Solve the following differential equation 1) y′′ + y = e^x + x^3 , y(0)=2, y′(0)=0 2) y′′ + y^′ − 2y = x + sin2x, y(0)=1, y′(0)=0 3) y′′ − y′ = xe^x , y(0)=2, y′(0)= 1 Thank you


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Question Number 197792 by Mastermind last updated on 28/Sep/23

$Solve the following differential equation$ $1) y^{″} + y = e^{x} + x^{3}, y (0) = 2, y^{'} (0) = 0$ $2) y^{″} + y^{^{'}} - 2 y = x + \sin 2 x, y (0) = 1, y^{'} (0) = 0$ $3) y^{″} - y^{'} = {xe}^{x}, y (0) = 2, y^{'} (0) = 1$ $Thank you$
	Answered by qaz last updated on 29/Sep/23

	$1) {(y^{'} - y)}^{'} + (y^{'} - y) = e^{x} + x^{3}$ $2) {(y^{'} - y)}^{'} + 2 (y^{'} - y) = x + \sin 2 x$ $3) y^{″} - y^{'} = {x e}^{x}$ $- - - - - - -$ $y^{'} + p (x) y = q (x) \Rightarrow y = e^{- \int p (x) d x} (C + \int q (x) e^{\int p (x) d x} d x)$
	Answered by Tokugami last updated on 30/Sep/23
	$Use Laplace Transform: 1. y′′+y=e^x +x^3 s^2 Y(s)−sy(0)−y′(0)+Y(s)=(1/(s−1))+(6/s^4 ) (s^2 +1)Y(s)−2s−0=(1/(s−1))+(6/s^4 ) Y(s)=(1/((s−1)(s^2 +1)))+(6/(s^4 (s^2 +1)))+((2s)/(s^2 +1)) Y(s)=((1/2)/(s−1))+((−(1/2)s−(1/2))/(s^2 +1))−(6/s^2 )+(6/s^4 )+(6/(s^2 +1))+2((s/(s^2 +1))) L^(−1) {Y(s)}=L^(−1) {(1/2)((1/(s−1)))+(3/2)((s/(s^2 +1)))+((11)/2)((1/(s^2 +1)))−(6/s^2 )+(6/s^4 )} y=(1/2)e^x +(3/2)cos(x)+((11)/2)sin(x)−6x+(1/(20))x^5 2. y′′+y′−2y=x+sin(2x), y(0)=1, y′(0)=0 s^2 Y(s)−sy(0)−y′(0)+sY(s)−y(0)−2Y(s)=(1/s^2 )+(2/(s^2 +4)) (s^2 +s−2)Y(s)−s−1=(1/s^2 )+(2/(s^2 +4)) (s^2 +s−2)Y(s)=(1/s^2 )+(2/(s^2 +4))+s+1 Y(s)=((s^5 +s^4 +4s^3 +7s^2 +4)/(s^2 (s^2 +4)(s−1)(s+2)))=((−1/2)/s^2 )−((1/4)/s)+((1/6)/(s+2))+((17/15)/(s−1))−(1/(20))(((s+6)/(s^2 +4))) L^(−1) {Y(s)}=L^(−1) {−(1/4)((1/s))+((17)/(15))((1/(s−1)))+(1/6)((1/(s+2)))−(1/2)((1/s^2 ))−(1/(20))((s/(s^2 +4)))−(3/(20))((2/(s^2 +4)))} y=−(1/4)−(1/2)x+((17)/(15))e^x +(1/6)e^(−2x) −(1/(20))cos(2x)−(3/(20))sin(2x) 3. y′′−y′=xe^x , y(0)=2, y′(0)=1 s^2 Y(s)−sy(0)−y′(0)−sY(s)+y(0)=(1/((s−1)^2 )) (s^2 −s)Y(s)−2s+1=(1/((s−1)^2 )) Y(s)=(1/(s(s−1)^3 ))+((2s−1)/(s(s−1))) Y(s)=((s^2 −3s+3)/((s−1)^3 ))+(1/(s−1))=(2/(s−1))−(1/((s−1)^2 ))+(1/((s−1)^3 )) Inverse Laplace: y=(2−x+(1/2)x^2 )e^x$
	$Use Laplace Transform :$ $1 . y^{″} + y = e^{x} + x^{3}$ $s^{2} Y (s) - s y (0) - y^{'} (0) + Y (s) = \frac{1}{s - 1} + \frac{6}{s^{4}}$ $(s^{2} + 1) Y (s) - 2 s - 0 = \frac{1}{s - 1} + \frac{6}{s^{4}}$ $Y (s) = \frac{1}{(s - 1) (s^{2} + 1)} + \frac{6}{s^{4} (s^{2} + 1)} + \frac{2 s}{s^{2} + 1}$ $Y (s) = \frac{1 / 2}{s - 1} + \frac{- \frac{1}{2} s - \frac{1}{2}}{s^{2} + 1} - \frac{6}{s^{2}} + \frac{6}{s^{4}} + \frac{6}{s^{2} + 1} + 2 (\frac{s}{s^{2} + 1})$ $L^{- 1} {Y (s)} = L^{- 1} {\frac{1}{2} (\frac{1}{s - 1}) + \frac{3}{2} (\frac{s}{s^{2} + 1}) + \frac{11}{2} (\frac{1}{s^{2} + 1}) - \frac{6}{s^{2}} + \frac{6}{s^{4}}}$ $y = \frac{1}{2} e^{x} + \frac{3}{2} \cos (x) + \frac{11}{2} \sin (x) - 6 x + \frac{1}{20} x^{5}$ $2 . y^{″} + y^{'} - 2 y = x + \sin (2 x), y (0) = 1, y^{'} (0) = 0$ $s^{2} Y (s) - s y (0) - y^{'} (0) + s Y (s) - y (0) - 2 Y (s) = \frac{1}{s^{2}} + \frac{2}{s^{2} + 4}$ $(s^{2} + s - 2) Y (s) - s - 1 = \frac{1}{s^{2}} + \frac{2}{s^{2} + 4}$ $(s^{2} + s - 2) Y (s) = \frac{1}{s^{2}} + \frac{2}{s^{2} + 4} + s + 1$ $Y (s) = \frac{s^{5} + s^{4} + 4 s^{3} + 7 s^{2} + 4}{s^{2} (s^{2} + 4) (s - 1) (s + 2)} = \frac{- 1 / 2}{s^{2}} - \frac{1 / 4}{s} + \frac{1 / 6}{s + 2} + \frac{17 / 15}{s - 1} - \frac{1}{20} (\frac{s + 6}{s^{2} + 4})$ $L^{- 1} {Y (s)} = L^{- 1} {- \frac{1}{4} (\frac{1}{s}) + \frac{17}{15} (\frac{1}{s - 1}) + \frac{1}{6} (\frac{1}{s + 2}) - \frac{1}{2} (\frac{1}{s^{2}}) - \frac{1}{20} (\frac{s}{s^{2} + 4}) - \frac{3}{20} (\frac{2}{s^{2} + 4})}$ $y = - \frac{1}{4} - \frac{1}{2} x + \frac{17}{15} e^{x} + \frac{1}{6} e^{- 2 x} - \frac{1}{20} \cos (2 x) - \frac{3}{20} \sin (2 x)$ $3 . y^{″} - y^{'} = {x e}^{x}, y (0) = 2, y^{'} (0) = 1$ $s^{2} Y (s) - s y (0) - y^{'} (0) - s Y (s) + y (0) = \frac{1}{{(s - 1)}^{2}}$ $(s^{2} - s) Y (s) - 2 s + 1 = \frac{1}{{(s - 1)}^{2}}$ $Y (s) = \frac{1}{s {(s - 1)}^{3}} + \frac{2 s - 1}{s (s - 1)}$ $Y (s) = \frac{s^{2} - 3 s + 3}{{(s - 1)}^{3}} + \frac{1}{s - 1} = \frac{2}{s - 1} - \frac{1}{{(s - 1)}^{2}} + \frac{1}{{(s - 1)}^{3}}$ $Inverse Laplace :$ $y = (2 - x + \frac{1}{2} x^{2}) e^{x}$
	Commented by deleteduser1 last updated on 03/Oct/23

	$I g u e s s {h e}^{'} s t a l k i n g a b o u t t h e ‘ ‘ {b o s s}^{″} a s p e c t .$
	Commented by Mastermind last updated on 30/Sep/23

	$Thank you so much my BOSS$
	Commented by necx122 last updated on 02/Oct/23

	$o n l y N i g e r i a n s s p e a k t h i s w a y$
	Commented by Mastermind last updated on 02/Oct/23

	$No, not only Nigerians . . . English is an official language$
	Commented by Mastermind last updated on 04/Oct/23

	$Oh! Ok$
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