Question

In figure, measures of some angles are shown. using the measures of angle X and angle y and hence show that line l|| line m.​

Answer 1

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)