Math Problem Statement
Solution
Let's proceed with solving problem (2) from the image.
Problem (2)
\frac{d^2x}{dt^2} + 2 \frac{dy}{dt} = 0, & x|_{t=0} = 0, \frac{dx}{dt}\bigg|_{t=0} = 1; \\ \frac{d^2y}{dt^2} - 2 \frac{dx}{dt} = 0, & y|_{t=0} = 1, \frac{dy}{dt}\bigg|_{t=0} = 1. \end{cases}$$ ### Solution 1. **First Equation:** $$\frac{d^2x}{dt^2} = -2 \frac{dy}{dt}$$ 2. **Second Equation:** $$\frac{d^2y}{dt^2} = 2 \frac{dx}{dt}$$ Take the derivative of the second equation with respect to $$t$$: $$\frac{d^3y}{dt^3} = 2 \frac{d^2x}{dt^2}$$ From the first equation, substitute $$\frac{d^2x}{dt^2} = -2 \frac{dy}{dt}$$: $$\frac{d^3y}{dt^3} = 2(-2 \frac{dy}{dt}) = -4 \frac{dy}{dt}$$ This is a third-order differential equation: $$\frac{d^3y}{dt^3} + 4 \frac{dy}{dt} = 0$$ ### Characteristic Equation Assume a solution of the form $$y = e^{rt}$$: $$r^3 + 4r = 0$$ $$r(r^2 + 4) = 0$$ $$r = 0, \pm 2i$$ Thus, the general solution is: $$y(t) = C_1 + C_2 \cos(2t) + C_3 \sin(2t)$$ ### Initial Conditions for $$y(t)$$ $$y(0) = 1 \implies C_1 + C_2 = 1$$ $$\frac{dy}{dt} = -2C_2 \sin(2t) + 2C_3 \cos(2t)$$ $$\frac{dy}{dt}\bigg|_{t=0} = 1 \implies 2C_3 = 1 \implies C_3 = \frac{1}{2}$$ ### Solution for $$x(t)$$ Since $$\frac{d^2x}{dt^2} = -2 \frac{dy}{dt}$$: $$\frac{d^2x}{dt^2} = -2(-2C_2 \sin(2t) + 2C_3 \cos(2t))$$ $$\frac{d^2x}{dt^2} = 4C_2 \sin(2t) - 4C_3 \cos(2t)$$ Integrate to find $$x(t)$$: $$\frac{dx}{dt} = -2C_2 \cos(2t) - 2C_3 \sin(2t) + C_4$$ Initial condition for $$\frac{dx}{dt}|_{t=0} = 1$$: $$1 = -2C_2 + C_4$$ Integrate again to find $$x(t)$$: $$x(t) = -C_2 \sin(2t) + C_3 \cos(2t) + C_4 t + C_5$$ Initial condition for $$x(0) = 0$$: $$0 = C_5$$ So, the solutions are: $$y(t) = C_1 + C_2 \cos(2t) + \frac{1}{2} \sin(2t)$$ \[ x(t) = -C_2 \sin(2t) + \frac{1}{2} \cos(2t) + CAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Differential Equations
Initial Value Problems
Characteristic Equation
Formulas
Differential equation solutions
Characteristic equation solutions
Theorems
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Suitable Grade Level
College Level
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