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Chapter Differential Equation ​

Answer 1

Given

$\tt \: \dfrac{dy}{dx} = \dfrac{ {x}^{2} + {y}^{2} }{2xy} - - - (i)$

Since, degree of each term of numerator and denominator is same, therefore given differential equation is homogeneous.

$\tt \: Put \: y \: = \: vx - - (ii) \: \: \: in \: (i) \: we \: get$

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

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

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

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

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

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

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

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

On integrating both sides, we get

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

$\tt \: log( {v}^{2} - 1) = - log(x) + log(c)$

$\boxed{\red{\because \int \: \dfrac{f'(x)}{f(x)} dx = log(f(x)) + c }}$

$\boxed{\because{ \purple{ \int\dfrac{1}{x} dx = log(x) + \: c}}}$

$\tt \: log |{v}^{2} - 1| + log |x| = log(c)$

$\tt \: log |({v}^{2} - 1)x| = log |c|$

$\tt \: ({v}^{2} - 1) x = \pm \: c$

On substituting the value of v, from equation (ii), we get

$\bf\implies \:(\dfrac{ {y}^{2} }{ {x}^{2} } - 1) \times x = \pm \: c$

$\bf\implies \:(\dfrac{ {y}^{2} - {x}^{2} }{ {x}^{2} } ) \times x = \pm \: c$

$\bf\implies \:(\dfrac{ {y}^{2} - {x}^{2} }{ {x} } ) = \pm \: c$

$\bf\implies \: {y}^{2} - {x}^{2} = \pm \: cx$

$\bf\implies \: {y}^{2} - {x}^{2} = \: cx \: \: or \: \bf\: {x}^{2} - {y}^{2} = \: cx$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

Answer 2

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