Question

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

13 Feb 2018
Answer :
Word Count : 441
We are tasked with solving the differential equation and showing that \( x(t) = a + (a' - a) \cos\left( \frac{\sqrt{g}}{b} t \right) \). The equation we are dealing with is: \[ \frac{d^2x}{dt^2} + \frac{g}{b}(x - a) = 0, \] with initial conditions \( x(0) = a' \) and \( \frac{dx}{dt}(0) = 0 \). ### Step 1: Simplifying the equation We begin by simplifying the differential equation: \[ \frac{d^2x}{dt^2} = -\frac{g}{b}(x - a). \] This is a second-order linear differential equation. We can rewrite it as: \[ \frac{d^2x}{dt^2} + \frac{g}{b}x = \frac{g}{b}a. \] ### Step 2: Homogeneous equation solution First, we solve the homogeneous equation (i.e.,
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If \frac{d^2x}{dt^2}+\frac{g}{b} (x-a) = 0, (a,b,g) being positive constants) and x = a′ and  \frac{dx}{dt} = 0when t = 0, show that  x = a + (a' - a )cos \left \{ \frac{\sqrt{g}}{b}t \right \}
 

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have the same normal form.
iii). Equation cos(x + y) p + sin(x + y)q = z2 + z is a quasi-linear equation.

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v). Every solution of the ordinary differential equation (D2 + 1)2 y = 0 is bounded on [0, ∞] .

Solve the following PDEs:

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