(i) y′′−4y′+5y=4sin^2 4x (ii) (x/2)+1 = (√(∣1−x^2 ∣))


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Question Number 120050 by bramlexs22 last updated on 29/Oct/20

$(i) y^{″} - 4 y^{'} + 5 y = 4 \sin^{2} 4 x$ $(i i) \frac{x}{2} + 1 = \sqrt{∣ 1 - x^{2} ∣}$
	Answered by bemath last updated on 29/Oct/20
	$(•) ((x+2)/2) = (√(∣1−x^2 ∣)) defined on x ≥−2 ⇔ squaring both sides (((x+2)^2 )/4) = ∣1−x^2 ∣ case(1) for −1≤x≤1 ⇒((x^2 +4x+4)/4)=1−x^2 ⇒x^2 +4x+4=4−4x^2 ⇒5x^2 +4x = 0 ; x(5x+4)=0 { ((x_1 =0)),((x_2 =−(4/5))) :} case(2) for x≤−1 ∪ x≥1⇒((x^2 +4x+4)/4)=x^2 −1 ⇒x^2 +4x+4=4x^2 −4 ; 3x^2 −4x−8=0 x = ((4 ± (√(16−4.3.(−8))))/6) = ((4 ± (√(112)))/6) x_(3.4) = ((4±4(√7))/6) = (2/3) ± ((2(√7))/3) The solution set is {−(4/5), ((2−2(√7))/3), 0, ((2+2(√7))/3) }$
	$(∙) \frac{x + 2}{2} = \sqrt{∣ 1 - x^{2} ∣} d e f i n e d o n x ⩾ - 2$ $\Leftrightarrow s q u a r i n g b o t h s i d e s$ $\frac{{(x + 2)}^{2}}{4} = ∣ 1 - x^{2} ∣$ $c a s e (1) f o r - 1 ⩽ x ⩽ 1 \Rightarrow \frac{x^{2} + 4 x + 4}{4} = 1 - x^{2}$ $\Rightarrow x^{2} + 4 x + 4 = 4 - 4 x^{2}$ $\Rightarrow 5 x^{2} + 4 x = 0; x (5 x + 4) = 0 {\begin{cases} x_{1} = 0 \\ x_{2} = - \frac{4}{5} \end{cases}$ $c a s e (2) f o r x ⩽ - 1 \cup x ⩾ 1 \Rightarrow \frac{x^{2} + 4 x + 4}{4} = x^{2} - 1$ $\Rightarrow x^{2} + 4 x + 4 = 4 x^{2} - 4; 3 x^{2} - 4 x - 8 = 0$ $x = \frac{4 \pm \sqrt{16 - 4.3 . (- 8)}}{6} = \frac{4 \pm \sqrt{112}}{6}$ $x_{3.4} = \frac{4 \pm 4 \sqrt{7}}{6} = \frac{2}{3} \pm \frac{2 \sqrt{7}}{3}$ $T h e s o l u t i o n s e t i s {- \frac{4}{5}, \frac{2 - 2 \sqrt{7}}{3}, 0, \frac{2 + 2 \sqrt{7}}{3}}$
	Answered by Lordose last updated on 29/Oct/20
	$i ((d^2 (y))/dx) − 4(dy/dx) + 5y = 4sin^2 4x y(x) = y_c (x) + y_p (x) y_c (x) = ((d^2 (y))/dx) − 4(dy/dx) +5y=0 set y = e^(kx) k^2 e^(kx) − 4ke^(kx) + 5e^(kx) = 0 e^(kx) (k^2 −4k+5)=0 e^(kx) ≠ 0 ∀ k∈R k^2 −4k+5=0 k=2±i where i=(√((−1))) y_1 (x)= c_1 e^((2+i)x) and y_2 (x) = c_2 e^((2+i)x) y_c (x)=y_1 (x) + y_2 (x) Recall: e^(x+iy) = e^x cosy + ie^x siny y_c (x) = c_1 (e^(2x) cos(x) + ie^(2x) sin(x)) + c_2 (e^(2x) cos(x)−ie^(2x) sin(x)) y_c (x) = (c_1 +c_2 )e^(2x) cos(x) + i(c_1 −c_2 )e^(2x) sin(x) let c_1 +c_2 = k_1 and i(c_1 −c_2 )=k_2 y_c (x) = k_1 e^(2x) cos(x) + k_2 e^(2x) sin(x) Next, We find y_p (x) by method of undetermined coefficient y_p (x) = ((d^2 y_p (x))/dx^2 ) − 4((dy_p (x))/dx) + 5y_p (x) = 2−2cos(8x) { 4sin^2 (4x)= 2−cos(8x)} The particular solution of (d^2 y/dx^2 ) − 4(dy/dx) + 5y = 2 is in the form y_(p1) (x)=a_1 And the particular solution of (d^2 y/dx) − 4(dy/dx) + 5y=−2cos(8x) is in the form y_(p2) (x)=a_2 cos(8x) + a_3 sin(8x) y_p (x) = y_(p1) (x) + y_(p2) (x) y_p (x) = a_1 + a_2 cos(8x) + a_3 sin(8x) ((dy_p (x))/dx) = −8a_2 sin(8x) + 8a_3 cos(8x) ((d^2 y_p (x))/dx) = −64a_2 cos(8x) − 64a_3 sin(8x) Sub for y_p (x) −64a_2 cos(8x)−64a_3 sin(8x)−4(8a_3 cos(8x)−8a_2 sin(8x))+5(a_1 +a_2 cos(8x)+a_3 sin(8x)) factorising, we have: 5a_1 + (−59a_2 −32a_3 )cos(8x)+(32a_2 −59a_3 )sin(8x)= 2−2cos(8x) 5a_1 = 2 59a_2 + 32a_3 = 2 32a_2 − 59a_3 = 0 Hence, a_1 = (2/5) , a_2 = ((118)/(4505)) , a_3 = ((64)/(4505)) y_p (x) = a_1 + a_2 cos(8x) + a_3 sin(8x) y_p (x) = (2/5) + ((118cos(8x))/(4505)) + ((64sin(8x))/(4505)) y(x) = y_c (x) + y_p (x) y(x) = k_1 e^(2x) cosx + k_2 e^(2x) sinx + ((118cos(8x))/(4505)) + ((64sin(8x))/(4505)) + (2/5)$
	$i$ $\frac{d^{2} (y)}{dx} - 4 \frac{dy}{dx} + 5 y = {4 \sin}^{2} 4 x$ $y (x) = y_{c} (x) + y_{p} (x)$ $y_{c} (x) = \frac{d^{2} (y)}{dx} - 4 \frac{dy}{dx} + 5 y = 0$ $set y = e^{kx}$ $k^{2} e^{kx} - {4 ke}^{kx} + {5 e}^{kx} = 0$ $e^{kx} (k^{2} - 4 k + 5) = 0$ $e^{kx} \neq 0 \forall k \in R$ $k^{2} - 4 k + 5 = 0$ $k = 2 \pm i where i = \sqrt{(- 1)}$ $y_{1} (x) = c_{1} e^{(2 + i) x} and y_{2} (x) = c_{2} e^{(2 + i) x}$ $y_{c} (x) = y_{1} (x) + y_{2} (x)$ $Recall : e^{x + iy} = e^{x} cosy + {ie}^{x} siny$ $y_{c} (x) = c_{1} (e^{2 x} \cos (x) + {ie}^{2 x} \sin (x)) + c_{2} (e^{2 x} \cos (x) - {ie}^{2 x} \sin (x))$ $y_{c} (x) = (c_{1} + c_{2}) e^{2 x} \cos (x) + i (c_{1} - c_{2}) e^{2 x} \sin (x)$ $let c_{1} + c_{2} = k_{1} and i (c_{1} - c_{2}) = k_{2}$ $y_{c} (x) = k_{1} e^{2 x} \cos (x) + k_{2} e^{2 x} \sin (x)$ $Next, We find y_{p} (x) by method of undetermined coefficient$ $y_{p} (x) = \frac{d^{2} y_{p} (x)}{{dx}^{2}} - 4 \frac{{dy}_{p} (x)}{dx} + {5 y}_{p} (x) = 2 - 2 \cos (8 x) {{4 \sin}^{2} (4 x) = 2 - \cos (8 x)}$ $The particular solution of$ $\frac{d^{2} y}{{dx}^{2}} - 4 \frac{dy}{dx} + 5 y = 2 is in the form y_{p 1} (x) = a_{1}$ $And the particular solution of \frac{d^{2} y}{dx} - 4 \frac{dy}{dx} + 5 y = - 2 \cos (8 x) is in the form$ $y_{p 2} (x) = a_{2} \cos (8 x) + a_{3} \sin (8 x)$ $y_{p} (x) = y_{p 1} (x) + y_{p 2} (x)$ $y_{p} (x) = a_{1} + a_{2} \cos (8 x) + a_{3} \sin (8 x)$ $\frac{{dy}_{p} (x)}{dx} = - {8 a}_{2} \sin (8 x) + {8 a}_{3} \cos (8 x)$ $\frac{d^{2} y_{p} (x)}{dx} = - {64 a}_{2} \cos (8 x) - {64 a}_{3} \sin (8 x)$ $Sub for y_{p} (x)$ $- {64 a}_{2} \cos (8 x) - {64 a}_{3} \sin (8 x) - 4 ({8 a}_{3} \cos (8 x) - {8 a}_{2} \sin (8 x)) + 5 (a_{1} + a_{2} \cos (8 x) + a_{3} \sin (8 x))$ $factorising, we have :$ ${5 a}_{1} + (- {59 a}_{2} - {32 a}_{3}) \cos (8 x) + ({32 a}_{2} - {59 a}_{3}) \sin (8 x) = 2 - 2 \cos (8 x)$ ${5 a}_{1} = 2$ ${59 a}_{2} + {32 a}_{3} = 2$ ${32 a}_{2} - {59 a}_{3} = 0$ $Hence, a_{1} = \frac{2}{5}, a_{2} = \frac{118}{4505}, a_{3} = \frac{64}{4505}$ $y_{p} (x) = a_{1} + a_{2} \cos (8 x) + a_{3} \sin (8 x)$ $y_{p} (x) = \frac{2}{5} + \frac{118 \cos (8 x)}{4505} + \frac{64 \sin (8 x)}{4505}$ $y (x) = y_{c} (x) + y_{p} (x)$ $y (x) = k_{1} e^{2 x} cosx + k_{2} e^{2 x} sinx + \frac{118 \cos (8 x)}{4505} + \frac{64 \sin (8 x)}{4505} + \frac{2}{5}$
	Answered by Bird last updated on 29/Oct/20
	$y^(′′) −4y^′ +5y =4×((1−cos(8x))/2) ⇒ y^(′′) −4y^(′ ) +5y =2−2cos(8x) h→r^2 −4r+5=0→Δ^′ =4−5=−1 ⇒r_1 =2+i and r_2 =2−i y_h =ae^((2+i)x) +be^((2−i)x) =e^(2x) {αcosx +βsinx} =α e^(2x) cosx +β e^(2x) sinx =αu_1 +βu_2 W(u_1 ,u_2 )= determinant (((e^(2x) cosx e^(2x) sinx)),((e^(2x) (2cosx−sinx) e^(2x) (2sinx +cosx) ))) =e^(4x) (2sinx+cosx)cosx−e^(4x) (2cosx−sinx)sinx =e^(4x) {2sinx cosx+cos^2 x−2cosxsinx+sin^2 x} =e^(4x ) ≠0 W_1 = determinant (((o e^(2x) sinx)),((2−2cos(8x) e^(2x) (2sinx+cosx)))) =−2e^(2x) (1−cos(8x)sinx W_2 = determinant (((e^(2x) cosx 0)),((e^(2x) (2cosx−sinx) 2−2cos(8x)))) =2e^(2x) (1−cos(8x))cosx V_1 =∫ (W_1 /W)dx =−∫((2e^(2x) (1−cos)8x)sinx)/e^(4x) )dx =−2 ∫ e^(−2x) (1−cos(8x))sinx dx =−2∫ e^(−2x) sinx dx+2∫ e^(−2x) cos(8x)sinx dx ∫ e^(−2x) sinx dx =Im(∫ e^(−2x+ix) dx) ∫ e^((−2+i)x) dx =(1/(−2+i))e^((−2+i)x) =((−1(2+i)e^(−2x) (cosx +isinx))/5) =−(1/5)e^(−2x) (2cosx+2isinx+icosx −sinx) ⇒ ∫ e^(−2x) sinx dx =−(1/5)e^(−2x) {2sinx +cosx} we have cos(8x)sinx =cos(8x)cos((π/2)−x) =(1/2){cos(7x+(π/2))+cos(9x−(π/2))) =(1/2){−sin(7x)+sin(9x)} ⇒ 2∫ e^(−2x) cos(8x)sinx dx =∫ e^(−2x) sin(9x)dx−∫ e^(−2x) sin(7x)dx =....(eszy to find) V_2 =∫ (W_2 /W)dx =∫ ((2e^(2x) (1−cos(8x)cosx)/e^(4x) )dx =∫ 2 e^(−2x) cosx dx−2∫ e^(2x) cos(8x)cosxdx =....(use same way...) ⇒ y_p =u_1 v_1 +u_2 v_2 and general solution is y =y_p +y_h$
	$y^{^{″}} - 4 y^{^{'}} + 5 y = 4 \times \frac{1 - c o s (8 x)}{2} \Rightarrow$ $y^{^{″}} - 4 y^{^{'}} + 5 y = 2 - 2 c o s (8 x)$ $h \to r^{2} - 4 r + 5 = 0 \to Δ^{^{'}} = 4 - 5 = - 1$ $\Rightarrow r_{1} = 2 + i a n d r_{2} = 2 - i$ $y_{h} = {a e}^{(2 + i) x} + {b e}^{(2 - i) x}$ $= e^{2 x} {α c o s x + β s i n x}$ $= α e^{2 x} c o s x + β e^{2 x} s i n x = α u_{1} + β u_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} e^{2 x} c o s x e^{2 x} s i n x \\ e^{2 x} (2 c o s x - s i n x) e^{2 x} (2 s i n x + c o s x) \end{matrix} \|$ $= e^{4 x} (2 s i n x + c o s x) c o s x - e^{4 x} (2 c o s x - s i n x) s i n x$ $= e^{4 x} {2 s i n x c o s x + {c o s}^{2} x - 2 c o s x s i n x + {s i n}^{2} x}$ $= e^{4 x} \neq 0$ $W_{1} = \| \begin{matrix} o e^{2 x} s i n x \\ 2 - 2 c o s (8 x) e^{2 x} (2 s i n x + c o s x) \end{matrix} \|$ $= - 2 e^{2 x} (1 - c o s (8 x) s i n x$ $W_{2} = \| \begin{matrix} e^{2 x} c o s x 0 \\ e^{2 x} (2 c o s x - s i n x) 2 - 2 c o s (8 x) \end{matrix} \|$ $= 2 e^{2 x} (1 - c o s (8 x)) c o s x$ $V_{1} = \int \frac{W_{1}}{W} d x = - \int \frac{2 e^{2 x} (1 - c o s) 8 x) s i n x}{e^{4 x}} d x$ $= - 2 \int e^{- 2 x} (1 - c o s (8 x)) s i n x d x$ $= - 2 \int e^{- 2 x} s i n x d x + 2 \int e^{- 2 x} c o s (8 x) s i n x d x$ $\int e^{- 2 x} s i n x d x = I m (\int e^{- 2 x + i x} d x)$ $\int e^{(- 2 + i) x} d x = \frac{1}{- 2 + i} e^{(- 2 + i) x}$ $= \frac{- 1 (2 + i) e^{- 2 x} (c o s x + i s i n x)}{5}$ $= - \frac{1}{5} e^{- 2 x} (2 c o s x + 2 i s i n x + i c o s x$ $- s i n x) \Rightarrow$ $\int e^{- 2 x} s i n x d x = - \frac{1}{5} e^{- 2 x} {2 s i n x + c o s x}$ $w e h a v e c o s (8 x) s i n x$ $= c o s (8 x) c o s (\frac{π}{2} - x)$ $= \frac{1}{2} {c o s (7 x + \frac{π}{2}) + c o s (9 x - \frac{π}{2}))$ $= \frac{1}{2} {- s i n (7 x) + s i n (9 x)} \Rightarrow$ $2 \int e^{- 2 x} c o s (8 x) s i n x d x$ $= \int e^{- 2 x} s i n (9 x) d x - \int e^{- 2 x} s i n (7 x) d x$ $= . . . . (e s z y t o f i n d)$ $V_{2} = \int \frac{W_{2}}{W} d x = \int \frac{2 e^{2 x} (1 - c o s (8 x) c o s x}{e^{4 x}} d x$ $= \int 2 e^{- 2 x} c o s x d x - 2 \int e^{2 x} c o s (8 x) c o s x d x$ $= . . . . (u s e s a m e w a y . . .) \Rightarrow$ $y_{p} = u_{1} v_{1} + u_{2} v_{2} a n d g e n e r a l s o l u t i o n$ $i s y = y_{p} + y_{h}$
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