Chapter 5: Problem 1
Man bestimme die allgemeine Lösung folgender Differentialgleichungen a) \(y^{\prime \prime}+7 y^{\prime}-8 y=0\); b) \(y^{\prime \prime}+8 y^{\prime}+16 y=0\) c) \(y^{\prime \prime}+8 y^{\prime}+25 y=0\); d) \(y^{\prime \prime}+8 y^{\prime}=0\); e) \(y^{\prime \prime}+8 y=0\); f) \(y^{\prime \prime}-4 y^{\prime}+4 y=0\).
Short Answer
Step by step solution
Determine the Characteristic Equation for Part a
Solve the Characteristic Equation for Part a
Write the General Solution for Part a
Determine the Characteristic Equation for Part b
Solve the Characteristic Equation for Part b
Write the General Solution for Part b
Determine the Characteristic Equation for Part c
Solve the Characteristic Equation for Part c
Write the General Solution for Part c
Determine the Characteristic Equation for Part d
Solve the Characteristic Equation for Part d
Write the General Solution for Part d
Determine the Characteristic Equation for Part e
Solve the Characteristic Equation for Part e
Write the General Solution for Part e
Determine the Characteristic Equation for Part f
Solve the Characteristic Equation for Part f
Write the General Solution for Part f
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
\[ y'' + ay' + by = 0 \]To find the characteristic equation, we assume a solution of the form \( y = e^{rt} \). By substituting this form into the differential equation, we transform the problem into a simpler algebraic form called the characteristic equation:
\[ ar^2 + br + c = 0 \]The roots of this polynomial equation, often solved using the quadratic formula, reveal critical information:
- Two distinct real roots
- A double, or repeated, real root
- A pair of complex roots
General Solution
- If the roots \( r_1 \) and \( r_2 \) are distinct and real, the general solution is:
\[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \] where \( C_1 \) and \( C_2 \) are arbitrary constants. - If there is a double root \( r \), the solution is of the form:
\[ y(t) = (C_1 + C_2 t) e^{r t} \] This extra \( t \) term represents the impact of the repeated root. - If complex roots \( r = ext{a} \pm ext{bi} \) are present, the general solution is:
\[ y(t) = e^{at} (C_1 \cos(bt) + C_2 \sin(bt)) \] which beautifully combines exponential decay or growth with oscillatory behavior.
Complex Roots
- Start with the solution structure:
\[ y(t) = e^{at} (C_1 \cos(bt) + C_2 \sin(bt)) \] - This form is due to Euler's formula, which relates complex exponentials to trigonometric functions.
- The exponential part, \( e^{at} \), represents the decay or growth function, while the sine and cosine terms introduce oscillation.
Double Roots
- When a double root \( r \) is present, the general solution is:
\[ y(t) = (C_1 + C_2 t) e^{r t} \] - The term \( C_2 t \) takes into account the multiplicity (repeated nature) of the root, ensuring all solutions are accounted for.