Math, asked by BrainlicaLDoll, 2 months ago

Question:
\sf{Find\:the\:particular\:solution\:of\:the\:differential\:equation-}
\sf{x{e}^{\frac{y}{x}}-{\sf ysin(\frac{y}{x})}+{\sf x\frac{dy}{dx}sin(\frac{y}{x})=0\:for\:x\:=\:1,\:y\:=0}}​​

Answers

Answered by mathdude500
9

Identities Used :-

 \boxed{ \pink{ \displaystyle \int \:\dfrac{1}{x}  = logx \:  +  \: c}}

 \boxed{ \pink{ \displaystyle \int \:  {e}^{ax}sinbx  \: dx = \dfrac{ {e}^{ax} }{ {a}^{2} +  {b}^{2}}(a \: sinbx - b \: cosbx)  \:  +  \: c}}

 \pink{ \boxed{ log(1) = 0}}

Given Differential equation is

\rm :\longmapsto\:\sf{x{e}^{\frac{y}{x}}-{\sf ysin(\frac{y}{x})}+{\sf x\frac{dy}{dx}sin(\frac{y}{x})=0\:}}

can be rewritten as

\rm :\longmapsto\:\sf{x{e}^{\frac{y}{x}}-{\sf ysin(\frac{y}{x})} =  -  \: {\sf x\frac{dy}{dx}sin(\frac{y}{x})\:}}

\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ysin\bigg(\dfrac{y}{x}\bigg) - x {\bigg(e\bigg) }^{\dfrac{y}{x} }}{xsin\bigg(\dfrac{y}{x}\bigg)} -  -  - (1)

Its a homogeneous differential equation.

So,

\red{\rm :\longmapsto\:\bf\:Put \: y \:  =  \: vx}

On differentiating w. r. t. x, we get

\red{\rm :\longmapsto\:\bf\: \: \dfrac{dy}{dx}  \:  =  \:\dfrac{d}{dx} vx} -  -  - (2)

\red{\rm :\longmapsto\:\bf\: \: \dfrac{dy}{dx}  \:  =  \:v\dfrac{d}{dx} x + x\dfrac{d}{dx}v}

\red{\rm :\longmapsto\:\bf\: \: \dfrac{dy}{dx}  \:  =  \:v + x\dfrac{dv}{dx}} -  -  - (3)

Substituting the values of equation (2), (3) in equation (1),

\rm :\longmapsto\:v + x\dfrac{dv}{dx} = \dfrac{vxsin\bigg(\dfrac{vx}{x}\bigg) - x {\bigg(e\bigg) }^{\dfrac{vx}{x} }}{xsin\bigg(\dfrac{vx}{x}\bigg)}

\rm :\longmapsto\:v + x\dfrac{dv}{dx} =\dfrac{vsinv -  {e}^{v} }{sinv}

\rm :\longmapsto\: x\dfrac{dv}{dx} =\dfrac{vsinv -  {e}^{v} }{sinv} - v

\rm :\longmapsto\: x\dfrac{dv}{dx} =\dfrac{vsinv -  {e}^{v} - vsinv}{sinv}

\rm :\longmapsto\: x\dfrac{dv}{dx} =\dfrac{-{e}^{v}}{sinv}

\rm :\longmapsto\: {e}^{ - v} sinvdv = -  \:  \dfrac{dx}{x}

On integrating both sides, we get

\rm :\longmapsto\: \int {e}^{ - v} sinvdv =  -  \:   \displaystyle\int\dfrac{dx}{x}

\rm :\longmapsto\:\dfrac{ {e}^{ - v} }{ {( - 1)}^{2} +  {(1)}^{2}}( - sinv - cosv) =  - logx + c

\rm :\longmapsto\:\dfrac{ {e}^{ - v} }{2}( sinv  + cosv) =  logx  -  c

On substituting the value of v, from equation (2), we get

\rm :\longmapsto\:\ \:\dfrac{1}{2}  {\bigg(e\bigg) }^{-\dfrac{y}{x} } \bigg( sin\bigg(\dfrac{y}{x}\bigg)  + cos\bigg(\dfrac{y}{x}\bigg) \bigg) =  logx  -  c -  -  (4)

Now,

Given that, y = 0 and x = 1, we get

\rm :\longmapsto\:\ \:\dfrac{1}{2}  {\bigg(e\bigg) }^{-\dfrac{0}{1} }( sin0  + cos0) =  log1  -  c -  -  (4)

\bf\implies \:c =  -  \: \dfrac{1}{2}

So, equation (4) can be rewritten as

\rm :\longmapsto\:\ \:\dfrac{1}{2}  {\bigg(e\bigg) }^{-\dfrac{y}{x} } \bigg( sin\bigg(\dfrac{y}{x}\bigg)  + cos\bigg(\dfrac{y}{x}\bigg) \bigg) =  logx + \dfrac{1}{2}

\rm :\longmapsto\:\ \:{\bigg(e\bigg) }^{-\dfrac{y}{x} } \bigg( sin\bigg(\dfrac{y}{x}\bigg)  + cos\bigg(\dfrac{y}{x}\bigg) \bigg) =  2logx + 1

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