Reduce to first order and solve , showing each step in detail. 1. y′′ +(y′)^3 siny=0 2. y′′=1+(y′)^2


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Question Number 193371 by gloriousman last updated on 11/Jun/23

$Reduce to first order and solve,$ $showing each step in detail .$ $1 . y^{″} + {(y^{'})}^{3} siny = 0$ $2 . y^{″} = 1 + {(y^{'})}^{2}$
	Answered by witcher3 last updated on 11/Jun/23
	$y′′+(y′)^3 sin(y)=0....(A) first we do not know how to starte caue just y′ cos(y(x))=z z′=y′′cos(y)−y′^2 sin(y)...E cosy=(z/(y′))..if y′=0⇒y=c solution suppose ∃y#constante and y∈C_2 [a,b]⇒∃I⊂[a,b] y′≠0 this justify division by y′ E⇔z′=((zy′′)/(y′))−y′^2 sin(y) ⇔z′y′−zy^′ ′−y′′=−y′^3 sin(y)−y′′=0byA ⇔z′y′−y′′(z−1)=0 ⇒((z′y′−y′′(z−1))/((z−1)^2 ))=0 ⇔(d/dx)(((−y′)/((z−1))))=0 ⇒−((y′)/(z−1))=c y′=c(z−1)⇒z=((y′)/c)+1=ay′+1 ⇒y′cos(y)−ay′=1⇒sin(y)−ay=x+c ⇒ { ((y′′+y′^3 sin(y)=0)),((sin(y)−ay=x+c)) :}$
	$y^{″} + {(y^{'})}^{3} \sin (y) = 0 . . . . (A)$ $first we do not know how to starte caue$ $just y^{'} \cos (y (x)) = z$ $z^{'} = y^{″} \cos (y) - y^{' 2} \sin (y) . . . E$ $You can't use 'macro parameter character #' in math mode$ $and y \in C_{2} [a, b] \Rightarrow \exists I \subset [a, b] y^{'} \neq 0$ $this justify division by y^{'}$ $E \Leftrightarrow z^{'} = \frac{{zy}^{″}}{y^{'}} - y^{' 2} \sin (y)$ $Prime causes double exponent: use braces to clarify$ $\Leftrightarrow z^{'} y^{'} - y^{″} (z - 1) = 0$ $\Rightarrow \frac{z^{'} y^{'} - y^{″} (z - 1)}{{(z - 1)}^{2}} = 0$ $\Leftrightarrow \frac{d}{dx} (\frac{- y^{'}}{(z - 1)}) = 0$ $\Rightarrow - \frac{y^{'}}{z - 1} = c$ $y^{'} = c (z - 1) \Rightarrow z = \frac{y^{'}}{c} + 1 = {ay}^{'} + 1$ $\Rightarrow y^{'} \cos (y) - {ay}^{'} = 1 \Rightarrow \sin (y) - ay = x + c$ $\Rightarrow$ ${\begin{cases} y^{″} + y^{' 3} \sin (y) = 0 \\ \sin (y) - ay = x + c \end{cases}$
	Answered by witcher3 last updated on 11/Jun/23
	$2,y′=z⇔z′=1+z^2 ⇒∫(dz/(z^2 +1))=∫1=x+c=arctan(z) ⇒z=tan(x+c),∀x∈R−{(π/2)+kπ−c} z=y′=tan(x+c)⇒y=−ln∣cos(x+c)∣+c_1 c,c_1 ∈R$
	$2, y^{'} = z \Leftrightarrow z^{'} = 1 + z^{2} \Rightarrow \int \frac{dz}{z^{2} + 1} = \int 1 = x + c = \arctan (z)$ $\Rightarrow z = \tan (x + c), \forall x \in R - {\frac{π}{2} + k π - c}$ $z = y^{'} = \tan (x + c) \Rightarrow y = - \ln ∣ \cos (x + c) ∣ + c_{1}$ $c, c_{1} \in R$
	Answered by aleks041103 last updated on 12/Jun/23

	$1 . y^{″} + {(y^{'})}^{3} s i n (y) = 0$ $\Rightarrow - \frac{y^{″}}{{(y^{'})}^{2}} + (- s i n (y)) y^{'} = 0$ $\Rightarrow {(\frac{1}{y^{'}} + c o s (y))}^{'} = 0 \Rightarrow \frac{1}{y^{'}} + c o s (y) = a = c o n s t .$ $\Rightarrow y^{'} = \frac{1}{a - c o s (y)}$ $\Rightarrow (a - c o s (y)) d y = d x$ $\Rightarrow a y - s i n (y) + b = x, a, b = c o n s t .$ $2 . y^{″} = 1 + {(y^{'})}^{2}$ $\Rightarrow \frac{y^{″}}{1 + {(y^{'})}^{2}} = {(a c t a n (y^{'}))}^{'} = 1$ $\Rightarrow a r c t a n (y^{'}) = x + a, a = c o n s t .$ $\Rightarrow y^{'} = \frac{d y}{d x} = t a n (x + a)$ $\Rightarrow d y = t a n (x + a) d x$ $\Rightarrow y = \int \frac{s i n (x + a)}{c o s (x + a)} d x = - \int \frac{d (c o s (x + a))}{c o s (x + a)}$ $\Rightarrow y = - l n (c o s (x + a)) + b, a, b = c o n s t .$
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