solve the differential equation (dy^2 /dx^(2 ) )+y=0


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Question Number 128903 by Engr_Jidda last updated on 11/Jan/21

$s o l v e t h e d i f f e r e n t i a l e q u a t i o n$ $\frac{{d y}^{2}}{{d x}^{2}} + y = 0$
	Answered by mindispower last updated on 11/Jan/21
	$X^2 +1=0⇒X∈{i,−i} y=ae^(it) +be^(−it) {(ae^(it) +be^(−it) )∣(a,b)∈C^2 }={acos(t)+bsin(t)} S=acos(t)+bsin(t) 2nd y=Σa_n x^n ⇒y′′=Σ_(n≥0) (n+1)(n+2)a_(n+2) x^n ⇒(n+1)(n+2)a_(n+2) +a_n =0 ⇒a_(n+2) =−(a_n /((n+1)(n+2))) a_(2k+2) =((−a_(2k) )/((2k+1)(2k+2)))=(((−1)^k a_0 )/((2k+2)!)) a_(2k+1) =(((−1)^k )/((2k+1)!))a_1 y=Σa_n x^n =Σ_(k=0) ^∞ a_(2k) x^(2k) +Σ_(k=0) ^∞ a_(2k+1) x^(2k+1) =a_0 Σ_(k≥0) (((−1)^k )/((2k+2)!))x^(2k) +a_1 Σ_(k≥0) (((−1)^k )/((2k+1)!))x^(2k+1) =a_0 cos(x)+a_1 sin(x)$
	$X^{2} + 1 = 0 \Rightarrow X \in {i, - i}$ $y = {a e}^{i t} + {b e}^{- i t}$ ${({a e}^{i t} + {b e}^{- i t}) ∣ (a, b) \in C^{2}} = {a c o s (t) + b s i n (t)}$ $S = a c o s (t) + b s i n (t)$ $2 n d$ $y = Σ a_{n} x^{n} \Rightarrow y^{″} = \sum_{n ⩾ 0} (n + 1) (n + 2) a_{n + 2} x^{n}$ $\Rightarrow (n + 1) (n + 2) a_{n + 2} + a_{n} = 0$ $\Rightarrow a_{n + 2} = - \frac{a_{n}}{(n + 1) (n + 2)}$ $a_{2 k + 2} = \frac{- a_{2 k}}{(2 k + 1) (2 k + 2)} = \frac{{(- 1)}^{k} a_{0}}{(2 k + 2)!}$ $a_{2 k + 1} = \frac{{(- 1)}^{k}}{(2 k + 1)!} a_{1}$ $y = Σ a_{n} x^{n} = \underset{k = 0}{\sum^{\infty}} a_{2 k} x^{2 k} + \underset{k = 0}{\sum^{\infty}} a_{2 k + 1} x^{2 k + 1}$ $= a_{0} \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{(2 k + 2)!} x^{2 k} + a_{1} \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{(2 k + 1)!} x^{2 k + 1}$ $= a_{0} c o s (x) + a_{1} s i n (x)$
	Answered by Olaf last updated on 11/Jan/21

	$r^{2} + 1 = 0 \Rightarrow r = \pm i$ $y = A e^{i x} + B e^{- i x}$ $or y = a \cos x + b \sin x$
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