Determine whether the given function is a solution to the given differential equation. d²e de d₁² 9=5e²t-3e5t -0-+30-5e St dt The function 0-5e2t-3e5t d²e de de ▼a solution to the differential equation2-0+30= -5 est, because when 5e2t-3 e ³t is substituted for 0. is substituted for dt and is substituted for d²e g₁² the two sides of the differential equation equivalent on any intervals of t.
Determine whether the given function is a solution to the given differential equation. d²e de d₁² 9=5e²t-3e5t -0-+30-5e St dt The function 0-5e2t-3e5t d²e de de ▼a solution to the differential equation2-0+30= -5 est, because when 5e2t-3 e ³t is substituted for 0. is substituted for dt and is substituted for d²e g₁² the two sides of the differential equation equivalent on any intervals of 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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Question
![**Determine whether the given function is a solution to the given differential equation.**
Equation: \( \theta = 5 e^{2t} - 3 e^{5t} \)
To verify if \( \theta = 5 e^{2t} - 3 e^{5t} \) is a solution to the differential equation:
\[ \dfrac{d^2 \theta}{dt^2} - \dfrac{d\theta}{dt} + 30\theta = -5 e^{5t} \]
**Step-By-Step Solution**
1. **The function \( \theta = 5 e^{2t} - 3 e^{5t} \)**
To begin, we must compute the first and second derivatives of \( \theta \).
2. **First derivative of \(\theta\)**
\[ \dfrac{d \theta}{dt} = \dfrac{d}{dt} (5 e^{2t} - 3 e^{5t}) \]
\[ \dfrac{d \theta}{dt} = 10 e^{2t} - 15 e^{5t} \]
3. **Second derivative of \(\theta\)**
\[ \dfrac{d^2 \theta}{dt^2} = \dfrac{d}{dt} (10 e^{2t} - 15 e^{5t}) \]
\[ \dfrac{d^2 \theta}{dt^2} = 20 e^{2t} - 75 e^{5t} \]
4. **Substitute \(\theta\), \(\dfrac{d\theta}{dt}\), and \(\dfrac{d^2 \theta}{dt^2}\) into the differential equation**
\[ \dfrac{d^2 \theta}{dt^2} - \dfrac{d\theta}{dt} + 30 \theta = -5 e^{5t} \]
Substituting:
\[ (20 e^{2t} - 75 e^{5t}) - (10 e^{2t} - 15 e^{5t}) + 30 (5 e^{2t} - 3 e^{5t}) \]
Simplifying:
\[ 20 e^{2t} - 75 e](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F13d2fa52-7679-4dbf-8ccf-e0669bc97f10%2F7c07e003-3cb5-4732-b1db-613e5450ae88%2Ft1y1lfi_processed.png&w=3840&q=75)
Transcribed Image Text:**Determine whether the given function is a solution to the given differential equation.**
Equation: \( \theta = 5 e^{2t} - 3 e^{5t} \)
To verify if \( \theta = 5 e^{2t} - 3 e^{5t} \) is a solution to the differential equation:
\[ \dfrac{d^2 \theta}{dt^2} - \dfrac{d\theta}{dt} + 30\theta = -5 e^{5t} \]
**Step-By-Step Solution**
1. **The function \( \theta = 5 e^{2t} - 3 e^{5t} \)**
To begin, we must compute the first and second derivatives of \( \theta \).
2. **First derivative of \(\theta\)**
\[ \dfrac{d \theta}{dt} = \dfrac{d}{dt} (5 e^{2t} - 3 e^{5t}) \]
\[ \dfrac{d \theta}{dt} = 10 e^{2t} - 15 e^{5t} \]
3. **Second derivative of \(\theta\)**
\[ \dfrac{d^2 \theta}{dt^2} = \dfrac{d}{dt} (10 e^{2t} - 15 e^{5t}) \]
\[ \dfrac{d^2 \theta}{dt^2} = 20 e^{2t} - 75 e^{5t} \]
4. **Substitute \(\theta\), \(\dfrac{d\theta}{dt}\), and \(\dfrac{d^2 \theta}{dt^2}\) into the differential equation**
\[ \dfrac{d^2 \theta}{dt^2} - \dfrac{d\theta}{dt} + 30 \theta = -5 e^{5t} \]
Substituting:
\[ (20 e^{2t} - 75 e^{5t}) - (10 e^{2t} - 15 e^{5t}) + 30 (5 e^{2t} - 3 e^{5t}) \]
Simplifying:
\[ 20 e^{2t} - 75 e
![## Differential Equations: Existence, Uniqueness, and Solutions
### Consideration of Differential Equation
Consider the differential equation:
\[ 25 \frac{dy}{dx} - 36x = 0, \]
which has a one-parameter family of implicit solutions \( 36x^2 - 25y^2 = C \) for every constant \( C \). Complete parts (a) through (c) below.
### (a) Existence and Uniqueness of Solution
**Question:**
Does the Existence and Uniqueness of Solution Theorem imply the existence of a unique solution to the differential equation that satisfies \( y(x_0) = 0 \)? Select the correct choice below and, if necessary, fill in the answer box within your choice.
- **A.** The theorem implies the existence of a unique solution for all values of \( x_0 \).
- **B.** The theorem implies the existence of a unique solution only for values of \( x_0 \) in the interval \( \_\_\_ \).
- *(Type your answer in interval notation.)*
- **C.** The theorem does not imply the existence of a unique solution for any values of \( x_0 \).
### (b) Neighborhood Solution Non-existence
**Question:**
Show that when \( x_0 \ne 0 \), the differential equation can't possibly have a solution in a neighborhood of \( x = x_0 \) that satisfies \( y(x_0) = 0 \).
Substituting the initial condition into the differential equation gives \( 25 \frac{dy}{dx} - 36x = 0 \). This equation can be simplified to \( \_\_\_ \) which \( \_\_\_ \) the given condition that \( \_\_\_ \).
### (c) Distinct Solutions when y(0) = 0
**Question:**
Show that there are two distinct solutions to the differential equation satisfying \( y(0) = 0 \). Refer to the graph showing several members of the solution family \( 36x^2 - 25y^2 = C \).
The initial condition \( y(0) = 0 \) leads to the solution(s) with \( C = \_\_\_ \), which gives two explicit solutions \( y = \_\_\_ \), that satisfy both the differential equation and](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F13d2fa52-7679-4dbf-8ccf-e0669bc97f10%2F7c07e003-3cb5-4732-b1db-613e5450ae88%2Fcylizwi_processed.png&w=3840&q=75)
Transcribed Image Text:## Differential Equations: Existence, Uniqueness, and Solutions
### Consideration of Differential Equation
Consider the differential equation:
\[ 25 \frac{dy}{dx} - 36x = 0, \]
which has a one-parameter family of implicit solutions \( 36x^2 - 25y^2 = C \) for every constant \( C \). Complete parts (a) through (c) below.
### (a) Existence and Uniqueness of Solution
**Question:**
Does the Existence and Uniqueness of Solution Theorem imply the existence of a unique solution to the differential equation that satisfies \( y(x_0) = 0 \)? Select the correct choice below and, if necessary, fill in the answer box within your choice.
- **A.** The theorem implies the existence of a unique solution for all values of \( x_0 \).
- **B.** The theorem implies the existence of a unique solution only for values of \( x_0 \) in the interval \( \_\_\_ \).
- *(Type your answer in interval notation.)*
- **C.** The theorem does not imply the existence of a unique solution for any values of \( x_0 \).
### (b) Neighborhood Solution Non-existence
**Question:**
Show that when \( x_0 \ne 0 \), the differential equation can't possibly have a solution in a neighborhood of \( x = x_0 \) that satisfies \( y(x_0) = 0 \).
Substituting the initial condition into the differential equation gives \( 25 \frac{dy}{dx} - 36x = 0 \). This equation can be simplified to \( \_\_\_ \) which \( \_\_\_ \) the given condition that \( \_\_\_ \).
### (c) Distinct Solutions when y(0) = 0
**Question:**
Show that there are two distinct solutions to the differential equation satisfying \( y(0) = 0 \). Refer to the graph showing several members of the solution family \( 36x^2 - 25y^2 = C \).
The initial condition \( y(0) = 0 \) leads to the solution(s) with \( C = \_\_\_ \), which gives two explicit solutions \( y = \_\_\_ \), that satisfy both the differential equation and
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