Question-97866

Question Number 97866 by me2love2math last updated on 10/Jun/20

Commented by me2love2math last updated on 10/Jun/20

1 2 5 a n d 6

Commented by bobhans last updated on 10/Jun/20

(2i)2xy (dy/dx) + y^2 = e^(2x) 2xy dy + (y^2 −e^(2x) ) dx = 0 exact if (∂N/∂x) = (∂M/∂y) { (((∂N/∂x) = 2y)),(((∂M/∂y) = 2y)) :} so it is diff equation exact F(x,y) = ∫ M(x,y) dx + g(y) where g′(y)=2xy−(∂/∂y) (∫ (y^2 −e^(2x) )dx) g′(y) = 2xy −2yx = 0 ⇒g(y) = k F(x,y) = ∫ (y^2 −e^(2x) ) dx + k F(x,y) = xy^2 −(1/2)e^(2x) + k solution is xy^2 −(1/2)e^(2x) +k = C or xy^2 −(1/2)e^(2x) = C_1

(2 i) 2 xy \frac{dy}{dx} + y^{2} = e^{2 x}

2 xy dy + (y^{2} - e^{2 x}) dx = 0

exact if \frac{\partial N}{\partial x} = \frac{\partial M}{\partial y}

{\begin{cases} \frac{\partial N}{\partial x} = 2 y \\ \frac{\partial M}{\partial y} = 2 y \end{cases} so it is diff equation exact

F (x, y) = \int M (x, y) dx + g (y)

where g^{'} (y) = 2 xy - \frac{\partial}{\partial y} (\int (y^{2} - e^{2 x}) dx)

g^{'} (y) = 2 xy - 2 yx = 0 \Rightarrow g (y) = k

F (x, y) = \int (y^{2} - e^{2 x}) dx + k

F (x, y) = {xy}^{2} - \frac{1}{2} e^{2 x} + k

solution is {xy}^{2} - \frac{1}{2} e^{2 x} + k = C

or {xy}^{2} - \frac{1}{2} e^{2 x} = C_{1}

Commented by bemath last updated on 10/Jun/20

(1ii) m^2 +7m+49=0 m = −7,−7 y_h = Ae^(−7x) + Bx e^(−7x) particular solution y_p = Ce^(5x) ⇔ { ((y′= 5Ce^(5x) )),((y′′=25Ce^(5x) )) :} ⇔25Ce^(5x) + 7(5Ce^(5x) )+49(Ce^(5x) )=4e^(5x) ⇔(25+35+49)C = 4 C = (4/(109)) ∴ generall solution y_g = Ae^(−7x) +Bx e^(−7x) + (4/(109)) e^(5x)

(1 ii) m^{2} + 7 m + 49 = 0

m = - 7, - 7

y_{h} = {Ae}^{- 7 x} + Bx e^{- 7 x}

particular solution

y_{p} = {Ce}^{5 x} \Leftrightarrow {\begin{cases} y^{'} = {5 Ce}^{5 x} \\ y^{″} = {25 Ce}^{5 x} \end{cases}

\Leftrightarrow {25 Ce}^{5 x} + 7 ({5 Ce}^{5 x}) + 49 ({Ce}^{5 x}) = {4 e}^{5 x}

\Leftrightarrow (25 + 35 + 49) C = 4

C = \frac{4}{109}

∴ generall solution

y_{g} = {Ae}^{- 7 x} + Bx e^{- 7 x} + \frac{4}{109} e^{5 x}

Commented by me2love2math last updated on 10/Jun/20

Thanks all. well appreciated....remain 5

T h a n k s a l l . w e l l a p p r e c i a t e d \dots . r e m a i n 5

Answered by john santu last updated on 10/Jun/20

(1) homogenous solution λ^2 −5λ+6 = 0 λ = 2,3 ⇒y_h = C_1 e^(2x) + C_2 e^(3x) particular solution y_p = a sin 4x + b cos 4x y′=4a cos 4x− 4b sin 4x y′′=−16a sin 4x −16b cos 4x ⇔ { ((22a+20b=2)),((−20a−10b=0)) :} we get b = −2a ∧a=−(1/9) generall solution y_g = C_1 e^(2x) + C_2 e^(3x) −(1/9)sin 4x +(2/9)cos 4x

(1) homogenous solution

λ^{2} - 5 λ + 6 = 0

λ = 2, 3 \Rightarrow y_{h} = C_{1} e^{2 x} + C_{2} e^{3 x}

particular solution

y_{p} = a \sin 4 x + b \cos 4 x

y^{'} = 4 a \cos 4 x - 4 b \sin 4 x

y^{″} = - 16 a \sin 4 x - 16 b \cos 4 x

\Leftrightarrow {\begin{cases} 22 a + 20 b = 2 \\ - 20 a - 10 b = 0 \end{cases}

we get b = - 2 a \land a = - \frac{1}{9}

generall solution

y_{g} = C_{1} e^{2 x} + C_{2} e^{3 x} - \frac{1}{9} \sin 4 x + \frac{2}{9} \cos 4 x

Answered by smridha last updated on 10/Jun/20

(6)s^2 x(s)−sx(0)−x^′ (0)−4x(s)=((2s)/(s^2 +4)) x(s)[s^2 −4]−3s−4=((2s)/(s^2 +4)) x(s)=((2s)/((s^2 −4)(s^2 +4))) +(4/(s^2 −2^2 ))+((3s)/(s^2 −2^2 )) x(s)=(s/4)[(1/((s^2 −2^2 )))−(1/((s^2 +2^2 )))]+2[(2/(s^2 −2^2 ))] +3[(s/(s^2 −2^2 ))] now take L^(−1) both sides we get x(t)=(1/4)cosh(2t)−(1/4)cos2t+2sinh(2t) 3cosh(2t) so the solution: x(t)=((13)/4)cosh(2t)−(1/4)cos(2t)+2sinh(2t)

(6) s^{2} x (s) - s x (0) - x^{^{'}} (0) - 4 x (s) = \frac{2 s}{s^{2} + 4}

x (s) [s^{2} - 4] - 3 s - 4 = \frac{2 s}{s^{2} + 4}

x (s) = \frac{2 s}{(s^{2} - 4) (s^{2} + 4)} + \frac{4}{s^{2} - 2^{2}} + \frac{3 s}{s^{2} - 2^{2}}

x (s) = \frac{s}{4} [\frac{1}{(s^{2} - 2^{2})} - \frac{1}{(s^{2} + 2^{2})}] + 2 [\frac{2}{s^{2} - 2^{2}}]

+ 3 [\frac{s}{s^{2} - 2^{2}}]

n o w t a k e L^{- 1} b o t h s i d e s w e g e t

x (t) = \frac{1}{4} c o s h (2 t) - \frac{1}{4} c o s 2 t + 2 s i n h (2 t)

3 c o s h (2 t)

s o t h e s o l u t i o n :

x (t) = \frac{13}{4} c o s h (2 t) - \frac{1}{4} c o s (2 t) + 2 s i n h (2 t)

Commented by smridha last updated on 10/Jun/20

or ans should be like that x(t)=((13)/8)[e^(2t) +e^(−2t) ]+[e^(2t) −e^(−2t) ]−(1/4)cos(2t) =((21)/8)e^(2t) +(5/8)e^(−2t) −(1/4)cos(2t)

o r a n s s h o u l d b e l i k e t h a t

x (t) = \frac{13}{8} [e^{2 t} + e^{- 2 t}] + [e^{2 t} - e^{- 2 t}] - \frac{1}{4} c o s (2 t)

= \frac{21}{8} e^{2 t} + \frac{5}{8} e^{- 2 t} - \frac{1}{4} c o s (2 t)

Answered by mathmax by abdo last updated on 10/Jun/20

y^(′′) −5y^′ +6 =2sin(4x) (he)→y^(′′) −5y^′ +6 =0→r^2 −5r +6 =0 Δ =25−24 =1 ⇒r_1 =((5+1)/2) =3 and r_2 =((5−1)/2) =2 ⇒y_h =αe^(3x) +βe^(2x) =α u_1 +β u_2 W(u_1 ,u_2 ) = determinant (((u_1 u_2 )),((u_1 ^′ u_2 ^′ )))= determinant (((e^(3x) e^(2x) )),((3e^(3x) 2e^(2x) )))=−e^(5x) W_1 = determinant (((0 e^(2x) )),((2sin(4x) 2e^(2x) )))=−2e^(2x) sin(4x) W_2 = determinant (((e^(3x) 0)),((3e^(3x) 2sin(4x))))=2e^(3x) sin(4x) v_1 =∫ (w_1 /w)dx =∫ ((−2e^(2x) sin(4x))/(−e^(5x) )) =2∫ e^(−3x) sin(4x)dx =2Im( ∫ e^(−3x+4ix) dx) we have ∫ e^((−3+4i)x) dx =(1/(−3+4i)) e^((−3+4i)x) =−(1/(3−4i))e^(−3x) (cos(4x)+isin(4x)) =−((3+4i)/(25))e^(−3x) (cos(4x)+isin(4x)) =−(1/(25))e^(−3x) (3cos(4x)+3isin(4x)+4icos(4x)−4sin(4x)) ⇒ v_1 =−(2/(25))e^(−3x) (3sin(4x)+4cos(4x)) v_2 =∫ (w_2 /w)dx =∫ ((2e^(3x) sin(4x))/(−e^(5x) ))dx =−2 ∫ e^(−2x) sin(4x)dx =−2 Im( ∫ e^((−2+4i)x) dx) and ∫ e^((−2+4i)x) dx=(1/(−2+4i)) e^((−2+4i)x) =−(1/(2−4i))e^(−2x) (cos(4x)+isin(4x)) =−((2+4i)/(20)) e^(−2x) {cos(4x)+isin(4x)} =−(e^(−2x) /(20)){ 2cos(4x)+2isin(4x)+4icos(4x)−4sin(4x)} ⇒ v_2 =(e^(−2x) /(10))( 2sin(4x)+4cos(4x)} =(e^(−2x) /5)(sin(4x)+2cos(4x)) so y_p =u_1 v_(1 ) +u_2 v_2 =−(2/(25)){3sin(4x)+4cos(4x)} +(1/5)(sin(4x)+2cos(4x)) =(−(6/(25)) +(1/5))sin(4x) +(−(8/(25)) +(2/5))cos(4x) =−(1/(25))sin(4x) +(2/(25)) cos(4x) the general solution is y =y_h +y_p

y^{^{″}} - {5 y}^{^{'}} + 6 = 2 \sin (4 x)

(he) \to y^{^{″}} - {5 y}^{^{'}} + 6 = 0 \to r^{2} - 5 r + 6 = 0

Δ = 25 - 24 = 1 \Rightarrow r_{1} = \frac{5 + 1}{2} = 3 and r_{2} = \frac{5 - 1}{2} = 2 \Rightarrow y_{h} = α e^{3 x} + β e^{2 x}

= α u_{1} + β u_{2}

W (u_{1}, u_{2}) = | \begin{matrix} u_{1} u_{2} \\ u_{1}^{^{'}} u_{2}^{^{'}} \end{matrix} | = | \begin{matrix} e^{3 x} e^{2 x} \\ {3 e}^{3 x} {2 e}^{2 x} \end{matrix} | = - e^{5 x}

W_{1} = | \begin{matrix} 0 e^{2 x} \\ 2 \sin (4 x) {2 e}^{2 x} \end{matrix} | = - {2 e}^{2 x} \sin (4 x)

W_{2} = | \begin{matrix} e^{3 x} 0 \\ {3 e}^{3 x} 2 \sin (4 x) \end{matrix} | = {2 e}^{3 x} \sin (4 x)

v_{1} = \int \frac{w_{1}}{w} dx = \int \frac{- {2 e}^{2 x} \sin (4 x)}{- e^{5 x}} = 2 \int e^{- 3 x} \sin (4 x) dx

= 2 Im (\int e^{- 3 x + 4 ix} dx) we have \int e^{(- 3 + 4 i) x} dx = \frac{1}{- 3 + 4 i} e^{(- 3 + 4 i) x}

= - \frac{1}{3 - 4 i} e^{- 3 x} (\cos (4 x) + isin (4 x)) = - \frac{3 + 4 i}{25} e^{- 3 x} (\cos (4 x) + isin (4 x))

= - \frac{1}{25} e^{- 3 x} (3 \cos (4 x) + 3 isin (4 x) + 4 icos (4 x) - 4 \sin (4 x)) \Rightarrow

v_{1} = - \frac{2}{25} e^{- 3 x} (3 \sin (4 x) + 4 \cos (4 x))

v_{2} = \int \frac{w_{2}}{w} dx = \int \frac{{2 e}^{3 x} \sin (4 x)}{- e^{5 x}} dx = - 2 \int e^{- 2 x} \sin (4 x) dx

= - 2 Im (\int e^{(- 2 + 4 i) x} dx) and \int e^{(- 2 + 4 i) x} dx = \frac{1}{- 2 + 4 i} e^{(- 2 + 4 i) x}

= - \frac{1}{2 - 4 i} e^{- 2 x} (\cos (4 x) + isin (4 x)) = - \frac{2 + 4 i}{20} e^{- 2 x} {\cos (4 x) + isin (4 x)}

= - \frac{e^{- 2 x}}{20} {2 \cos (4 x) + 2 isin (4 x) + 4 icos (4 x) - 4 \sin (4 x)} \Rightarrow

v_{2} = \frac{e^{- 2 x}}{10} (2 \sin (4 x) + 4 \cos (4 x)} = \frac{e^{- 2 x}}{5} (\sin (4 x) + 2 \cos (4 x)) so

y_{p} = u_{1} v_{1} + u_{2} v_{2} = - \frac{2}{25} {3 \sin (4 x) + 4 \cos (4 x)} + \frac{1}{5} (\sin (4 x) + 2 \cos (4 x))

= (- \frac{6}{25} + \frac{1}{5}) \sin (4 x) + (- \frac{8}{25} + \frac{2}{5}) \cos (4 x)

= - \frac{1}{25} \sin (4 x) + \frac{2}{25} \cos (4 x) the general solution is

y = y_{h} + y_{p}

Answered by mathmax by abdo last updated on 10/Jun/20

4) let solve by laplace transform y^(′′) +3y^′ +2y =sin(2x) ⇒L(y^(′′) )+3L(y^′ )+2L(y) =L(sin(2x)) ⇒ x^2 L(y)−xy(0)−y^′ (o) +3(xL(y)−y(0))+2L(y) =L(sin(2x)) ⇒ (x^2 +3x+2)L(y) −(x+3)y(o)−y^′ (0) =L(sin(2x)) L(sin(2x) =∫_0 ^∞ sin(2t)e^(−xt) dt =Im(∫_0 ^∞ e^(2it−xt) dt) ∫_0 ^∞ e^((−x+2i)t) dt =[(1/(−x+2i)) e^((−x+2i)t) ]_(t=0) ^∞ =((−1)/(−x+2i)) =(1/(x−2i)) =((x+2i)/(x^2 +4)) ⇒ L(sin(2x))=(2/(x^2 +4)) (e)⇒(x^2 +3x+2)L(y) =y(o)(x+3)+y^′ (0)+(2/(x^2 +4)) ⇒ L(y) =((x+3)/(x^2 +3x+2))y(0) +((y^′ (0))/(x^2 +3x+2)) +(2/((x^2 +3x+2)(x^2 +4))) ⇒ y(x) =y(0)L^(−1) (((x+3)/(x^2 +3x+2)))+y^′ (0)L^(−1) ((1/(x^2 +3x+2)))+2L^(−1) ((1/((x^2 +3x+2)(x^2 +4)))) f(x) =((x+3)/(x^2 +3x+2))=((x+3)/((x+1)(x+2))) =(x+3)((1/(x+1))−(1/(x+2))) =((x+3)/(x+1))−((x+3)/(x+2)) =1+(2/(x+1))−(1+(1/(x+2))) =(2/(x+1))−(1/(x+2)) ⇒ L^(−1) (f) =2e^(−x) −e^(−2x) g(x) =(1/(x^2 +3x+2)) =(1/(x+1))−(1/(x+2)) ⇒L^(−1) (g) =e^(−x) −e^(−2x) h(x) =(1/((x^2 +3x+2)(x^2 +4))) =(1/((x+1)(x+2)(x−2i)(x+2i))) =(a/(x+1)) +(b/(x+2)) +(c/(x−2i)) +(d/(x+2i)) a =(1/((−1−2i)(−1+2i))) =(1/((1+2i)(1−2i))) =(1/5) b =((−1)/((−2−2i)(−2+2i))) =(1/((2+2i)(2−2i))) =(1/(4×2))=(1/8) c =(1/((2i+1)(2i+2)4i)) =(1/(4i(−4+4i+2i+2))) =(1/(4i(−2+6i))) =(1/(−8i−24)) =((−1)/(8i+24)) d =(1/((−2i+1)(−2i+2)(−4i))) =.... L^(−1) (h) =a e^(−x) +b^(−2x) +c e^(2ix) +d e^(−2ix) →at form a e^(−x) +be^(−2x) +αcos(2x) +βsin(2x) so the general solution is y(x) =ae^(−x) +b e^(−2x) +αcos(2x) +βsin(2x)

4) let solve by laplace transform

y^{^{″}} + {3 y}^{^{'}} + 2 y = \sin (2 x) \Rightarrow L (y^{^{″}}) + 3 L (y^{^{'}}) + 2 L (y) = L (\sin (2 x)) \Rightarrow

x^{2} L (y) - xy (0) - y^{^{'}} (o) + 3 (xL (y) - y (0)) + 2 L (y) = L (\sin (2 x)) \Rightarrow

(x^{2} + 3 x + 2) L (y) - (x + 3) y (o) - y^{^{'}} (0) = L (\sin (2 x))

L (\sin (2 x) = \int_{0}^{\infty} \sin (2 t) e^{- xt} dt = Im (\int_{0}^{\infty} e^{2 it - xt} dt)

\int_{0}^{\infty} e^{(- x + 2 i) t} dt = {[\frac{1}{- x + 2 i} e^{(- x + 2 i) t}]}_{t = 0}^{\infty} = \frac{- 1}{- x + 2 i} = \frac{1}{x - 2 i} = \frac{x + 2 i}{x^{2} + 4} \Rightarrow

L (\sin (2 x)) = \frac{2}{x^{2} + 4}

(e) \Rightarrow (x^{2} + 3 x + 2) L (y) = y (o) (x + 3) + y^{^{'}} (0) + \frac{2}{x^{2} + 4} \Rightarrow

L (y) = \frac{x + 3}{x^{2} + 3 x + 2} y (0) + \frac{y^{^{'}} (0)}{x^{2} + 3 x + 2} + \frac{2}{(x^{2} + 3 x + 2) (x^{2} + 4)} \Rightarrow

y (x) = y (0) L^{- 1} (\frac{x + 3}{x^{2} + 3 x + 2}) + y^{^{'}} (0) L^{- 1} (\frac{1}{x^{2} + 3 x + 2}) + {2 L}^{- 1} (\frac{1}{(x^{2} + 3 x + 2) (x^{2} + 4)})

f (x) = \frac{x + 3}{x^{2} + 3 x + 2} = \frac{x + 3}{(x + 1) (x + 2)} = (x + 3) (\frac{1}{x + 1} - \frac{1}{x + 2})

= \frac{x + 3}{x + 1} - \frac{x + 3}{x + 2} = 1 + \frac{2}{x + 1} - (1 + \frac{1}{x + 2}) = \frac{2}{x + 1} - \frac{1}{x + 2} \Rightarrow

L^{- 1} (f) = {2 e}^{- x} - e^{- 2 x}

g (x) = \frac{1}{x^{2} + 3 x + 2} = \frac{1}{x + 1} - \frac{1}{x + 2} \Rightarrow L^{- 1} (g) = e^{- x} - e^{- 2 x}

h (x) = \frac{1}{(x^{2} + 3 x + 2) (x^{2} + 4)} = \frac{1}{(x + 1) (x + 2) (x - 2 i) (x + 2 i)}

= \frac{a}{x + 1} + \frac{b}{x + 2} + \frac{c}{x - 2 i} + \frac{d}{x + 2 i}

a = \frac{1}{(- 1 - 2 i) (- 1 + 2 i)} = \frac{1}{(1 + 2 i) (1 - 2 i)} = \frac{1}{5}

b = \frac{- 1}{(- 2 - 2 i) (- 2 + 2 i)} = \frac{1}{(2 + 2 i) (2 - 2 i)} = \frac{1}{4 \times 2} = \frac{1}{8}

c = \frac{1}{(2 i + 1) (2 i + 2) 4 i} = \frac{1}{4 i (- 4 + 4 i + 2 i + 2)} = \frac{1}{4 i (- 2 + 6 i)} = \frac{1}{- 8 i - 24} = \frac{- 1}{8 i + 24}

d = \frac{1}{(- 2 i + 1) (- 2 i + 2) (- 4 i)} = \dots .

L^{- 1} (h) = a e^{- x} + b^{- 2 x} + c e^{2 ix} + d e^{- 2 ix} \to at form a e^{- x} + {be}^{- 2 x} + α \cos (2 x)

+ β \sin (2 x) so the general solution is

y (x) = {ae}^{- x} + b e^{- 2 x} + α \cos (2 x) + β \sin (2 x)

Answered by mathmax by abdo last updated on 10/Jun/20

{ ((x^′ =x−2y ⇒ ((x^′ ),(y^′ ) )=A ((x),(y) ))),((y^′ =−2x +y)) :} with A = (((1 −2)),((−2 1)) ) let X = ((x),(h) ) e ⇒X^′ =A X ⇒ X (t)=k e^(A(t−t_0 )) let find e^(tA) p_c (A) =det(A−xI) = determinant (((1−x −2)),((−2 1−x)))=(1−x)^2 −4 =(1−x−2)(1−x+2) =(−x−1)(3−x) =(x+1)(x−3) so the values are λ_1 =−1 and λ_2 =3 x^n =p_c (x)q +u_n x +v_n ⇒(−1)^n =−u_n +v_n and 3^n =3u_n +v_n ⇒ 4u_n =3^n −(−1)^n ⇒ u_n =(((3)^n −(−1)^n )/4) v_n =u_n +(−1)^n =((3^n −(−1)^n )/4) +(−1)^n =((3^n −(−1)^n +4(−1)^n )/4) =((3^n +3(−1)^n )/4) A^n =u_n A +v_n I=((3^n −(−1)^n )/4) (((1 −2)),((−2 1)) ) +((3^n +3(−1)^n )/4) (((1 0)),((0 1)) ) =(1/4) (((3^n −(−1)^n −2(3^n −(−1)^n ))),((−2(3^n −(−1)^n 3^n −(−1)^n )) ) +(1/4) (((3^n +3(−1)^n 0 )),((0 3^n +3(−1)^n )) ) =(1/4) (((2.3^n +2(−1)^n −2.3^n +2(−1)^n )),((−2.3^n +2(−1)^n 2.3^n +2(−1)^n )) ) e^(tA ) =Σ_(n=0) ^∞ (t^n /(n!)) A^n =(1/4)Σ_(n=0) ^∞ (t^n /(n!)) (((2.3^n +2(−1)^n −2.3^n +2(−1)^n )),((−2.3^n +2(−1)^n 2.3^n +2(−1)^n )) ) =(1/2) Σ_(n=0) ^∞ (t^n /(n!)) (((3^n +(−1)^n −3^n +(−1)^n )),((−3^n +(−1)^n 3^n +(−1)^n )) ) =(1/2) ((( Σ_(n=0) ^∞ (((3t)^n )/(n!)) +Σ_(n=0) ^∞ (((−t)^n )/(n!)) −Σ_(n=0) ^∞ (((3t)^n )/(n!))+Σ_(n=0) ^∞ (((−t)^n )/(n!)))),((...... ....)) ) =(1/2) (((e^(3t) +e^(−t) −e^(3t) +e^(−t) )),((−e^(3t) +e^(−t) e^(3t) +e^(−t) )) ) X(t) =k e^(A(t−t_0 ))

{\begin{cases} x^{^{'}} = x - 2 y \Rightarrow (\begin{array}{c} x^{^{'}} \\ y^{^{'}} \end{array}) = A (\begin{array}{c} x \\ y \end{array}) \\ y^{^{'}} = - 2 x + y \end{cases} with A = (\begin{matrix} 1 - 2 \\ - 2 1 \end{matrix})

let X = (\begin{matrix} x \\ h \end{matrix}) e \Rightarrow X^{^{'}} = A X \Rightarrow

X (t) = k e^{A (t - t_{0})} let find e^{tA}

p_{c} (A) = \det (A - xI) = | \begin{matrix} 1 - x - 2 \\ - 2 1 - x \end{matrix} | = {(1 - x)}^{2} - 4 = (1 - x - 2) (1 - x + 2)

= (- x - 1) (3 - x) = (x + 1) (x - 3) so the values are λ_{1} = - 1 and λ_{2} = 3

x^{n} = p_{c} (x) q + u_{n} x + v_{n} \Rightarrow {(- 1)}^{n} = - u_{n} + v_{n} and 3^{n} = {3 u}_{n} + v_{n} \Rightarrow

{4 u}_{n} = 3^{n} - {(- 1)}^{n} \Rightarrow u_{n} = \frac{{(3)}^{n} - {(- 1)}^{n}}{4}

v_{n} = u_{n} + {(- 1)}^{n} = \frac{3^{n} - {(- 1)}^{n}}{4} + {(- 1)}^{n} = \frac{3^{n} - {(- 1)}^{n} + 4 {(- 1)}^{n}}{4} = \frac{3^{n} + 3 {(- 1)}^{n}}{4}

A^{n} = u_{n} A + v_{n} I = \frac{3^{n} - {(- 1)}^{n}}{4} (\begin{matrix} 1 - 2 \\ - 2 1 \end{matrix}) + \frac{3^{n} + 3 {(- 1)}^{n}}{4} (\begin{matrix} 1 0 \\ 0 1 \end{matrix})

= \frac{1}{4} (\begin{matrix} 3^{n} - {(- 1)}^{n} - 2 (3^{n} - {(- 1)}^{n}) \\ - 2 (3^{n} - {(- 1)}^{n} 3^{n} - {(- 1)}^{n} \end{matrix}) + \frac{1}{4} (\begin{matrix} 3^{n} + 3 {(- 1)}^{n} 0 \\ 0 3^{n} + 3 {(- 1)}^{n} \end{matrix})

= \frac{1}{4} (\begin{matrix} 2 . 3^{n} + 2 {(- 1)}^{n} - 2 . 3^{n} + 2 {(- 1)}^{n} \\ - 2 . 3^{n} + 2 {(- 1)}^{n} 2 . 3^{n} + 2 {(- 1)}^{n} \end{matrix})

e^{tA} = \sum_{n = 0}^{\infty} \frac{t^{n}}{n!} A^{n} = \frac{1}{4} \sum_{n = 0}^{\infty} \frac{t^{n}}{n!} (\begin{matrix} 2 . 3^{n} + 2 {(- 1)}^{n} - 2 . 3^{n} + 2 {(- 1)}^{n} \\ - 2 . 3^{n} + 2 {(- 1)}^{n} 2 . 3^{n} + 2 {(- 1)}^{n} \end{matrix})

= \frac{1}{2} \sum_{n = 0}^{\infty} \frac{t^{n}}{n!} (\begin{matrix} 3^{n} + {(- 1)}^{n} - 3^{n} + {(- 1)}^{n} \\ - 3^{n} + {(- 1)}^{n} 3^{n} + {(- 1)}^{n} \end{matrix})

= \frac{1}{2} (\begin{matrix} \sum_{n = 0}^{\infty} \frac{{(3 t)}^{n}}{n!} + \sum_{n = 0}^{\infty} \frac{{(- t)}^{n}}{n!} - \sum_{n = 0}^{\infty} \frac{{(3 t)}^{n}}{n!} + \sum_{n = 0}^{\infty} \frac{{(- t)}^{n}}{n!} \\ \dots \dots \dots . \end{matrix})

= \frac{1}{2} (\begin{matrix} e^{3 t} + e^{- t} - e^{3 t} + e^{- t} \\ - e^{3 t} + e^{- t} e^{3 t} + e^{- t} \end{matrix})

X (t) = k e^{A (t - t_{0})}

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