Question

(x2 + y2) dy – (x2 + xy) dx = 0​

Answer 1

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

Given Differential equation is

$\rm \: ( {x}^{2} + {y}^{2} )dy \: - \: ( {x}^{2} + xy)dx \: = \: 0$

can be rewritten as

$\rm \: ( {x}^{2} + {y}^{2} )dy \: = \: ( {x}^{2} + xy)dx \:$

$\rm \: \dfrac{dy}{dx} = \dfrac{ {x}^{2} + xy}{ {x}^{2} + {y}^{2} }$

Let assume that

$\rm \: f(x,y) = \dfrac{ {x}^{2} + xy}{ {x}^{2} + {y}^{2} }$

So,

$\rm \: f(kx,ky) = \dfrac{ {k}^{2} {x}^{2} + {k}^{2} xy}{ {k}^{2}{x}^{2} + {k}^{2} {y}^{2} }$

$\rm \: f(kx,ky) = \dfrac{ {k}^{2}( {x}^{2} + xy)}{ {k}^{2}({x}^{2} + {y}^{2} )}$

$\rm \: f(kx,ky) = \dfrac{ {x}^{2} + xy}{{x}^{2} + {y}^{2}}$

$\rm\implies \:f(x,y) = f(kx,ky)$

$\rm\implies \:f(x,y) \: is \: homogenous \: function \: of \: degree \: 0$

So, it means

$\rm \: \dfrac{dy}{dx} = \dfrac{ {x}^{2} + xy}{ {x}^{2} + {y}^{2} } \: is \: homogeneous$

To evaluate this Differential equation, we substitute

$\rm \: y \: = \: vx$

So, on substituting,

$\rm \: \dfrac{d}{dx} (vx) = \dfrac{ {x}^{2} + x(vx)}{ {x}^{2} + {(vx)}^{2} }$

$\rm \: v\dfrac{d}{dx}x \: + \: x\dfrac{d}{dx}v = \dfrac{ {x}^{2} + {vx}^{2} }{ {x}^{2} + {v}^{2} {x}^{2} }$

$\rm \: v \: + \: x\dfrac{dv}{dx} = \dfrac{ {x}^{2}(1 + v) }{ {x}^{2}(1 + {v}^{2})}$

$\rm \: v \: + \: x\dfrac{dv}{dx} = \dfrac{1 + v}{1 + {v}^{2}}$

$\rm \: \: x\dfrac{dv}{dx} = \dfrac{1 + v}{1 + {v}^{2}} - v$

$\rm \: \: x\dfrac{dv}{dx} = \dfrac{1 + v - v - {v}^{3} }{1 + {v}^{2}}$

$\rm \: \: x\dfrac{dv}{dx} = \dfrac{1 - {v}^{3} }{1 + {v}^{2}}$

$\rm \: \: \dfrac{1 + {v}^{2} }{1 - {v}^{3}} dv \: = \dfrac{dx}{x}$

On integrating both sides,

$\rm \: \:\displaystyle\int\rm \dfrac{1 + {v}^{2} }{1 - {v}^{3}} dv \: = \displaystyle\int\rm \dfrac{dx}{x}$

Let assume that,

$\rm \: \:I = \displaystyle\int\rm \dfrac{1 + {v}^{2} }{1 - {v}^{3}} dv \:$

Let

$\rm \: \: \dfrac{1 + {v}^{2} }{(1 - v)(1 + v + {v}^{2}) } \: = \dfrac{a}{1 - v} + \dfrac{bv + c}{1 + v + {v}^{2} }$

$\rm \: 1 + {v}^{2} = a(1 + v + {v}^{2}) + (bv + c)(1 - v)$

Put v = 1

$\rm \: 1 + 1 = a(1 + 1 + 1) + 0$

$\rm \: 2 = 3a$

$\rm\implies \:a = \dfrac{2}{3}$

Put v = 0,

$\rm \: 1 + 0 = a(1 + 0 + 0) + c(1 - 0)$

$\rm \: 1 = a + c$

$\rm \: 1 = \dfrac{2}{3} + c$

$\rm\implies \:c = \dfrac{1}{3}$

Put v = - 1,

$\rm \: 1 + 1 = a(1 - 1 + 1) + ( - b + c)(1 + 1)$

$\rm \: 2 = a - 2b + 2c$

$\rm \: 2 = \dfrac{2}{3} - 2b + \dfrac{2}{3}$

$\rm \: 2 = \dfrac{4}{3} - 2b$

$\rm \: 1 = \dfrac{2}{3} - b$

$\rm \: b = \dfrac{2}{3} - 1$

$\rm\implies \: \: b = - \: \dfrac{1}{3}$

On substituting the values,

$\rm \: \: \dfrac{1 + {v}^{2} }{(1 - v)(1 + v + {v}^{2}) } \: = \dfrac{2}{3(1 - v)} + \dfrac{ - v + 1}{3(1 + v + {v}^{2}) }$

$\rm \: \: \dfrac{1 + {v}^{2} }{1 - {v}^{3} } \: = \dfrac{2}{3(1 - v)} + \dfrac{ - v + 1}{3(1 + v + {v}^{2}) }$

$\rm \: \: \dfrac{1 + {v}^{2} }{1 - {v}^{3} } \: = \dfrac{2}{3(1 - v)} - \dfrac{2 v - 2}{6(1 + v + {v}^{2}) }$

$\rm \: \: \dfrac{1 + {v}^{2} }{1 - {v}^{3} } \: = - \dfrac{2}{3( v - 1)} - \dfrac{2 v + 1 - 3}{6(1 + v + {v}^{2}) }$

$\rm \: \: \dfrac{1 + {v}^{2} }{1 - {v}^{3} } \: = - \dfrac{2}{3( v - 1)} - \dfrac{2 v + 1}{6(1 + v + {v}^{2}) } + \dfrac{1}{6(1 + v + {v}^{2} )}$

On integrating both sides, we get

$\rm \displaystyle\int\rm \dfrac{1 + {v}^{2} }{1 - {v}^{3} }dv = -\displaystyle\int\rm \dfrac{2}{3( v - 1)}dv - \displaystyle\int\rm \dfrac{2 v + 1}{6(1 + v + {v}^{2}) }dv + \displaystyle\int\rm \dfrac{1}{6(1 + v + {v}^{2} )} dv$

$\rm \: = - \dfrac{2}{3}log |v - 1| - \dfrac{1}{6}log |1 + v + {v}^{2} | + \dfrac{1}{6}\displaystyle\int\rm \frac{dv}{ {v}^{2} + v + \frac{1}{4} - \frac{1}{4} + 1}$

$\rm \: = - \dfrac{2}{3}log |v - 1| - \dfrac{1}{6}log |1 + v + {v}^{2} | + \dfrac{1}{6}\displaystyle\int\rm \frac{dv}{ {(v + \frac{1}{2} )}^{2} + \frac{3}{4}}$

$\rm \: = - \dfrac{2}{3}log |v - 1| - \dfrac{1}{6}log |1 + v + {v}^{2} | + \dfrac{1}{6}\displaystyle\int\rm \frac{dv}{ {(v + \frac{1}{2} )}^{2} + {( \frac{ \sqrt{3} }{2} )}^{2} }$

$\rm \: = - \dfrac{2}{3}log |v - 1| - \dfrac{1}{6}log |1 + v + {v}^{2} | + \dfrac{1}{3 \sqrt{3} } {tan}^{ - 1} \dfrac{2v + 1}{ \sqrt{3} }$

So,

On substituting the value,

$\rm \: \:\displaystyle\int\rm \dfrac{1 + {v}^{2} }{1 - {v}^{3}} dv \: = \displaystyle\int\rm \dfrac{dx}{x}$

$\rm \: - \dfrac{2}{3}log |v - 1| - \dfrac{1}{6}log |1 + v + {v}^{2} | + \dfrac{1}{3 \sqrt{3} } {tan}^{ - 1} \dfrac{2v + 1}{ \sqrt{3} } = log |x| + c$

where,

$\rm \: v = \dfrac{y}{x}$

is the solution of required differential equation.