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Question Number 120050 by bramlexs22 last updated on 29/Oct/20
(i) y′′−4y′+5y=4sin^2 4x (ii) (x/2)+1 = (√(∣1−x^2 ∣))
$$\:\left({i}\right)\:{y}''−\mathrm{4}{y}'+\mathrm{5}{y}=\mathrm{4sin}\:^{\mathrm{2}} \mathrm{4}{x} \\ $$$$\:\left({ii}\right)\:\frac{{x}}{\mathrm{2}}+\mathrm{1}\:=\:\sqrt{\mid\mathrm{1}−{x}^{\mathrm{2}} \mid}\: \\ $$
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) }
$$\left(\bullet\right)\:\frac{{x}+\mathrm{2}}{\mathrm{2}}\:=\:\sqrt{\mid\mathrm{1}−{x}^{\mathrm{2}} \mid}\:{defined}\:{on}\:{x}\:\geqslant−\mathrm{2}\: \\ $$$$\Leftrightarrow\:{squaring}\:{both}\:{sides} \\ $$$$\frac{\left({x}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}\:=\:\mid\mathrm{1}−{x}^{\mathrm{2}} \mid\: \\ $$$${case}\left(\mathrm{1}\right)\:{for}\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1}\:\Rightarrow\frac{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}}{\mathrm{4}}=\mathrm{1}−{x}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}=\mathrm{4}−\mathrm{4}{x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{5}{x}^{\mathrm{2}} +\mathrm{4}{x}\:=\:\mathrm{0}\:;\:{x}\left(\mathrm{5}{x}+\mathrm{4}\right)=\mathrm{0}\:\begin{cases}{{x}_{\mathrm{1}} =\mathrm{0}}\\{{x}_{\mathrm{2}} =−\frac{\mathrm{4}}{\mathrm{5}}}\end{cases} \\ $$$${case}\left(\mathrm{2}\right)\:{for}\:{x}\leqslant−\mathrm{1}\:\cup\:{x}\geqslant\mathrm{1}\Rightarrow\frac{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}}{\mathrm{4}}={x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}\:;\:\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{8}=\mathrm{0} \\ $$$${x}\:=\:\frac{\mathrm{4}\:\pm\:\sqrt{\mathrm{16}−\mathrm{4}.\mathrm{3}.\left(−\mathrm{8}\right)}}{\mathrm{6}}\:=\:\frac{\mathrm{4}\:\pm\:\sqrt{\mathrm{112}}}{\mathrm{6}} \\ $$$${x}_{\mathrm{3}.\mathrm{4}} \:=\:\frac{\mathrm{4}\pm\mathrm{4}\sqrt{\mathrm{7}}}{\mathrm{6}}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\pm\:\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{3}} \\ $$$${The}\:{solution}\:{set}\:{is}\:\left\{−\frac{\mathrm{4}}{\mathrm{5}},\:\frac{\mathrm{2}−\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{3}},\:\mathrm{0},\:\frac{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{3}}\:\right\} \\ $$
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)
$$\boldsymbol{\mathrm{i}}\: \\ $$$$\frac{\mathrm{d}^{\mathrm{2}} \left(\mathrm{y}\right)}{\mathrm{dx}}\:−\:\mathrm{4}\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{5y}\:=\:\mathrm{4sin}^{\mathrm{2}} \mathrm{4x} \\ $$$$\mathrm{y}\left(\mathrm{x}\right)\:=\:\mathrm{y}_{\mathrm{c}} \left(\mathrm{x}\right)\:+\:\mathrm{y}_{\mathrm{p}} \left(\mathrm{x}\right) \\ $$$$\mathrm{y}_{\mathrm{c}} \left(\mathrm{x}\right)\:=\:\frac{\mathrm{d}^{\mathrm{2}} \left(\mathrm{y}\right)}{\mathrm{dx}}\:−\:\mathrm{4}\frac{\mathrm{dy}}{\mathrm{dx}}\:+\mathrm{5y}=\mathrm{0} \\ $$$$\mathrm{set}\:\mathrm{y}\:=\:\mathrm{e}^{\mathrm{kx}} \\ $$$$\mathrm{k}^{\mathrm{2}} \mathrm{e}^{\mathrm{kx}} \:−\:\mathrm{4ke}^{\mathrm{kx}} \:+\:\mathrm{5e}^{\mathrm{kx}} \:=\:\mathrm{0} \\ $$$$\mathrm{e}^{\mathrm{kx}} \left(\mathrm{k}^{\mathrm{2}} −\mathrm{4k}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\mathrm{e}^{\mathrm{kx}} \:\neq\:\mathrm{0}\:\forall\:\mathrm{k}\in\mathbb{R} \\ $$$$\mathrm{k}^{\mathrm{2}} −\mathrm{4k}+\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{k}=\mathrm{2}\pm\boldsymbol{\mathrm{i}}\:\mathrm{where}\:\mathrm{i}=\sqrt{\left(−\mathrm{1}\right)} \\ $$$$\mathrm{y}_{\mathrm{1}} \left(\mathrm{x}\right)=\:\mathrm{c}_{\mathrm{1}} \mathrm{e}^{\left(\mathrm{2}+\mathrm{i}\right)\mathrm{x}} \:\:\mathrm{and}\:\mathrm{y}_{\mathrm{2}} \left(\mathrm{x}\right)\:=\:\mathrm{c}_{\mathrm{2}} \mathrm{e}^{\left(\mathrm{2}+\mathrm{i}\right)\mathrm{x}} \\ $$$$\mathrm{y}_{\mathrm{c}} \left(\mathrm{x}\right)=\mathrm{y}_{\mathrm{1}} \left(\mathrm{x}\right)\:+\:\mathrm{y}_{\mathrm{2}} \left(\mathrm{x}\right) \\ $$$$\mathrm{Recall}:\:\mathrm{e}^{\mathrm{x}+\mathrm{iy}} \:=\:\mathrm{e}^{\mathrm{x}} \mathrm{cosy}\:+\:\mathrm{ie}^{\mathrm{x}} \mathrm{siny} \\ $$$$\mathrm{y}_{\mathrm{c}} \left(\mathrm{x}\right)\:=\:\mathrm{c}_{\mathrm{1}} \left(\mathrm{e}^{\mathrm{2x}} \mathrm{cos}\left(\mathrm{x}\right)\:+\:\mathrm{ie}^{\mathrm{2x}} \mathrm{sin}\left(\mathrm{x}\right)\right)\:+\:\mathrm{c}_{\mathrm{2}} \left(\mathrm{e}^{\mathrm{2x}} \mathrm{cos}\left(\mathrm{x}\right)−\mathrm{ie}^{\mathrm{2x}} \mathrm{sin}\left(\mathrm{x}\right)\right) \\ $$$$\mathrm{y}_{\mathrm{c}} \left(\mathrm{x}\right)\:=\:\left(\mathrm{c}_{\mathrm{1}} +\mathrm{c}_{\mathrm{2}} \right)\mathrm{e}^{\mathrm{2x}} \mathrm{cos}\left(\mathrm{x}\right)\:+\:\mathrm{i}\left(\mathrm{c}_{\mathrm{1}} −\mathrm{c}_{\mathrm{2}} \right)\mathrm{e}^{\mathrm{2x}} \mathrm{sin}\left(\mathrm{x}\right) \\ $$$$\mathrm{let}\:\mathrm{c}_{\mathrm{1}} +\mathrm{c}_{\mathrm{2}} =\:\mathrm{k}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{i}\left(\mathrm{c}_{\mathrm{1}} −\mathrm{c}_{\mathrm{2}} \right)=\mathrm{k}_{\mathrm{2}} \\ $$$$\mathrm{y}_{\mathrm{c}} \left(\mathrm{x}\right)\:=\:\mathrm{k}_{\mathrm{1}} \mathrm{e}^{\mathrm{2x}} \mathrm{cos}\left(\mathrm{x}\right)\:+\:\mathrm{k}_{\mathrm{2}} \mathrm{e}^{\mathrm{2x}} \mathrm{sin}\left(\mathrm{x}\right) \\ $$$$\boldsymbol{\mathrm{Next}},\:\mathrm{We}\:\mathrm{find}\:\mathrm{y}_{\mathrm{p}} \left(\mathrm{x}\right)\:\mathrm{by}\:\mathrm{method}\:\mathrm{of}\:\mathrm{undetermined}\:\mathrm{coefficient} \\ $$$$\mathrm{y}_{\mathrm{p}} \left(\mathrm{x}\right)\:=\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}_{\mathrm{p}} \left(\mathrm{x}\right)}{\mathrm{dx}^{\mathrm{2}} }\:−\:\mathrm{4}\frac{\mathrm{dy}_{\mathrm{p}} \left(\mathrm{x}\right)}{\mathrm{dx}}\:+\:\mathrm{5y}_{\mathrm{p}} \left(\mathrm{x}\right)\:=\:\mathrm{2}−\mathrm{2cos}\left(\mathrm{8x}\right)\:\left\{\:\mathrm{4sin}^{\mathrm{2}} \left(\mathrm{4x}\right)=\:\mathrm{2}−\mathrm{cos}\left(\mathrm{8x}\right)\right\} \\ $$$$\mathrm{The}\:\mathrm{particular}\:\mathrm{solution}\:\mathrm{of} \\ $$$$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:−\:\mathrm{4}\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{5y}\:=\:\mathrm{2}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{y}_{\mathrm{p1}} \left(\mathrm{x}\right)=\mathrm{a}_{\mathrm{1}} \\ $$$$\mathrm{And}\:\mathrm{the}\:\mathrm{particular}\:\mathrm{solution}\:\mathrm{of}\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}}\:−\:\mathrm{4}\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{5y}=−\mathrm{2cos}\left(\mathrm{8x}\right)\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form} \\ $$$$\mathrm{y}_{\mathrm{p2}} \left(\mathrm{x}\right)=\mathrm{a}_{\mathrm{2}} \mathrm{cos}\left(\mathrm{8x}\right)\:+\:\mathrm{a}_{\mathrm{3}} \mathrm{sin}\left(\mathrm{8x}\right) \\ $$$$\mathrm{y}_{\mathrm{p}} \left(\mathrm{x}\right)\:=\:\mathrm{y}_{\mathrm{p1}} \left(\mathrm{x}\right)\:+\:\mathrm{y}_{\mathrm{p2}} \left(\mathrm{x}\right) \\ $$$$\mathrm{y}_{\mathrm{p}} \left(\mathrm{x}\right)\:=\:\mathrm{a}_{\mathrm{1}} \:+\:\mathrm{a}_{\mathrm{2}} \mathrm{cos}\left(\mathrm{8x}\right)\:+\:\mathrm{a}_{\mathrm{3}} \mathrm{sin}\left(\mathrm{8x}\right) \\ $$$$\frac{\mathrm{dy}_{\mathrm{p}} \left(\mathrm{x}\right)}{\mathrm{dx}}\:=\:−\mathrm{8a}_{\mathrm{2}} \mathrm{sin}\left(\mathrm{8x}\right)\:+\:\mathrm{8a}_{\mathrm{3}} \mathrm{cos}\left(\mathrm{8x}\right) \\ $$$$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}_{\mathrm{p}} \left(\mathrm{x}\right)}{\mathrm{dx}}\:=\:−\mathrm{64a}_{\mathrm{2}} \mathrm{cos}\left(\mathrm{8x}\right)\:−\:\mathrm{64a}_{\mathrm{3}} \mathrm{sin}\left(\mathrm{8x}\right) \\ $$$$\mathrm{Sub}\:\mathrm{for}\:\mathrm{y}_{\mathrm{p}} \left(\mathrm{x}\right) \\ $$$$−\mathrm{64a}_{\mathrm{2}} \mathrm{cos}\left(\mathrm{8x}\right)−\mathrm{64a}_{\mathrm{3}} \mathrm{sin}\left(\mathrm{8x}\right)−\mathrm{4}\left(\mathrm{8a}_{\mathrm{3}} \mathrm{cos}\left(\mathrm{8x}\right)−\mathrm{8a}_{\mathrm{2}} \mathrm{sin}\left(\mathrm{8x}\right)\right)+\mathrm{5}\left(\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} \mathrm{cos}\left(\mathrm{8x}\right)+\mathrm{a}_{\mathrm{3}} \mathrm{sin}\left(\mathrm{8x}\right)\right) \\ $$$$\mathrm{factorising},\:\mathrm{we}\:\mathrm{have}: \\ $$$$\mathrm{5a}_{\mathrm{1}} \:+\:\left(−\mathrm{59a}_{\mathrm{2}} −\mathrm{32a}_{\mathrm{3}} \right)\mathrm{cos}\left(\mathrm{8x}\right)+\left(\mathrm{32a}_{\mathrm{2}} −\mathrm{59a}_{\mathrm{3}} \right)\mathrm{sin}\left(\mathrm{8x}\right)=\:\mathrm{2}−\mathrm{2cos}\left(\mathrm{8x}\right)\: \\ $$$$\mathrm{5a}_{\mathrm{1}} \:=\:\mathrm{2}\: \\ $$$$\mathrm{59a}_{\mathrm{2}} \:+\:\mathrm{32a}_{\mathrm{3}} \:=\:\mathrm{2} \\ $$$$\mathrm{32a}_{\mathrm{2}} \:−\:\mathrm{59a}_{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$\mathrm{Hence},\:\mathrm{a}_{\mathrm{1}} \:=\:\frac{\mathrm{2}}{\mathrm{5}}\:,\:\mathrm{a}_{\mathrm{2}} \:=\:\frac{\mathrm{118}}{\mathrm{4505}}\:,\:\mathrm{a}_{\mathrm{3}} \:=\:\frac{\mathrm{64}}{\mathrm{4505}} \\ $$$$\mathrm{y}_{\mathrm{p}} \left(\mathrm{x}\right)\:=\:\mathrm{a}_{\mathrm{1}} \:+\:\mathrm{a}_{\mathrm{2}} \mathrm{cos}\left(\mathrm{8x}\right)\:+\:\mathrm{a}_{\mathrm{3}} \mathrm{sin}\left(\mathrm{8x}\right) \\ $$$$\mathrm{y}_{\mathrm{p}} \left(\mathrm{x}\right)\:=\:\frac{\mathrm{2}}{\mathrm{5}}\:+\:\frac{\mathrm{118cos}\left(\mathrm{8x}\right)}{\mathrm{4505}}\:+\:\frac{\mathrm{64sin}\left(\mathrm{8x}\right)}{\mathrm{4505}} \\ $$$$\mathrm{y}\left(\mathrm{x}\right)\:=\:\mathrm{y}_{\mathrm{c}} \left(\mathrm{x}\right)\:+\:\mathrm{y}_{\mathrm{p}} \left(\mathrm{x}\right) \\ $$$$\mathrm{y}\left(\mathrm{x}\right)\:=\:\mathrm{k}_{\mathrm{1}} \mathrm{e}^{\mathrm{2x}} \mathrm{cosx}\:+\:\mathrm{k}_{\mathrm{2}} \mathrm{e}^{\mathrm{2x}} \mathrm{sinx}\:+\:\frac{\mathrm{118cos}\left(\mathrm{8x}\right)}{\mathrm{4505}}\:+\:\frac{\mathrm{64sin}\left(\mathrm{8x}\right)}{\mathrm{4505}}\:+\:\frac{\mathrm{2}}{\mathrm{5}} \\ $$
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
$${y}^{''} −\mathrm{4}{y}^{'} \:+\mathrm{5}{y}\:=\mathrm{4}×\frac{\mathrm{1}−{cos}\left(\mathrm{8}{x}\right)}{\mathrm{2}}\:\Rightarrow \\ $$$${y}^{''} −\mathrm{4}{y}^{'\:} +\mathrm{5}{y}\:=\mathrm{2}−\mathrm{2}{cos}\left(\mathrm{8}{x}\right) \\ $$$${h}\rightarrow{r}^{\mathrm{2}} −\mathrm{4}{r}+\mathrm{5}=\mathrm{0}\rightarrow\Delta^{'} =\mathrm{4}−\mathrm{5}=−\mathrm{1} \\ $$$$\Rightarrow{r}_{\mathrm{1}} =\mathrm{2}+{i}\:{and}\:{r}_{\mathrm{2}} =\mathrm{2}−{i} \\ $$$${y}_{{h}} ={ae}^{\left(\mathrm{2}+{i}\right){x}} \:+{be}^{\left(\mathrm{2}−{i}\right){x}} \\ $$$$={e}^{\mathrm{2}{x}} \left\{\alpha{cosx}\:+\beta{sinx}\right\} \\ $$$$=\alpha\:{e}^{\mathrm{2}{x}} {cosx}\:+\beta\:{e}^{\mathrm{2}{x}} {sinx}\:=\alpha{u}_{\mathrm{1}} \:+\beta{u}_{\mathrm{2}} \\ $$$${W}\left({u}_{\mathrm{1}} ,{u}_{\mathrm{2}} \right)=\begin{vmatrix}{{e}^{\mathrm{2}{x}} {cosx}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}^{\mathrm{2}{x}} {sinx}}\\{{e}^{\mathrm{2}{x}} \left(\mathrm{2}{cosx}−{sinx}\right)\:\:\:\:{e}^{\mathrm{2}{x}} \left(\mathrm{2}{sinx}\:+{cosx}\right)\:\:\:\:\:}\end{vmatrix} \\ $$$$={e}^{\mathrm{4}{x}} \left(\mathrm{2}{sinx}+{cosx}\right){cosx}−{e}^{\mathrm{4}{x}} \left(\mathrm{2}{cosx}−{sinx}\right){sinx} \\ $$$$={e}^{\mathrm{4}{x}} \left\{\mathrm{2}{sinx}\:{cosx}+{cos}^{\mathrm{2}} {x}−\mathrm{2}{cosxsinx}+{sin}^{\mathrm{2}} {x}\right\} \\ $$$$={e}^{\mathrm{4}{x}\:} \neq\mathrm{0} \\ $$$${W}_{\mathrm{1}} =\begin{vmatrix}{{o}\:\:\:\:\:\:\:\:\:\:\:\:{e}^{\mathrm{2}{x}} {sinx}}\\{\mathrm{2}−\mathrm{2}{cos}\left(\mathrm{8}{x}\right)\:\:{e}^{\mathrm{2}{x}} \left(\mathrm{2}{sinx}+{cosx}\right)}\end{vmatrix} \\ $$$$=−\mathrm{2}{e}^{\mathrm{2}{x}} \left(\mathrm{1}−{cos}\left(\mathrm{8}{x}\right){sinx}\right. \\ $$$${W}_{\mathrm{2}} =\begin{vmatrix}{{e}^{\mathrm{2}{x}} {cosx}\:\:\:\:\:\:\:\:\mathrm{0}}\\{{e}^{\mathrm{2}{x}} \left(\mathrm{2}{cosx}−{sinx}\right)\:\:\:\mathrm{2}−\mathrm{2}{cos}\left(\mathrm{8}{x}\right)}\end{vmatrix} \\ $$$$=\mathrm{2}{e}^{\mathrm{2}{x}} \left(\mathrm{1}−{cos}\left(\mathrm{8}{x}\right)\right){cosx} \\ $$$${V}_{\mathrm{1}} =\int\:\frac{{W}_{\mathrm{1}} }{{W}}{dx}\:=−\int\frac{\left.\mathrm{2}{e}^{\mathrm{2}{x}} \left(\mathrm{1}−{cos}\right)\mathrm{8}{x}\right){sinx}}{{e}^{\mathrm{4}{x}} }{dx} \\ $$$$=−\mathrm{2}\:\int\:{e}^{−\mathrm{2}{x}} \left(\mathrm{1}−{cos}\left(\mathrm{8}{x}\right)\right){sinx}\:{dx} \\ $$$$=−\mathrm{2}\int\:{e}^{−\mathrm{2}{x}} {sinx}\:{dx}+\mathrm{2}\int\:{e}^{−\mathrm{2}{x}} {cos}\left(\mathrm{8}{x}\right){sinx}\:{dx} \\ $$$$\int\:{e}^{−\mathrm{2}{x}} {sinx}\:{dx}\:={Im}\left(\int\:{e}^{−\mathrm{2}{x}+{ix}} {dx}\right) \\ $$$$\int\:{e}^{\left(−\mathrm{2}+{i}\right){x}} {dx}\:=\frac{\mathrm{1}}{−\mathrm{2}+{i}}{e}^{\left(−\mathrm{2}+\boldsymbol{{i}}\right)\boldsymbol{{x}}} \\ $$$$=\frac{−\mathrm{1}\left(\mathrm{2}+\boldsymbol{{i}}\right)\boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{x}}} \left(\boldsymbol{{cosx}}\:+\boldsymbol{{isinx}}\right)}{\mathrm{5}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{5}}\boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{x}}} \left(\mathrm{2}\boldsymbol{{cosx}}+\mathrm{2}\boldsymbol{{isinx}}+\boldsymbol{{icosx}}\right. \\ $$$$\left.−\boldsymbol{{sinx}}\right)\:\Rightarrow \\ $$$$\int\:{e}^{−\mathrm{2}{x}} {sinx}\:{dx}\:=−\frac{\mathrm{1}}{\mathrm{5}}{e}^{−\mathrm{2}{x}} \left\{\mathrm{2}{sinx}\:+{cosx}\right\} \\ $$$${we}\:{have}\:{cos}\left(\mathrm{8}{x}\right){sinx} \\ $$$$={cos}\left(\mathrm{8}{x}\right){cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\mathrm{7}{x}+\frac{\pi}{\mathrm{2}}\right)+{cos}\left(\mathrm{9}{x}−\frac{\pi}{\mathrm{2}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{−{sin}\left(\mathrm{7}{x}\right)+{sin}\left(\mathrm{9}{x}\right)\right\}\:\Rightarrow \\ $$$$\mathrm{2}\int\:{e}^{−\mathrm{2}{x}} {cos}\left(\mathrm{8}{x}\right){sinx}\:{dx} \\ $$$$=\int\:{e}^{−\mathrm{2}{x}} \:{sin}\left(\mathrm{9}{x}\right){dx}−\int\:{e}^{−\mathrm{2}{x}} {sin}\left(\mathrm{7}{x}\right){dx} \\ $$$$=….\left({eszy}\:{to}\:{find}\right) \\ $$$${V}_{\mathrm{2}} =\int\:\frac{{W}_{\mathrm{2}} }{{W}}{dx}\:=\int\:\frac{\mathrm{2}{e}^{\mathrm{2}{x}} \left(\mathrm{1}−{cos}\left(\mathrm{8}{x}\right){cosx}\right.}{{e}^{\mathrm{4}{x}} }{dx} \\ $$$$=\int\:\mathrm{2}\:{e}^{−\mathrm{2}{x}} {cosx}\:{dx}−\mathrm{2}\int\:{e}^{\mathrm{2}{x}} {cos}\left(\mathrm{8}{x}\right){cosxdx} \\ $$$$=….\left({use}\:{same}\:{way}…\right)\:\Rightarrow \\ $$$${y}_{{p}} ={u}_{\mathrm{1}} {v}_{\mathrm{1}} +{u}_{\mathrm{2}} {v}_{\mathrm{2}} \:{and}\:{general}\:{solution} \\ $$$${is}\:{y}\:={y}_{{p}} \:+{y}_{{h}} \\ $$

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