Question

The differential equation y′′=0 has one of the following two parameter families as its general solution: yyyy=C1ex+C2e−x=C1cos(x)+C2sin(x)=C1tan(x)+C2sec(x)=C1+C2x Find the solution such that y(0)=2 and y′(0)=10. y(x)=

Answer 1

Answer:

y_c = 2 + 10*x

Step-by-step explanation:

Given:

                                                y'' = 0

Find:

- The solution to ODE such that y(0) = 2, y'(0) = 10

Solution:

- Assuming a solution y = Ce^(mt)

So,                                y' = C*me^(mt)

                                    y'' = C*m^2e^(mt)

- Back substitute into given ODE, we get:

                                    y'' = C*m^2e^(mt) = 0

                                    e^(mt) can not be equal to zero

- Hence,                       m^2 = 0

                                     m = 0 , 0 - (repeated roots)

- The complimentary function for repeated roots is:

                                    y_c = (C1 + C2*x)*e^(m*t)

                                    y_c = C1 + C2*x  

- Evaluate @ y(0) = 2

                                    2 = C1 + C2*0

                                    C1 = 2

-Evaluate @ y'(0) = 10

                                    y'(t) = C2 = 10

Hence,                         y_c = 2 + 10*x