Question

Find the general solution of the differential equation :-

$↦{ \boxed{ \mathsf{( x - \: y) \frac{DY}{DX } = x + 2y}}}$
​

Answer 1

GIVEN :–

$\\ \implies\bf( x - y) \dfrac{dy}{dx} = x + 2y$

TO FIND :–

• Value of differential equation = ?

SOLUTION :–

$\\ \implies\bf( x - y) \dfrac{dy}{dx} = x + 2y$

$\\ \implies\bf\dfrac{dy}{dx} = \dfrac{x + 2y}{x - y}$

• It's a Homogeneous equation. So let's put y=vx . And –

$\\ \implies\bf\dfrac{dy}{dx} = v + x \dfrac{dv}{dx}$

• So –

$\\ \implies\bf v + x \dfrac{dv}{dx} =\dfrac{x + 2(vx)}{x - (vx)}$

$\\ \implies\bf v + x \dfrac{dv}{dx} =\dfrac{x(1+ 2v)}{x(1 -v)}$

$\\ \implies\bf v + x \dfrac{dv}{dx} =\dfrac{1+ 2v}{1 -v}$

$\\ \implies\bf x \dfrac{dv}{dx} =\dfrac{1+ 2v}{1 -v} - v$

$\\ \implies\bf x \dfrac{dv}{dx} =\dfrac{1+ 2v - v(1 -v)}{1 -v}$

$\\ \implies\bf x \dfrac{dv}{dx} =\dfrac{1+ 2v - v + {v}^{2} }{1 -v}$

$\\ \implies\bf x \dfrac{dv}{dx} =\dfrac{1+ v + {v}^{2} }{1 -v}$

$\\ \implies\bf \dfrac{1 -v}{1+ v + {v}^{2}}dv = \dfrac{dx}{x}$

• Integrate on both side –

$\\ \implies\bf \int\dfrac{1 -v}{1+ v + {v}^{2}}dv = \int\dfrac{dx}{x}$

$\\ \implies\bf \int\dfrac{1}{1+ v + {v}^{2}}dv - \int\dfrac{v}{1+ v + {v}^{2}}dv= \int\dfrac{dx}{x}$

$\\ \implies\bf \int\dfrac{1}{1+ v + {v}^{2}}dv - \dfrac{1}{2} \int\dfrac{2v + 1 - 1}{1+ v + {v}^{2}}dv= \int\dfrac{dx}{x}$

$\\ \implies\bf \int\dfrac{1}{1+ v + {v}^{2}}dv - \dfrac{1}{2} \int\dfrac{2v + 1}{1+ v + {v}^{2}}dv +\dfrac{1}{2}\int\dfrac{1}{1+ v + {v}^{2}}dv= \int\dfrac{dx}{x}$

$\\ \implies\bf \dfrac{3}{2} \int\dfrac{1}{1+ v + {v}^{2}}dv - \dfrac{1}{2} \int\dfrac{2v + 1}{1+ v + {v}^{2}}dv= \int\dfrac{dx}{x}$

• Let put –

$\\ \implies\bf \dfrac{3}{2} I_1- \dfrac{1}{2}I_2= \int\dfrac{dx}{x} \: \: \: \: \: \: \: - - - - eq.(1)$

• Now –

$\\ \implies\bf I_1 =\int\dfrac{1}{1+ v + {v}^{2}}dv$

$\\ \implies\bf I_1 =\int\dfrac{1}{{ \bigg(v +\dfrac{1}{2} \bigg)}^{2} + \dfrac{3}{4} }dv$

• Put v+1=√3t/2 –

$\\ \implies\bf dv = \dfrac{ \sqrt{3}}{2}dt$

• So –

$\\ \implies\bf I_1 =\dfrac{ \sqrt{3}}{2}\int\dfrac{1}{{ \bigg(\dfrac{ \sqrt{3}t}{2} \bigg)}^{2} + \dfrac{3}{4} }dt$

$\\ \implies\bf I_1 =\dfrac{ \sqrt{3}}{2}\int\dfrac{1}{\dfrac{3}{4} {t}^{2} + \dfrac{3}{4} }dt$

$\\ \implies\bf I_1 =\dfrac{ \sqrt{3}}{2}\int\dfrac{1}{\dfrac{3}{4}({t}^{2} +1)}dt$

$\\ \implies\bf I_1 =\dfrac{ \sqrt{3}}{2} \times \dfrac{4}{3}\int\dfrac{1}{{t}^{2} +1}dt$

$\\ \implies\bf I_1 =\dfrac{2}{ \sqrt{3} }\int\dfrac{1}{{t}^{2} +1}dt$

$\\ \implies\bf I_1 =\dfrac{2}{ \sqrt{3} } \tan^{ - 1} (t)$

• Now replace 't' –

$\\ \implies \large{ \boxed{\bf I_1 =\dfrac{2}{ \sqrt{3} } \tan^{ - 1} \bigg( \dfrac{2v + 1}{ \sqrt{3} } \bigg)}}$

• And –

$\\ \implies\bf I_2=\int\dfrac{2v + 1}{1+ v + {v}^{2}}dv$

• Put 1+v+v² = t –

$\\ \implies \bf (1 + 2v)dv = dt$

$\\ \implies\bf I_2=\int\dfrac{dt}{t}$

$\\ \implies\bf I_2= log(t)$

• Now replace 't' –

$\\ \implies\bf I_2= log(1 + v + {v}^{2} )$

• Put the values of $\tt I_1\:\:and\:\:I_2$ in eq.(1) –

$\\ \implies\bf \dfrac{3}{2} \times \dfrac{2}{ \sqrt{3} } \tan^{ - 1} \bigg( \dfrac{2v + 1}{ \sqrt{3} } \bigg)- \dfrac{1}{2}log(1 + v + {v}^{2} )= \int\dfrac{dx}{x}$

$\\ \implies\bf \sqrt{3} \tan^{ - 1} \bigg( \dfrac{2v + 1}{ \sqrt{3} } \bigg)- \dfrac{1}{2}log(1 + v + {v}^{2} )= log(x) + c$

• Now replace 'v' –

$\\ \implies\bf \sqrt{3} \tan^{ - 1} \bigg( \dfrac{2 \times \dfrac{y}{x} + 1}{ \sqrt{3} } \bigg)- \dfrac{1}{2}log \bigg(1 + \dfrac{y}{x} + { \bigg\{\dfrac{y}{x} \bigg\}}^{2} \bigg)= log(x) + c$

$\\ \implies \large \red{ \boxed{\bf \sqrt{3} \tan^{ - 1} \bigg( \dfrac{2y + x}{ \sqrt{3}x } \bigg)- \dfrac{1}{2}log \bigg( \dfrac{ {x}^{2} + xy + {y}^{2} }{ {x}^{2} } \bigg)= log(x) + c}}$