Question

Solve the initial value problems in Exercises 71–90.y⁽⁴⁾ = −sin t + cos t;y′′′(0) =7, y′′(0) = y′(0) = −1, y(0) = 0

Accepted Answer

Identify the given differential equation and initial conditions: $$y^{(4)}$$ = -\sin t + \cos t$$ with y$$'''(0) = 7$$, y$$''(0) = -1$$, y$$'(0) = -1$$, and y(0) = 0$. Integrate the differential equation step-by-step to find y(t$$)$$. Since y$$^{(4)}$$ is the fourth derivative of y$, integrate four times, introducing constants of integration C_1$, C_2$, C_3$, and C_4$ at each step. First integration: Integrate y$$^{(4)}$$ = -\sin t + \cos t$$ once to find $$y'''(t)$$:

$$y'''(t) = \int (-\sin t + \cos t) \, dt = \cos t + \sin t + C_1$$ Second integration: Integrate $$y'''(t) to $$find $$y''(t)$$:

$$y''(t) = \int (\cos t + \sin t + C_1) \, dt = \sin t - \cos t + C_1 t + C_2$$ Third integration: Integrate $$y''(t) to $$find $$y'(t)$$:

$$y'(t) = \int (\sin t - \cos t + C_1 t + C_2) \, dt = -\cos t - \sin t + \frac{C_1 t^2}{2} + C_2 t + C_3$$

Fourth integration: Integrate $$y'(t) to $$find $$y(t)$$:

$$y(t) = \int \left(-\cos t - \sin t + \frac{C_1 t^2}{2} + C_2 t + C_3\right) dt = -\sin t + \cos t + \frac{C_1 t^3}{6} + \frac{C_2 t^2}{2} + C_3 t + C_4$$ Use the initial conditions to form a system of equations for $$C_1$$, $$C_2$$, $$C_3$$, and $$C_4 by $$substituting $$t=0$$ into $$y(0)$$, $$y'(0)$$, $$y''(0)$$, and $$y'''(0)$$:

- From $$y(0) = 0$$,
- From $$y'(0) = -1$$,
- From $$y''(0) = -1$$,
- From $$y'''(0) = 7.

$$Solve this system to find the constants.

Solve the initial value problems in Exercises 71–90.
y⁽⁴⁾ = −sin t + cos t;
y′′′(0) =7, y′′(0) = y′(0) = −1, y(0) = 0

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Key Concepts

Solving Higher-Order Differential Equations

Initial Conditions and Their Role

Nonhomogeneous Differential Equations and Particular Solutions