Linear Simultaneous Equations

$\frac{d^2x}{dt^2} - 4\frac{dy}{dx} + 4x = y}$

$\frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 4y = 25x + 16\,e^t$

The equations may be written as:

Eliminating y from equations (1) and (2)

$(D^2 + 4D + 4)(D^2 - 4D + 4)\,x = 25x + 16\,e^t$ i.e. $(D^2 + 4)^2 - 16D^2 - 25\,x = 16\,e^t$

$\therefore\;\;\;\;(D^4 - 8D^2 - 9) = 16\,e^t$

$\therefore\;\;\;\;(D^2 - 9)(D^2 + 1) = 16\,e^t$

$\therefore\;\;\;x = A\,e^{3t} + B\,e^{-3t} + E\,cos\,t + F\,sin\,t + \frac{1}{(D^2\,-\,9)(D^2\,+\,1)}\times16\,e^t$

$\mathbf{Thus\;\;\;\;x = A\,e^{3t} + B\,e^{-3t} + E\,cos\,t + F\,sin\,t - e^t}$

From equation (1) $y = (D^2 - 4D + 4)[A\,e^{3t} + B\,e^{-3t} + E\,cos\,t + F\,sin\,t - e^t]$ Therefore $\matbhf{y = A\,e^{3t} + 25\,B\,e^{-3t} + E\,(3\,cos\,t + 4\,sin\,t) + F\,(3\,sin\,t\;-\,4\,cos\,t) - e^t}$

Solve the following Equations subject:

$2\frac{dx}{dt} + \frac{dy}{dt} + 90\,x = 45$

$\frac{dx}{dt} + 2\frac{dy}{dt} + 240\,y = 0$

Give that $x = y = 0$ at $t = 0$

Show also that $y$ is negative for all values of $t$ and that its minimum value is :

$y\,(min.) = -\,\frac{1}{12}\left(\frac{2}{9} \right)^{\frac{2}{7}}$

The equations may be written as:

Multiply equation (3) by $\displaystyle (2D + 240)$ Equation (4) by D and then subtract.

$\displaystyle \left[(2D + 90)(2D + 240) - D^2 \right]x = (2D + 240)\,45$ Therefore $(D^2 + 220\,D + 7200)x = 3600$ Or $(D + 180)(D + 40)\,x = 3600$ Therefore $x = A\,e^{-180\,t} + B\,e^{-40\,t} + \frac{1}{2}$ But when $t = 0$ $x = 0$ and so $B = (A + 1/2)$

From Equation (3)

$Dy = 45 - (2D + 90)\,x$

$= 45 - (2D + 90)\,\left[A\,e^{-180\,t} - (A + \frac{1}{2})\,e^{-40\,t} + \frac{1}{2} \right]$ Therefore $Dy = 45 - \left[- 270\,A\,e^{-180\,t} - 10(A + \frac{1}{2})\,e^{-40\,t} + 45 \right]$ Thus $y = \frac{270}{180}\timesA\,e^{-180\,t} - \frac{10}{40}\;(A + \frac{1}{2})\,e^{-40\,t} + C$

Putting $y = 0$ when $t = 0$

$0 = -\,\frac{3}{2}A - \frac{1}{4}(A + \frac{1}{2}) + C$ therefore $C = \frac{7}{4}A + \frac{1}{8}$ Thus $A = -\,\frac{1}{14}$ Therefore $C = \frac{7}{4}\,A + \frac{1}{8}$

The Solutions are :

$x = A\,e^{-180\,t} - (A + \frac{1}{2})\,e^{-40\,t} + \frac{1}{2}$

$y = -\,\frac{3}{2}\,A\,e^{-180\,t} - \frac{1}{4}(A + \frac{1}{2})\,e^{-40\,t} + \frac{7}{4}A + \frac{1}{8}$

Substitute these values into equation (17)

$-\,180\,A\,e^{-180\,t} + 40(A + \frac{1}{2})\,e^{-40\,t} + 180\;A\,e^{-180\,t} - 40(A + \frac{1}{2})\,e^{-40\,t} + 240\;\left(\frac{7}{4}\,A + \frac{1}{8} \right) = 0}$ Therefore $\frac{7}{4}\,A + \frac{1}{8} = 0$

Substituting these values into equations (29) and (30), the required solutions are:

$x = -\,\frac{1}{14}\,e^{-180\,t} - \frac{3}{7}\,e^{-40\,t}\,+\;\frac{1}{2}$

$y = \frac{3}{28}\,e^{-180\,t} - \frac{3}{28}\,e^{-40\,t}$

Hence $y$ is negative for all values of $t > 0$

To find the minimum value of $y Dy = 0$

$Dy = \frac{3}{28}\left( -180\,e^{-180\,t} + 40\,e^{-40\,t}\;\right) = 0$ i.e. $e^{180\,t} = \frac{9}{2}$ Or $e^{-20} = \left( \frac{2}{9} \right)^{\frac{1}{7}}$

Therefore $\therefore\;\;\;\;y = \frac{3}{8}\left(\frac{2}{9} \right)^{\frac{9}{7}} = -\,\frac{3}{8}\left(\frac{2}{9}} \right)^{\frac{9}{7}} = -\,\frac{1}{12}\left(\frac{2}{9} \right)^{\frac{2}{7}}$

Apply the Laplace Transform and find $X(s)$ and $Y(s)$ :

$2\frac{dx}{dt} + \frac{dy}{dt} + 90\,x = 45$

$\frac{dx}{dt} + 2\frac{dy}{dt} + 240\,y = 0$

Give that $x = y = 0$ at $t = 0$

The equations may be written as:

$2\mathhf{L\{\frac{dx}{dt}\}} + \mathhf{L\{\frac{dy}{dt}\}} + \mathhf{L\{90\,x\}} = \mathhf{L\{45\}}$

$\mathhf{L\{\frac{dx}{dt}\}} + 2\mathhf{L\{\frac{dy}{dt}\}} + \mathhf{L\{240\,y\}} = 0$ Hence $2(sX(s)-x(0)) + (sY(s)-y(0)) + 90\,X(s) = \frac{45}{s}$

$(sX(s)-x(0)) + 2(sY(s)-y(0)) + 240\,Y(s) = 0$ Therefore

Then we multiply the equation (5) with $(s+120)$ :

We subtract equation (6) from equation (7) : $((2s+90)(s+120)-s)X(s)=(s+120)\frac{45}{s}$

Therefore $X(s)=\frac{45(s+120)}{s(2s^2+329s+10800)}$ From equation (6) we get : $Y(s)=\frac{2s+240}{s}X(s)$

Solve the following simultaneous equations.

$\frac{dx}{dt} + 2x + y = 0$

$\frac{dy}{dt} + x + 2y = 0$

Given that $x = 1$ and $y = 0$ when $t = 0$

Re-writing using the Laplace transform :

$sX(s)-x(0)+2X(s)+Y(s)=0$ $sY(s)-y(0)+X(s)+2Y(s)=0$

Hence

If we multiply the equation (9) with (s+2)

If we subtract from equation (10) the equation (8) $(s^2+4s+3)Y(s)=-1$ Hence $Y(s)=\frac{-1}{s^2+4s+3}$ Replacing $Y(s)$ in the equation (9) $X(s)=\frac{s+2}{s^2+4s+3}$

Applying the inverse Laplace transform $x(t)=\frac{1}{2}(e^{-t}+e^{-3t})$ $y(t)=\frac{1}{2}(-e^{-t}+e^{-3t})$

Find $x(t)$ and $y(t)$ : $\frac{dx}{dt}+2\frac{dy}{dt}+x-y=5sin(t)$ $2\frac{dx}{dt}+3\frac{dy}{dy}+x-y=e^t$ Initial conditions $x(0)=0$ and $y(0)=0$

The equations may be written as : $(s+1)X(s)+(2s-1)Y(s)=\frac{5}{s^2+1}$ $(2s+1)X(s)+(3s-1)Y(s)=\frac{1}{s-1}$ Solving the system we get :

$X(s)=\frac{5s+5}{s^2+1}-\frac{1}{s-1}-\frac{1}{s-2}-\frac{3}{s}$ $Y(s)=\frac{-5}{s^2+1}+\frac{2}{s-1}+\frac{1}{s-2}-\frac{3}{s}$

Hence $x(t)=5cost(t)+5sin(t)-e^t-e^{2t}-3$ \plot{xy,[[1:10, sin(x)]]} $y(t)=-5sin(t)+2e^t+e^{2t}-3$