solve-by-Laplace-transform-y-5y-2y-x-2-cosx-with-y-o-1-and-y-0-2-

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) \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 \dots .) = \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 \dots be continued \dots