solve by Laplace transform y^(′′) +5y^′ +2y =x^2 cosx with y(o)=1 and y^′ (0) =2


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
Question Number 95212 by mathmax by abdo last updated on 24/May/20

$solve by Laplace transform$ $y^{^{″}} + {5 y}^{^{'}} + 2 y = x^{2} cosx with y (o) = 1 and y^{^{'}} (0) = 2$
	Answered by mathmax by abdo last updated on 24/May/20
	$(e) ⇒L(y^(′′) )+5L(y^′ )+2L(y) =L(x^2 cosx) ⇒ x^2 L(y)−xy(0)−y^′ (0) +5(xL(y)−y(0))+2L(y)=L(x^2 cosx) ⇒ (x^2 +5x+2)L(y) −x−2−5 =L(x^2 cosx) but L(x^2 cosx) =∫_0 ^∞ t^2 cost e^(−xt) dt =Re( ∫_0 ^∞ t^2 e^(−xt+it) dt) =Re(∫_0 ^∞ t^2 e^((−x+i)t) dt) by parts ∫_0 ^∞ t^2 e^((−x+i)t) dt =[(t^2 /(−x+i)) e^((−x+i)t) ]_0 ^∞ −∫_0 ^∞ ((2t)/((−x+i)))e^((−x+i)t) [dt =(2/(x−i)) ∫_0 ^∞ t e^((−x+i)t) dt =(2/(x−i)){ [(t/(−x+i))e^((−x+i)t) ]_0 ^∞ −∫_0 ^∞ (1/(−x+i))e^((−x+i)t) dt =(2/(x−i)){ (1/(x−i))[(1/(−x+i))e^((−x+i)t) ]_0 ^∞ } =−(2/((x−i)^3 )){−1} =(2/((x−i)^3 )) =((2(x+i)^3 )/((x^2 +1)^3 )) =((2( x^3 −3x +3ix^2 −i^3 ))/((x^2 +1)^3 )) ⇒Re(∫....) =((2x^3 −6x)/((x^2 +1)^3 )) (e) ⇒(x^2 +5x+2)L(y) =x+7 +((2x^3 −6x)/((x^2 +1)^3 )) ⇒ L(y) =((x+7)/(x^2 +5x+2)) +((2x^3 −6x)/((x^2 +5x+2)(x^2 +1)^3 )) ⇒ y =L^(−1) (((x+7)/(x^2 +5x+2))) +L^(−1) (((2x^3 −6x)/((x^2 +5x+2)(x^2 +1)^3 ))) rest decomposition of those fractions ...be continued...$
	$(e) \Rightarrow L (y^{^{″}}) + 5 L (y^{^{'}}) + 2 L (y) = L (x^{2} cosx) \Rightarrow$ $x^{2} L (y) - xy (0) - y^{^{'}} (0) + 5 (xL (y) - y (0)) + 2 L (y) = L (x^{2} cosx) \Rightarrow$ $(x^{2} + 5 x + 2) L (y) - x - 2 - 5 = L (x^{2} cosx) but$ $L (x^{2} cosx) = \int_{0}^{\infty} t^{2} cost e^{- xt} dt = Re (\int_{0}^{\infty} t^{2} e^{- xt + it} dt)$ $= Re (\int_{0}^{\infty} t^{2} e^{(- x + i) t} dt) by parts$ $\int_{0}^{\infty} t^{2} e^{(- x + i) t} dt = {[\frac{t^{2}}{- x + i} e^{(- x + i) t}]}_{0}^{\infty} - \int_{0}^{\infty} \frac{2 t}{(- x + i)} e^{(- x + i) t} [dt$ $= \frac{2}{x - i} \int_{0}^{\infty} t e^{(- x + i) t} dt = \frac{2}{x - i} {{[\frac{t}{- x + i} e^{(- x + i) t}]}_{0}^{\infty} - \int_{0}^{\infty} \frac{1}{- x + i} e^{(- x + i) t} dt$ $= \frac{2}{x - i} {\frac{1}{x - i} {[\frac{1}{- x + i} e^{(- x + i) t}]}_{0}^{\infty}} = - \frac{2}{{(x - i)}^{3}} {- 1} = \frac{2}{{(x - i)}^{3}}$ $= \frac{2 {(x + i)}^{3}}{{(x^{2} + 1)}^{3}} = \frac{2 (x^{3} - 3 x + {3 ix}^{2} - i^{3})}{{(x^{2} + 1)}^{3}} \Rightarrow Re (\int . . . .) = \frac{{2 x}^{3} - 6 x}{{(x^{2} + 1)}^{3}}$ $(e) \Rightarrow (x^{2} + 5 x + 2) L (y) = x + 7 + \frac{{2 x}^{3} - 6 x}{{(x^{2} + 1)}^{3}} \Rightarrow$ $L (y) = \frac{x + 7}{x^{2} + 5 x + 2} + \frac{{2 x}^{3} - 6 x}{(x^{2} + 5 x + 2) {(x^{2} + 1)}^{3}} \Rightarrow$ $y = L^{- 1} (\frac{x + 7}{x^{2} + 5 x + 2}) + L^{- 1} (\frac{{2 x}^{3} - 6 x}{(x^{2} + 5 x + 2) {(x^{2} + 1)}^{3}})$ $rest decomposition of those fractions . . . be continued . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum