re^{rt} \), and \( y''(t) = r^2e^{rt} \) into the differential equation.
- You get the characteristic equation: \( 4r^2 + 20r + 25 = 0 \).
2. **Solve the Characteristic Equation:**
- Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) if necessary.
- In this equation, \( a = 4 \), \( b = 20 \), and \( c = 25 \).
3. **Find the Roots:**
- Depending on the discriminant (\( b^2 - 4ac \)), you may have:
- Two distinct real roots
- One repeated real root
- Two complex roots
4. **General Solution:**
- If roots are real and distinct: \( y(t) = c_1e^{r_1t} + c_2e^{r_2t} \)
- If roots are real and repeated: \( y(t) = (c_1 + c_2t)e^{rt} \)
- If roots are complex: \( y(t) = e^{\alpha t}(c_1\cos(\beta t) + c_2\sin(\beta t)) \), where roots are \( \alpha \pm \beta i \).
**Note:** Ensure each step is clearly explained to help understand the process of solving differential equations with constant coefficients.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad35cd49-88f1-4759-9e4c-232c0c792d3e%2F31af797b-4f79-4a5d-9db8-85e15d3e7d91%2Fsfoc5jy_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the general solution of the given differential equation.
\[ 4y'' + 20y' + 25y = 0 \]
\( y(t) = \) [Your Solution Here]
**Explanation:**
This is a second-order linear homogeneous differential equation with constant coefficients. To solve this type of equation, follow these steps:
1. **Characteristic Equation:**
- Assume a solution of the form \( y(t) = e^{rt} \).
- Substitute \( y(t) = e^{rt} \), \( y'(t) = re^{rt} \), and \( y''(t) = r^2e^{rt} \) into the differential equation.
- You get the characteristic equation: \( 4r^2 + 20r + 25 = 0 \).
2. **Solve the Characteristic Equation:**
- Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) if necessary.
- In this equation, \( a = 4 \), \( b = 20 \), and \( c = 25 \).
3. **Find the Roots:**
- Depending on the discriminant (\( b^2 - 4ac \)), you may have:
- Two distinct real roots
- One repeated real root
- Two complex roots
4. **General Solution:**
- If roots are real and distinct: \( y(t) = c_1e^{r_1t} + c_2e^{r_2t} \)
- If roots are real and repeated: \( y(t) = (c_1 + c_2t)e^{rt} \)
- If roots are complex: \( y(t) = e^{\alpha t}(c_1\cos(\beta t) + c_2\sin(\beta t)) \), where roots are \( \alpha \pm \beta i \).
**Note:** Ensure each step is clearly explained to help understand the process of solving differential equations with constant coefficients.
![**Problem Statement:**
Find the general solution of the given differential equation.
\[ 25y'' - 25y' - 6y = 0 \]
\( y(t) = \) [ ]
**Explanation:**
This is a second-order linear homogeneous differential equation. To solve it, we look for solutions of the form \( y(t) = e^{rt} \), where \( r \) is a constant to be determined. Substituting \( y(t) = e^{rt} \) into the differential equation gives a characteristic equation, which can be solved for \( r \). The roots of this equation determine the form of the general solution. Depending on whether the roots are real and distinct, repeated, or complex conjugates, the general solution will take different forms.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad35cd49-88f1-4759-9e4c-232c0c792d3e%2F31af797b-4f79-4a5d-9db8-85e15d3e7d91%2Fwba6uf2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the general solution of the given differential equation.
\[ 25y'' - 25y' - 6y = 0 \]
\( y(t) = \) [ ]
**Explanation:**
This is a second-order linear homogeneous differential equation. To solve it, we look for solutions of the form \( y(t) = e^{rt} \), where \( r \) is a constant to be determined. Substituting \( y(t) = e^{rt} \) into the differential equation gives a characteristic equation, which can be solved for \( r \). The roots of this equation determine the form of the general solution. Depending on whether the roots are real and distinct, repeated, or complex conjugates, the general solution will take different forms.
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