Question

Solve the initial-value problem dy/dx = sec² t - sin t, y(pi/4) = 1

A. y'=cosx+sinx y=sinx-cosx+C 1=sin(pi)-cos(pi)+C C=0 y=sinx-cosx
B. y'=cosx+sinx y=sinx-cosx+C 1=sin(pi)-cos(pi)+C C=0 y=cosx-cosx
C. y'=cosx+sinx y=cosx-cosx+C 1=sin(pi)-cos(pi)+C C=0 y=sinx-cosx
D. y'=cosx+sinx y=sinx-sinx+C 1=sin(pi)-cos(pi)+C C=0 y=sinx-cosx

Answer 1

To solve the initial-value problem dy/dx = sec² t - sin t, with y(π/4) = 1, we integrate the equation and solve for the constant C, with the final solution being y = tan t + cos t + 1 - (√2/2 +1).

Step-by-step explanation:

To solve the initial-value problem dy/dx = sec² t - sin t, y(π/4) = 1, we must integrate the right side of the equation to find y. Upon doing so, we get:

1. ∫ydx = ∫(sec² t - sin t)dt
2. y = tan t + cos t + C
3. Since y(π/4) = 1, we substitute t = π/4 and y = 1 to find C.
4. 1 = tan(π/4) + cos(π/4) + C => 1 = 1 + √2/2 + C
5. C = 1 - (√2/2 +1)
6. Therefore, y = tan t + cos t + 1 - (√2/2 +1)

Note that the variable 't' is used instead of 'x' in the given differential equation, which suggests it may be a function of another variable. It is important to remain consistent with variable names when solving differential equations.