y ′′ + y′ = sin x cos 2x


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Question Number 86003 by jagoll last updated on 26/Mar/20

$y^{″} + y^{'} = \sin x \cos 2 x$
Commented by mathmax by abdo last updated on 26/Mar/20
$let y^′ =z (e)⇒z^′ +z =sinx cos2x (he)→z^′ +z=0 ⇒(z^′ /z)=−1 ⇒ln∣z∣=−x +λ ⇒z =K e^(−x) mvc method z^′ =K^′ e^(−x) −K e^(−x) (e) ⇒K^′ e^(−x) −K e^(−x) +K e^(−x) =sinx cos(2x) ⇒ K^′ (x) =e^x sin(x)cos(2x) ⇒K(x) =∫_. ^x e^t sint cos(2t)dt sint cos(2t) =cos((π/2)−t)cos(2t) =(1/2){cos((π/2)+t)+cos((π/2)−3t)} =−(1/2)sint +(1/2)sin(3t) ⇒K(x) =−(1/2)∫^x e^t sint dt+(1/2)∫^x e^t sin(3t)dt ∫ e^t sint dt =Im(∫ e^(t+it) dt) =Im(∫ e^((1+i)t) dt) and ∫ e^((1+i)t) dt =(1/(1+i))e^((1+i)t) =((1−i)/2) e^t { cost +isint} =(e^t /2){cost +isint −icost +sint} ⇒∫^x e^t sint dt=(e^t /2)(sint−cost) +c_0 ∫ e^t sin(3t)dt =Im(∫ e^((1+3i)t) dt) and ∫ e^((1+3i)t) dt =(1/(1+3i))e^((1+3i)t) +c_1 =((1−3i)/(10))e^t {cos(3t)+isin(3t)} =(e^t /(10)){ cos(3t)+isin(3t)−3i cos(3t)+3 sin(3t)} ⇒∫^x e^t sin(3t)dt =(e^x /(10)){sin(3x)−3cos(3x)} ⇒ K(x)=−(1/4)e^x ( sinx−cosx) +(e^x /(20))(sin(3x)−3cos(3x)) +C z(x)=K(x) e^(−x) =−(1/4)(sinx−cosx)+(1/(20))(sin(3x)−3cos(3x))+Ce^(−x) y^′ =z ⇒y(x) =∫^x z(u)du +λ =∫^x {−(1/4)(sinu−cosu)+(1/(20))(sin(3u)−3cos(3u)) +C e^(−u) }du +λ y(x)=(1/4)(cosx +sinx) +(1/(20))(−(1/3) sin(3x)−sin(3x))−C e^(−x) +λ$
$l e t y^{^{'}} = z (e) \Rightarrow z^{^{'}} + z = s i n x c o s 2 x$ $(h e) \to z^{^{'}} + z = 0 \Rightarrow \frac{z^{^{'}}}{z} = - 1 \Rightarrow l n ∣ z ∣= - x + λ \Rightarrow z = K e^{- x}$ $m v c m e t h o d z^{^{'}} = K^{^{'}} e^{- x} - K e^{- x}$ $(e) \Rightarrow K^{^{'}} e^{- x} - K e^{- x} + K e^{- x} = s i n x c o s (2 x) \Rightarrow$ $K^{^{'}} (x) = e^{x} s i n (x) c o s (2 x) \Rightarrow K (x) = \int_{.}^{x} e^{t} s i n t c o s (2 t) d t$ $s i n t c o s (2 t) = c o s (\frac{π}{2} - t) c o s (2 t) = \frac{1}{2} {c o s (\frac{π}{2} + t) + c o s (\frac{π}{2} - 3 t)}$ $= - \frac{1}{2} s i n t + \frac{1}{2} s i n (3 t) \Rightarrow K (x) = - \frac{1}{2} \int^{x} e^{t} s i n t d t + \frac{1}{2} \int^{x} e^{t} s i n (3 t) d t$ $\int e^{t} s i n t d t = I m (\int e^{t + i t} d t) = I m (\int e^{(1 + i) t} d t) a n d$ $\int e^{(1 + i) t} d t = \frac{1}{1 + i} e^{(1 + i) t} = \frac{1 - i}{2} e^{t} {c o s t + i s i n t}$ $= \frac{e^{t}}{2} {c o s t + i s i n t - i c o s t + s i n t} \Rightarrow \int^{x} e^{t} s i n t d t = \frac{e^{t}}{2} (s i n t - c o s t) + c_{0}$ $\int e^{t} s i n (3 t) d t = I m (\int e^{(1 + 3 i) t} d t) a n d$ $\int e^{(1 + 3 i) t} d t = \frac{1}{1 + 3 i} e^{(1 + 3 i) t} + c_{1} = \frac{1 - 3 i}{10} e^{t} {c o s (3 t) + i s i n (3 t)}$ $= \frac{e^{t}}{10} {c o s (3 t) + i s i n (3 t) - 3 i c o s (3 t) + 3 s i n (3 t)}$ $\Rightarrow \int^{x} e^{t} s i n (3 t) d t = \frac{e^{x}}{10} {s i n (3 x) - 3 c o s (3 x)} \Rightarrow$ $K (x) = - \frac{1}{4} e^{x} (s i n x - c o s x) + \frac{e^{x}}{20} (s i n (3 x) - 3 c o s (3 x)) + C$ $z (x) = K (x) e^{- x} = - \frac{1}{4} (s i n x - c o s x) + \frac{1}{20} (s i n (3 x) - 3 c o s (3 x)) + {C e}^{- x}$ $y^{^{'}} = z \Rightarrow y (x) = \int^{x} z (u) d u + λ$ $= \int^{x} {- \frac{1}{4} (s i n u - c o s u) + \frac{1}{20} (s i n (3 u) - 3 c o s (3 u)) + C e^{- u}} d u + λ$ $y (x) = \frac{1}{4} (c o s x + s i n x) + \frac{1}{20} (- \frac{1}{3} s i n (3 x) - s i n (3 x)) - C e^{- x} + λ$
	Answered by john santu last updated on 26/Mar/20

	$y_{h} = C_{1} \cos x + C_{2} \sin x$ $y^{″} + y^{'} = \frac{1}{2} \sin 3 x - \frac{1}{2} \sin x$ $y_{p} = A \sin 3 x + B \cos 3 x + C x \sin x + D x \cos x$ $y^{^{'}} = 3 A \cos 3 x - 3 B \sin 3 x + C \sin x$ $+ C x \cos x + D \cos x - D x \sin x$ $y^{″} = - 9 A \sin 3 x - 9 B \cos 3 x + 2 C \cos x$ $- C x \sin x - 2 D \sin x - D x \cos x$ $\Rightarrow A = - \frac{1}{16}, B = 0, C = 0$ $D = \frac{1}{4} . \Rightarrow y_{p} = - \frac{1}{16} \sin 3 x + \frac{x}{4} \sin x$
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