Question

Find the general solution of the given system. dx dt = 8x − y dy dt = 5x + 4y x(t), y(t) =

Answer 1

Answer:

The general solution of given system is $x(t)=C_1e^{6t}\cos t+C_2e^{6t}\sin t$ and $y(t)=C_1e^{6t}\cos t+C_2e^{6t}\sin t$.

Step-by-step explanation:

The given differential equations are

$\frac{dx}{dt}=8x-y$       .... (1)

$\frac{dy}{dt}=5x+4y$      .... (2)

Differentiate both equations with respect to t.

$x''=8x'-y'$                       ...... (3)

$y''=5x'+4y'$                    ..... (4)

Using equation (3) and (2).

$x''=8x'-(5x+4y)$

$x''=8x'-5x-4y)$

$x''=8x'-5x-4(8x-x')$                   (Using 1)

$x''=8x'-5x-32x+4x')$

$x''=12x'-37x$

$x''-12x'+37x=0$

$r^2-12r+37=0$

Using Quadratic formula

$r=\frac{-b\pm \sqrt{b^2-4ac} }{2a}$

$r=\frac{12\pm \sqrt{144-148} }{2(1)}$

$r=6\pm i$

If $r=\alpha+i\beta$, then

$y(x)=e^{\alpha x}(C_1\cos \beta x+C_2\sin \beta x)$

Therefore x(t) is defined as

$x(t)=e^{6t}(C_1\cos t+C_2\sin t)$

Using equation (4) and (1).

$y''=5(8x-y)+4y'$

$y''=40x-5y+4y'$

$y''=40(\frac{y'-4y}{5})-5y+4y'$                     (Using 2)

$y''=8(y'-4y)-5y+4y'$

$y''=8y'-32y-5y+4y'$

$y''+12y'-37y=0$

$r^2-12r+37=0$

Using Quadratic formula

$r=\frac{-b\pm \sqrt{b^2-4ac} }{2a}$

$r=\frac{12\pm \sqrt{144-148} }{2(1)}$

$r=6\pm i$

Therefore y(t) is defined as

$y(t)=e^{6t}(C_1\cos t+C_2\sin t)$