Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

0. Functions

Introduction to Functions

Calculus

0. Functions

Introduction to Functions

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Multiple Choice

Given the boundary-value problem $y^{″} - 2 y^{'} + 2 y = 2 x - 2$ , with $y (0) = 0$ and $y (π) = π$ , which of the following is the correct solution for $y (x)$ ?

y (x) = x + 1 + e^{x} (C \cos x + D \sin x)

y (x) = x + 1 + e^{x} (\cos x + \sin x)

y (x) = x + 1 + e^{x} (\cos x - \sin x)

y (x) = x + e^{x} (A \cos x + B \sin x)

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Verified step by step guidance

Step 1: Start by analyzing the given differential equation y'' - 2y' + 2y = 2x - 2. This is a second-order linear differential equation with constant coefficients. The solution will consist of two parts: the complementary solution (y_c) and the particular solution (y_p).

Step 2: Solve for the complementary solution y_c by setting the right-hand side of the equation to zero: y'' - 2y' + 2y = 0. Assume a solution of the form y = e^{rx}, substitute it into the equation, and solve the characteristic equation r^2 - 2r + 2 = 0.

Step 3: Solve the characteristic equation r^2 - 2r + 2 = 0 using the quadratic formula r = (-b ± √(b^2 - 4ac)) / 2a. Here, a = 1, b = -2, and c = 2. Compute the discriminant (b^2 - 4ac) to determine the nature of the roots. Since the discriminant is negative, the roots will be complex: r = 1 ± i.

Step 4: Write the complementary solution y_c based on the complex roots r = 1 ± i. The general form for complex roots is y_c = e^{αx} (C \, \cos(βx) + D \, \sin(βx)), where α = 1 and β = 1. Thus, y_c = e^{x} (C \, \cos x + D \, \sin x).

Step 5: Find the particular solution y_p by guessing a solution based on the form of the non-homogeneous term (2x - 2). Since the right-hand side is a polynomial, guess y_p = Ax + B, substitute it into the original equation, and solve for A and B. Combine y_c and y_p to form the general solution y(x). Finally, apply the boundary conditions y(0) = 0 and y(π) = π to solve for the constants C and D.