Answer to Question #217227 in Differential Equations for Masande

Question

Question1

The conditions in a certain electrical circuit is represented by the following differential equation:

1/50Q’’ +Q’ +8q =50 Cos30t

1.      Determine an expression for q in term of t

2.      Determine an expression for the current  I        (hint I =dq/dt)

3.      Determine the amplitude and the frequency of the steady state current.

 

Question2

Find the general solution of the following DE using the method of undetermined coefficients :

Y’’ +2y’+5y = 34sinxcosx  

Answer 1

Question1

1) $\frac{1}{{50}}q'' + q' + 8q = 50\cos 30t$

$q'' + 50q' + 400q = 2500\cos 30t$

characteristic equation:

${k^2} + 50k + 400 = 0$

$D = 2500 - 1600 = 900$

${k_1} = \frac{{ - 50 - 30}}{2} = - 40$

${k_2} = \frac{{ - 50 + 30}}{2} = - 10$

Then the general solution of the homogeneous equation is

${q_0} = {C_1}{e^{ - 40t}} + {C_2}{e^{ - 10t}}$

We will seek a particular solution in the form

$\widetilde q = A\cos 30t + B\sin 30t \Rightarrow {\widetilde q^\prime } = - 30A\sin 30t + 30B\cos 30t \Rightarrow {\widetilde q^{\prime \prime }} = \\=- 900A\cos 30t - 900B\sin 30t$

Substitute the obtained values ​​into the original equation:

$- 900A\cos 30t - 900B\sin 30t - 1500A\sin 30t + 1500B\cos 30t + 400A\cos 30t + 400B\sin 30t = 2500\cos 30t$

$( - 900A + 1500B + 400A)\cos 30t + ( - 900B - 1500A + 400B)\sin 30t = 2500\cos 30t$

$\left\{ \begin{array}{l} - 500A + 1500B = 2500\\ - 1500A - 500B = 0 \end{array} \right. \Rightarrow$

$\Rightarrow \left\{ {\begin{matrix} {A = - \frac{1}{2}}\\ {B = \frac{3}{2}} \end{matrix}} \right.$

So,

$\widetilde q = - \frac{1}{2}\cos 30t + \frac{3}{2}\sin 30t$

$q(t) = {q_0} + \widetilde q = {C_1}{e^{ - 40t}} + {C_2}{e^{ - 10t}} - \frac{1}{2}\cos 30t + \frac{3}{2}\sin 30t$

Answer: $q(t) = {C_1}{e^{ - 40t}} + {C_2}{e^{ - 10t}} - \frac{1}{2}\cos 30t + \frac{3}{2}\sin 30t$

2)

$I(t) = \frac{{dq}}{{dt}} = {\left( {{C_1}{e^{ - 40t}} + {C_2}{e^{ - 10t}} - \frac{1}{2}\cos 30t + \frac{3}{2}\sin 30t} \right)^\prime } =\\= - 40{C_1}{e^{ - 40t}} - 10{C_2}{e^{ - 10t}} + 15\sin 30t + 45\cos 30t$

Answer: $I(t) = - 40{C_1}{e^{ - 40t}} - 10{C_2}{e^{ - 10t}} + 15\sin 30t + 45\cos 30t$

3) The steady state solution is $\widetilde q = - \frac{1}{2}\cos 30t + \frac{3}{2}\sin 30t = \sqrt {\frac{5}{2}} \sin \left( {30t - arctan\frac{1}{3}} \right)$

So, the amplitude is $A = \sqrt {\frac{5}{2}}$ , the frequency is $f =\frac{ 30}{2\pi}$

Answer: $A = \sqrt {\frac{5}{2}}$ , $f = \frac{30}{2 \pi}$

Question 2

$y'' + 2y' + 5y = 34\sin x\cos x$

$y'' + 2y' + 5y = 17\sin 2x$

characteristic equation:

${k^2} + 2k + 5 = 0$

$D = 4 - 20 = - 16$

${k_1} = \frac{{ - 2 - 4i}}{2} = - 1 - 2i$

${k_2} = \frac{{ - 2 + 4i}}{2} = - 1 + 2i$

Then the general solution of the homogeneous equation is

${y_0} = {e^{ - x}}\left( {{C_1}\cos 2x + {C_2}\sin 2x} \right)$

We will seek a particular solution in the form

$\widetilde y = A\sin 2x + B\cos 2x \Rightarrow {\widetilde y^\prime } = 2A\cos 2x - 2B\sin 2x \Rightarrow {\widetilde y^{\prime \prime }} = - 4A\sin 2x - 4B\cos 2x$

Substitute the obtained values ​​into the original equation:

$- 4A\sin 2x - 4B\cos 2x + 4A\cos 2x - 4B\sin 2x + 5A\sin 2x + 5B\cos 2x = 17\sin 2x$

$(A - 4B)\sin 2x + (4A + B)\cos 2x = 17\sin 2x$

$\left\{ \begin{array}{l} A - 4B = 17\\ 4A + B = 0 \end{array} \right. \Rightarrow A = 1,B = - 4$

So,

$\widetilde y = \sin 2x - 4\cos 2x$

$y = {y_0} + \widetilde y = {e^{ - x}}\left( {{C_1}\cos 2x + {C_2}\sin 2x} \right) + \sin 2x - 4\cos 2x$

Answer: $y = {e^{ - x}}\left( {{C_1}\cos 2x + {C_2}\sin 2x} \right) + \sin 2x - 4\cos 2x$