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

13. Intro to Differential Equations

Basics of Differential Equations

Calculus

13. Intro to Differential Equations

Basics of Differential Equations

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Multiple Choice

Solve the differential equation $y^{''} + 2 y^{'} = 2 x + 3 - e^{- 2 x}$ using the method of undetermined coefficients. Which of the following is a particular solution?

y_{p} (x) = 2 x + 3 + e^{- 2 x}

y_{p} (x) = a x^{2} + b x + c + d e^{- 2 x}

, where

a = 1

b = 0

c = 0

d = 0

y_{p} (x) = x^{2} + 2 x - e^{- 2 x}

y_{p} (x) = x + \frac{3}{2} - \frac{1}{2} e^{- 2 x}

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Verified step by step guidance

Step 1: Recognize that the given differential equation is a second-order linear non-homogeneous differential equation of the form y'' + 2y' = 2x + 3 - e^{-2x}. The solution will be the sum of the complementary solution (y_c) and a particular solution (y_p).

Step 2: Solve for the complementary solution y_c by solving the associated homogeneous equation y'' + 2y' = 0. The characteristic equation is r^2 + 2r = 0, which factors as r(r + 2) = 0. This gives roots r = 0 and r = -2. Thus, the complementary solution is y_c = C_1 + C_2 e^{-2x}, where C_1 and C_2 are constants.

Step 3: Use the method of undetermined coefficients to find a particular solution y_p. The non-homogeneous term on the right-hand side is 2x + 3 - e^{-2x}, which is a combination of a polynomial (2x + 3) and an exponential term (-e^{-2x}). Assume a particular solution of the form y_p = Ax + B + De^{-2x}, where A, B, and D are constants to be determined.

Step 4: Substitute y_p = Ax + B + De^{-2x} into the original differential equation. Compute the first derivative y_p' = A - 2De^{-2x} and the second derivative y_p'' = 4De^{-2x}. Substitute these into y'' + 2y' = 2x + 3 - e^{-2x} and collect like terms to form an equation for A, B, and D.

Step 5: Solve the resulting system of equations for A, B, and D by equating coefficients of like terms (x, constant, and e^{-2x}) on both sides of the equation. This will yield the values A = 1, B = 3/2, and D = -1/2. Thus, the particular solution is y_p(x) = x + 3/2 - (1/2)e^{-2x}.