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Question Number 181858 by KINMATICS last updated on 01/Dec/22

	Answered by hmr last updated on 01/Dec/22

	$T h e g e n e r a l s o l u t i o n o f$ $t h e l i n e a r n o n h o m o g e n e o u s$ $e q u a t i o n i s e q u a l t o :$ $(1) g e n e r a l s o l u t i o n o f t h e$ $c o r r e s p o n d i n g h o m o g e n e o u s e q u a t i o n$ $+$ $(2) p a r t i c u l a r s o l u t i o n o f t h e$ $n o n h o m o g e n e o u s e q u a t i o n .$ $(1)$ $y^{″} + 4 y = 0$ $\to r^{2} + 4 = 0 \to r = \pm 2 i$ $y_{h} (x) = c_{1} C o s (2 x) + c_{2} S i n (2 x)$ $(2)$ $y (x) = (A_{1} x + A_{0}) C o s (x) + (B_{1} x + B_{0}) S i n (x)$ $y^{'} (x) = (A_{1}) C o s (x) - (A_{1} x + A_{0}) S i n (x)$ $+ (B_{1}) S i n (x) + (B_{1} x + B_{0}) C o s (x)$ $\to$ $y^{'} (x) = (A_{1} + B_{1} x + B_{0}) C o s (x) + (- A_{1} x - A_{0} + B_{1}) S i n (x)$ $y^{″} (x) = (B_{1}) C o s (x) - (A_{1} + B_{1} x + B_{0}) S i n (x)$ $+ (- A_{1}) S i n (x) + (- A_{1} x - A_{0} + B_{1}) C o s (x)$ $\to$ $y^{″} (x) = (- A_{1} x - A_{0} + 2 B_{1}) C o s (x) - (2 A_{1} + B_{1} x + B_{0}) S i n (x)$ $∙ y^{″} + 4 y = x C o s (x)$ $\to (- A_{1} x - A_{0} + 2 B_{1}) C o s (x) - (2 A_{1} + B_{1} x + B_{0}) S i n (x)$ $+ 4 ((A_{1} x + A_{0}) C o s (x) + (B_{1} x + B_{0}) S i n (x))$ $= x C o s (x)$ $\to$ $(3 A_{1} x + 3 A_{0} + 2 B_{1}) C o s (x) + (- 2 A_{1} + 3 B_{1} x + 3 B_{0}) S i n (x)$ $= x C o s (x)$ $\to A_{0} = B_{1} = 0 A_{1} = \frac{1}{3} B_{0} = \frac{2}{9}$ $\to$ $y (x) = \frac{1}{3} x C o s (x) + \frac{2}{9} S i n (x)$ $t h u s t h e a n s w e r i s :$ $φ (x) = c_{1} C o s (2 x) + c_{2} S i n (2 x) + \frac{1}{3} x C o s (x) + \frac{2}{9} S i n (x)$
	Commented by hmr last updated on 01/Dec/22

	$φ (x) = c_{1} C o s (2 x) + c_{2} S i n (2 x) + \frac{1}{3} x C o s (x) + \frac{2}{9} S i n (x)$ $∙ φ (0) = 1$ $\to$ $c_{1} = 1$ $∙ ∙ φ (\frac{3 π}{4}) = \frac{- π \sqrt{2}}{8}$ $\to$ $- c_{2} - \frac{π \sqrt{2}}{8} + \frac{\sqrt{2}}{9} = \frac{- π \sqrt{2}}{8}$ $\to$ $c_{2} = \frac{\sqrt{2}}{9}$ $φ (x) = C o s (2 x) + \frac{\sqrt{2}}{9} S i n (2 x) + \frac{1}{3} x C o s (x) + \frac{2}{9} S i n (x)$ $1 - \frac{π}{3}$ $φ (π) + φ (- π) = (1 - \frac{π}{3}) + (1 + \frac{π}{3}) = 2$
	Answered by qaz last updated on 01/Dec/22

	$y_{p} = \frac{1}{D^{2} + 4} x \cos x = (x - \frac{2 D}{D^{2} + 4}) \frac{1}{D^{2} + 4} \cos x$ $= (x - \frac{2 D}{D^{2} + 4}) \frac{1}{3} \cos x = \frac{1}{3} x \cos x + \frac{2}{9} \sin x$ $\Rightarrow y = C_{1} \sin 2 x + C_{2} \cos 2 x + \frac{1}{3} x \cos x + \frac{2}{9} \sin x$ $y (0) = C_{2} = 1$ $y (\frac{3 π}{4}) = - C_{1} - \frac{π}{4 \sqrt{2}} + \frac{\sqrt{2}}{9} = - \frac{π}{4 \sqrt{2}} \Rightarrow C_{1} = \frac{\sqrt{2}}{9}$ $\Rightarrow y = \frac{\sqrt{2}}{9} \sin 2 x + \cos 2 x + \frac{1}{3} x \cos x + \frac{2}{9} \sin x$ $\Rightarrow y (π) + y (- π) = \frac{\sqrt{2}}{9}$
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