Reduce-to-first-order-and-solve-showing-each-step-in-detail-1-y-y-3-siny-0-2-y-1-y-2-

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 \dots . (A)

first we do not know how to starte caue

just y^{'} \cos (y (x)) = z

z^{'} = y^{″} \cos (y) - y^{' 2} \sin (y) \dots 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 \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 .