Math, asked by Coolboyjayant3674, 1 month ago

if y=3cos(log x)+4sin(log x0. show that x*2y2+xy1+y=0

Answers

Answered by mathdude500
4

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

Given that,

\rm :\longmapsto\:y = 3cos(logx) + 4sin(logx)

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

\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}\bigg[ 3cos(logx) + 4sin(logx)\bigg]

\rm :\longmapsto\:y_1 = 3\dfrac{d}{dx}cos(logx) + 4\dfrac{d}{dx}sin(logx)

We know,

\boxed{ \tt{ \: \dfrac{d}{dx}sinx = cosx \: }}

and

\boxed{ \tt{ \: \dfrac{d}{dx}cosx = -  \: sinx \: }}

So, using this, we get

\rm :\longmapsto\:y_1 =  - 3sin(logx)\dfrac{d}{dx}logx + 4cos(logx)\dfrac{d}{dx}logx

We know,

\boxed{ \tt{ \: \dfrac{d}{dx}logx \:  =  \:  \frac{1}{x} \: }}

So, using this, we get

\rm :\longmapsto\:y_1 =  - 3sin(logx) \times \dfrac{1}{x} + 4cos(logx) \times \dfrac{1}{x}

\rm :\longmapsto\:xy_1 =  - 3sin(logx)  + 4cos(logx)

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

\rm :\longmapsto\:\dfrac{d}{dx}xy_1 = \dfrac{d}{dx}\bigg[ - 3sin(logx)  + 4cos(logx)\bigg]

\rm :\longmapsto\:x\dfrac{d}{dx}y_1 + y_1\dfrac{d}{dx}x =  - 3[cos(logx)]\dfrac{1}{x}  - 4[sin(logx)]\dfrac{1}{x}

\rm :\longmapsto\:xy_2 + y_1 =  - \dfrac{1}{x}\bigg[3cos(logx) + 4sin(logx)\bigg]

\rm :\longmapsto\: {x}^{2} y_2 + xy_1  \: =  \:  -  \: y

\bf :\longmapsto\: {x}^{2} y_2 + xy_1  + y \: = \: 0

Additional Information :-

\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\  \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf  -  \: sinx \\ \\ \sf tanx & \sf  {sec}^{2}x \\ \\ \sf cotx & \sf  -  {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf  -  \: cosecx \: cotx\\ \\ \sf  \sqrt{x}  & \sf  \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf  {e}^{x}  & \sf  {e}^{x}  \end{array}} \\ \end{gathered}

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