Question

If d2x/dt2+g(x-a)/b=0, (a, b, gbeing positive constants) and x = a′ and dx/dt=0 when t=0, show that x=a+(a'-a)cos {g underoot t/b}

Answer 1

Answer on Question #76273 – Math – Differential Equations

Question

If $\frac{\mathrm{d}^2\mathbf{x}}{\mathrm{d}t^2} + \frac{\mathbf{g}(\mathbf{x} - \mathbf{a})}{\mathbf{b}} = 0$, (a, b, g being positive constants) and $\mathbf{x} = \mathbf{a}'$ and $\frac{dx}{dt} = 0$ when $t = 0$, show that $\mathbf{x} = \mathbf{a} + (\mathbf{a}' - \mathbf{a})\cos \left\{\sqrt{\left(\frac{\partial}{\partial t}\right)} t\right\}$

Solution

Given, differential equation:

$$\frac {\mathrm {d} ^ {2} \mathbf {x}}{\mathrm {d t} ^ {2}} + \frac {\mathbf {g} (\mathbf {x} - \mathbf {a})}{\mathbf {b}} = 0$$

Boundary conditions are given by, $\mathbf{x} = \mathbf{a}'$ and $\frac{dx}{dt} = 0$ at $t = 0$.

Now, solution of the equation (1) is

$$x - a = A \cos \left\{\sqrt {\left(\frac {\partial}{b}\right)} t \right\} + B \sin \left\{\sqrt {\left(\frac {\partial}{b}\right)} t \right\}$$

At $t = 0$, $x = a'$ then from equation (2) we get,

$$A = a ^ {\prime} - a$$

Now, take the derivative of equation (2) and we get,

$$\frac {d x}{d t} = \left[ - A \sin \left\{\sqrt {\left(\frac {\partial}{b}\right)} t \right\} + B \cos \left\{\sqrt {\left(\frac {\partial}{b}\right)} t \right\} \right] \left\{\sqrt {\left(\frac {\partial}{b}\right)} \right\}$$

At, $t = 0$, $\frac{dx}{dt} = 0$ then from equation (3) we get,

$$B = 0.$$

Now, put the value of A and B in equation (2) and we get,

$$x - a = \left(a ^ {\prime} - a\right) \cos \left\{\sqrt {\left(\frac {\partial}{b}\right)} t \right\}$$

or,

$$x = a + \left(a ^ {\prime} - a\right) \cos \left\{\sqrt {\left(\frac {\partial}{b}\right)} t \right\}$$

Answer: $x = a + (a' - a) \cos \left\{\sqrt{\left(\frac{\partial}{b}\right)} t\right\}$.

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