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Question Number 75639 by aliesam last updated on 14/Dec/19

	Answered by $@ty@m123 last updated on 14/Dec/19
	$x=ycos y ....(1) Differentiating w.r.t. x, 1={y(−sin y)+cos y}y′ (1/(y′))=cos y−ysin y ....(2) Multiplying (2) by y, (y/(y′))=ycos y−y^2 sin y (y/(y′))=x−y^2 sin y Differentiating w.r.t. x ((y′.y′−y.y′′)/((y′)^2 ))=1−(y^2 cos y.y′+2ysin y.y′) 1−((y.y′′)/((y′)^2 ))=1−y′.y(ycos y+2sin y) ((y.y′′)/((y′)^2 ))=y′.y(ycos y+2sin y) ((y′′)/((y′)^3 ))=ycos y+2sin y ((y′′)/((y′)^3 ))=x+2sin y {from (1)$
	$x = y \cos y . . . . (1)$ $D i f f e r e n t i a t i n g w . r . t . x,$ $1 = {y (- \sin y) + \cos y} y^{'}$ $\frac{1}{y^{'}} = \cos y - y \sin y . . . . (2)$ $M u l t i p l y i n g (2) b y y,$ $\frac{y}{y^{'}} = y \cos y - y^{2} \sin y$ $\frac{y}{y^{'}} = x - y^{2} \sin y$ $D i f f e r e n t i a t i n g w . r . t . x$ $\frac{y^{'} . y^{'} - y . y^{″}}{{(y^{'})}^{2}} = 1 - (y^{2} \cos y . y^{'} + 2 y \sin y . y^{'})$ $1 - \frac{y . y^{″}}{{(y^{'})}^{2}} = 1 - y^{'} . y (y \cos y + 2 \sin y)$ $\frac{y . y^{″}}{{(y^{'})}^{2}} = y^{'} . y (y \cos y + 2 \sin y)$ $\frac{y^{″}}{{(y^{'})}^{3}} = y \cos y + 2 \sin y$ $\frac{y^{″}}{{(y^{'})}^{3}} = x + 2 \sin y {f r o m (1)$
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