Answer to Question #200115 in Differential Equations for Raj Kumar

Question

If d²x/dt² + g(x-a)/b = 0, (a,b,g being positive constants) and x=a' ans dx/dt = 0 when t = 0, show that x = a + (a'-a) cos[(√g)t/b].

Answer 1

given

$\frac{d^2x}{st^2}+\frac{g}{b}(x-a)=0; \space x=a' ,\frac{dx}{dt}=0 \space when \space t=0 \\$

we can write it

$\frac{d^2x}{dt^2}+\frac{g}{b}x-\frac{ag}{b}=0\space \\ \frac{d^2x}{dt^2}+\frac{g}{b}x=\frac{ag}{b}\space \\$ .........[1]

Complementary Function

 Auxiliary equation is:

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

thus the C.F. solution is

$x = e^{0x}(Acos(\sqrt{\frac{g}{b}}t + Bsin(\sqrt{\frac{g}{b}}t))\\ x =(Acos(\sqrt{\frac{g}{b}}t + Bsin(\sqrt{\frac{g}{b}}t))\\$ ........................[CF]

Particular Solution

$\frac{d^2x}{dt^2}+\frac{g}{b}x=f(t)\space \\ where \space f(t)=\frac{ag}{b}\space \\$

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

x=C........[2]

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

Differentiating [2] wrt x twice we get:

x′ =0

x''=0

Substituting these results into the DE [1] we get:

$(0) + \frac{gc}{b} = \frac{ag}{b}\\ C=a$

thus the P.I. solution is

x=a ....................................[PI]

General Solution = CF + PI

$x =(Acos(\sqrt{\frac{g}{b}}t + Bsin(\sqrt{\frac{g}{b}}t))+a\\$

Next we apply the initial conditions:

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

So that:

a'=A cos(0)+B sin(0)+a

A=a'−a

And differentiating the above result wrt t

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

And again applying the initial conditions:

$0=−A\sqrt{\frac{g}{b}}sin0+B\sqrt{\frac{g}{b}}cos0\\ B=0$

And so the complete solution is:

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