Math, asked by yadavganesh5624, 5 months ago

THE NUMBER OF TIMES THAT A PARTICULAR ENTRY OCCUR IN GIVEN OBSERVATION GIVES US ITS

Answers

Answered by Anonymous
0

Answer:

✪QUESTION✪:-

Find the Solution of the given condition

\sf\dfrac{dy}{dx} - 3y \: cot \: x = sin2x,y = 2 \: when \: x = \dfrac{\pi}{2}

✪SOLUTION✪

•The given differential equation is a linear differential equation i.e, of the form \displaystyle\sf\dfrac{dy}{dx}+Py=Q

P=-3cot x and Q=sin 2 x

Here,⠀

\sf I.F.={ \sf e\:}^{ \displaystyle \sf\int \small P \: dx}={e}^{ \displaystyle \sf\int \small -3 \: cot \: x}

⠀⠀

\displaystyle \sf = {e}^{ - 3 \: log(sin \: x)} = {e}^{log(sin \: x) ^{ - 3} }=e

⠀⠀⠀⠀⠀⠀⠀⠀⠀\sf( Assuming\:0 < x < \pi)(Assuming0<x<π)

⠀⠀

\sf = {(sin \: x)}^{ - 3}=(sinx)

Hence,the solution of the given equation is given by

⠀⠀

\displaystyle \sf y{(sin \: x)}^{ - 3} = \int {(sin \: x)}^{ - 3} cos \: x \: dx+Cy(sinx)

= \displaystyle\sf2\int {(sin \: x)}^{ - 2} cos \: x \: dx=2∫(sinx)

\displaystyle \sf = \dfrac{2{(sin \: x)}^{ - 1} }{ - 1} + C=

⠀⠀\sf or \: y = - 2{(sin \: x)}^{3 - 1} + C \: {sin}^{3}xory=−2(sinx)

⠀⠀\sf or \: y = - 2{(sin \: x)}^{2} + C \: {sin}^{3}x .....(1)

When \sf x=\dfrac{\pi}{2},\:then\:y=2x=

⠀⠀

\sf \therefore2 = - 2{ sin}^{2} \bigg( \dfrac{ \pi}{2}\bigg) + C \:{sin}^{3} \bigg( \dfrac{\pi}{2} \bigg)

\implies \sf2 = - 2 + C⟹2=−2+C

⠀⠀

\implies \sf C=4⟹C=4</p><p></p><p>Hence the required solution is</p><p></p><p>[tex]\sf \: y = - 2{(sin \: x)}^{2} + 4{sin}^{3}x .......(from 1)

Answered by akshrajain30aug2007
0

Step-by-step explanation:

Frequency of a value(observations) is defined by the number of times the value occurs in a given data set. Therefore, the number of times a particular observation occurs is a given data set is called its frequency.

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