# If d^(2)x/dt^(2)+g/b(x-a)=0,(a,b,g being positive constants) and x=a' and dx/dt=0 when t=0,show that x=a+(a'-a)cos{√(g)/(b)t}?

Jul 3, 2018

$x \left(t\right) = \left(a ' - a\right) \cos \left(\sqrt{\frac{g}{b}} t\right) + a \setminus \setminus \setminus \setminus$

#### Explanation:

We have:

$\frac{{d}^{2} x}{{\mathrm{dt}}^{2}} + \frac{g}{b} \left(x - a\right) = 0$ with $x = a ' , \frac{\mathrm{dx}}{\mathrm{dt}} = 0$ when $t = 0$

We can write the equation as:

$\frac{{d}^{2} x}{{\mathrm{dt}}^{2}} + \frac{g}{b} x - \frac{a g}{b} = 0 \iff \frac{{d}^{2} x}{{\mathrm{dt}}^{2}} + \frac{g}{b} x = \frac{a g}{b} \ldots . . \left[A\right]$

This is a second order non-Homogeneous Differentiation Equation. The standard approach is to find a solution, ${y}_{c}$ of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, ${y}_{p}$ of the non-homogeneous equation.

Complementary Function

The homogeneous equation associated with [A] is

$x ' ' - 0 x ' + \frac{g}{b} = 0$

And it's associated Auxiliary equation is:

${m}^{2} + \frac{g}{b} = 0 \implies {m}^{2} = - \frac{g}{b}$

And as we are given that $a , b , g > 0$ then we have pure imaginary roots:

$m = \pm \sqrt{\frac{g}{b}}$

The roots of the auxiliary equation determine parts of the solution, which if linearly independent then the superposition of the solutions form the full general solution.

• Real distinct roots $m = \alpha , \beta , \ldots$ will yield linearly independent solutions of the form ${x}_{1} = A {e}^{\alpha t}$, ${x}_{2} = B {e}^{\beta t}$, ...
• Real repeated roots $m = \alpha$, will yield a solution of the form $x = \left(A t + B\right) {e}^{\alpha t}$ where the polynomial has the same degree as the repeat.
• Complex roots (which must occur as conjugate pairs) $m = p \pm q i$ will yield a pairs linearly independent solutions of the form $x = {e}^{p t} \left(A \cos \left(q x\right) + B \sin \left(q x\right)\right)$

Thus the solution of the homogeneous equation [A] is:

 x = e^(0x)(Acos(sqrt(g/b)t + Bsin(sqrt(g/b)t))
 \ \ = Acos(sqrt(g/b)t + Bsin(sqrt(g/b)t)

Particular Solution

In order to find a particular solution of the non-homogeneous equation:

$\frac{{d}^{2} x}{{\mathrm{dt}}^{2}} + \frac{g}{b} x = f \left(t\right) \setminus \setminus$ with $f \left(t\right) = \frac{a g}{b}$

So, we should probably look for a solution of the form:

$x = C$ ..... [B]

Where the constant $C$ is to be determined by direct substitution and comparison:

Differentiating [B] wrt $x$ twice we get:

$x ' \setminus \setminus = 0$
$x ' ' = 0$

Substituting these results into the DE [A] we get:

$\left(0\right) + \frac{g}{b} C = \frac{a g}{b} \implies C = a$

And so we form the Particular solution:

${x}_{p} = a$

General Solution

Which then leads to the GS of [A}

$x \left(t\right) = {x}_{c} + {x}_{p}$
 \ \ \ \ \ \ \ = Acos(sqrt(g/b)t + Bsin(sqrt(g/b)t) + a

Next we apply the initial conditions:

$x = a ' , \frac{\mathrm{dx}}{\mathrm{dt}} = 0$ when $t = 0$

So that:

$a ' = A \cos 0 + B \sin 0 + a \implies A = a ' - a$

And differentiating the above result wrt $t$

$x ' \left(t\right) = - A \sqrt{\frac{g}{b}} \sin \left(\sqrt{\frac{g}{b}} t\right) + B \sqrt{\frac{g}{b}} \cos \left(\sqrt{\frac{g}{b}} t\right)$

And again applying the initial conditions:

$0 = - A \sqrt{\frac{g}{b}} \sin 0 + B \sqrt{\frac{g}{b}} \cos 0 \implies B = 0$

And so the complete solution is:

$x \left(t\right) = \left(a ' - a\right) \cos \left(\sqrt{\frac{g}{b}} t\right) + a \setminus \setminus \setminus \setminus$ QED

Jul 3, 2018

The Problem:

• $x ' ' + \frac{g}{b} \left(x - a\right) = 0 q \quad q \quad a , b , g > 0$

• ${x}_{o} = a ' q \quad x {'}_{o} = 0$

Show that: $x = a + \left(a ' - a\right) \cos \sqrt{\frac{g}{b}} t$

Substitute :

• $\boldsymbol{\xi = x - a} q \quad \xi ' = x ' q \quad \xi ' ' = x ' '$

• ${\xi}_{o} = a ' - a q \quad \xi {'}_{o} = 0$

This makes it homogeneous , which means we can race to a solution.

• $\xi ' ' + {\omega}^{2} \xi = 0 q \quad q \quad {\omega}^{2} = \frac{g}{b}$

This is the Harmonic Oscillator, which can be solved in any number of ways, but always with well-known solution:

• $\left\{\begin{matrix}\xi = A \cos \omega t + B \sin \omega t \\ \xi ' = - \omega A \sin \omega t + \omega B \cos \omega t\end{matrix}\right.$

Applying the IV's at $t = 0$:

• $\left\{\begin{matrix}a ' - a = A \\ 0 = \omega B\end{matrix}\right.$

So:

$\xi = \left(a ' - a\right) \cos \sqrt{\frac{g}{b}} t$

Therefore:

$x = a + \left(a ' - a\right) \cos \sqrt{\frac{g}{b}} t$