Question

د Solve the following simultaneous equations. dx dy = 3x + 2y, + 5x + 3y = 0 dt di​

Answer 1

Appropriate Question :-

Solve the following simultaneous equations

$\rm \: \dfrac{dx}{dt} = 3x + 2y$

and

$\rm \: \dfrac{dy}{dt} = 5x + 3y$

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm \: \dfrac{d}{dt} = D$

So, above equations can be rewritten as

$\rm \: Dx = 3x + 2y$

$\rm \: Dx - 3x - 2y = 0$

$\rm \: (D - 3)x - 2y = 0 - - - (1)$

and

$\rm \: Dy = 5x + 3y$

$\rm \: Dy - 5x - 3y = 0$

$\rm \: (D - 3)y - 5x = 0 - - - (2)$

Now, multiply equation (1) by D - 3 and (2) by 2, we get

$\rm \: 2(D - 3)y - 10x = 0 - - - (3)$

and

$\rm \: (D - 3)^{2} x - 2(D - 3)y = 0 - - - (4)$

On adding equation (3) and (4), we get

$\rm \: {(D - 3)}^{2}x - 10x = 0$

$\rm \: ( {D}^{2} + 9 - 6D - 10)x = 0$

$\rm \: ( {D}^{2} - 6D - 1)x = 0$

So, its Auxiliary equation is

$\rm \: {m}^{2} - 6m - 1 = 0$

$\rm \: m = \dfrac{6 \: \pm \: \sqrt{36 + 4} }{2}$

$\rm \: m = \dfrac{6 \: \pm \: \sqrt{40} }{2}$

$\rm \: m = \dfrac{6 \: \pm \: 2 \sqrt{10} }{2}$

$\rm\implies \:m = 3 \: \pm \: \sqrt{10}$

So, Solution is given by

$\rm\implies \:\boxed{\rm{  \:\rm \: x = c_1 {e}^{(3 + \sqrt{10}) \: t } + c_2 {e}^{(3 - \sqrt{10}) \: t} }}$

or

$\rm\implies \:\boxed{\rm{  \:\rm \: x = {e}^{3t} \bigg( c_1 {e}^{(\sqrt{10}) \: t } + c_2 {e}^{( - \sqrt{10}) \: t} \bigg)}}$

Now, from equation (1) we have

$\rm \: 2y = (D - 3)x$

$\rm \: 2y = Dx - 3x$

$\rm \: 2y = D(c_1 {e}^{(3 + \sqrt{10}) \: t } + c_2 {e}^{(3 - \sqrt{10}) \: t} ) - 3(c_1 {e}^{(3 + \sqrt{10}) \: t } + c_2 {e}^{(3 - \sqrt{10}) \: t} )$

$\rm \: 2y = c_1 {e}^{(3 + \sqrt{10}) \: t }(3 + \sqrt{10}) + c_2 {e}^{(3 - \sqrt{10}) \: t}(3 - \sqrt{10}) - 3c_1 {e}^{(3 + \sqrt{10}) \: t } - 3c_2 {e}^{(3 - \sqrt{10}) \: t}$

$\rm \: 2y = c_1 {e}^{(3 + \sqrt{10}) \: t }(3 + \sqrt{10} - 3) + c_2 {e}^{(3 - \sqrt{10}) \: t}(3 - \sqrt{10} - 3)$

$\rm \: 2y = \sqrt{10} \: c_1 {e}^{(3 + \sqrt{10}) \: t } \: - \sqrt{10} \: c_2 {e}^{(3 - \sqrt{10}) \: t}$

$\rm\implies \:y = \dfrac{ \sqrt{10} }{2}\bigg(c_1 {e}^{(3 + \sqrt{10}) \: t } - c_2 {e}^{(3 - \sqrt{10}) \: t} \bigg)$

or

$\rm\implies \:y = \dfrac{ \sqrt{10} }{2} \: {e}^{3t} \bigg(c_1 {e}^{(\sqrt{10}) \: t } - c_2 {e}^{(- \sqrt{10}) \: t} \bigg)$