Question

please fast Answer ✌️✌️​

Answer 1

GIVEN :–

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

TO FIND :–

• General solution = ?

SOLUTION :–

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

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

• It's an zero order Homogeneous Differential equation.

$\\\bf { \huge{.}} \: \: Now \:\:put \: \: { \underline{ \bf \: y = vx }} : -$

• Differentiate with respect to 'x' –

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

• So that –

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

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

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

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

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

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

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

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

• Now Apply integration on both sides –

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

• We should write this as –

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

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

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

$\\\implies \bf I_{1} -I_{2}= log(x) + c \:\:\:\: ----eq.(1)$

• Here –

$\\ \implies \bf \: I_{1}= \dfrac{3}{2} \int \dfrac{dv}{ {v}^{2} + v + 1}$

• We should write this as –

$\\ \implies \bf \: I_{1}= \dfrac{3}{2}\int \dfrac{dv}{ {v}^{2} + v + \frac{1}{4} + \frac{3}{4} }$

$\\ \implies \bf \: I_{1}= \dfrac{3}{2}\int \dfrac{dv}{ { \bigg(v + \dfrac{1}{2} \bigg)}^{2} + \bigg(\dfrac{ \sqrt{3} }{2} \bigg)^{2}}$

• We know that –

$\\ \implies \bf \: \int \dfrac{dx}{{x}^{2} + a^{2}} = \dfrac{1}{a} { \tan}^{ - 1} \bigg( \dfrac{x}{a} \bigg) + c$

• So that –

$\\ \implies \bf \: I_{1}= \dfrac{3}{2}\frac{1}{\bigg(\dfrac{ \sqrt{3} }{2} \bigg)} { \tan}^{ - 1}\bigg( \dfrac{v + \frac{1}{2}}{\frac{ \sqrt{3} }{2}} \bigg)$

$\\ \implies \bf \: I_{1}= \dfrac{3}{2} \times \dfrac{2}{ \sqrt{3} } { \tan}^{ - 1}\bigg( \dfrac{2v +1}{\sqrt{3}} \bigg)$

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

• And –

$\\ \implies \bf \: I_{2}= \dfrac{1}{2}\int \dfrac{(2v + 1)dv}{ {v}^{2} + v + 1}$

• Put v² + v + 1 = t :–

• Differentiate with respect to 't' –

$\\ \implies \bf \:(2v + 1) \dfrac{dv}{dt} =1$

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

• So that –

$\\ \implies \bf \: I_{2}= \dfrac{1}{2}\int \dfrac{dt}{t}$

$\\ \implies \bf \: I_{2}= \dfrac{1}{2} log(t)$

$\\ \implies \bf \: I_{2}= \dfrac{1}{2} log({v}^{2} + v + 1)$

• Now put the values in eq.(1) –

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

• Now replace 'v' –

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

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

Answer 2

$\Huge\rm\red{SEE\: Attachment}$