solve-by-laplace-transform-y-3y-2y-e-x-withy-0-1-and-y-0-2-

Question Number 95694 by mathmax by abdo last updated on 27/May/20

solve by laplace transform y^{^{″}} + {3 y}^{^{'}} + 2 y = e^{- x} withy (0) = 1 and y^{^{'}} (0) = 2

Answered by mathmax by abdo last updated on 27/May/20

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

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

(x^{2} + 3 x + 2) L (y) - x - 2 - 3 = L (e^{- x}) \Rightarrow (x^{2} + 3 x + 2) L (y) = x + 5 + L (e^{- x})

L (e^{- x}) = \int_{0}^{\infty} e^{- t} e^{- xt} dt = \int_{0}^{\infty} e^{- (1 + x) t} dt = {[- \frac{1}{1 + x} e^{- (1 + x) t}]}_{0}^{\infty}

= \frac{1}{x + 1} so (e) \Rightarrow (x^{2} + 3 x + 2) L (y) = x + 5 + \frac{1}{x + 1} \Rightarrow

L (y) = \frac{x + 5}{x^{2} + 3 x + 2} + \frac{1}{(x + 1) (x^{2} + 3 x + 2)} \Rightarrow y (x) = L^{- 1} (\frac{x + 5}{x^{2} + 3 x + 2}) + L^{- 1} (\frac{1}{(x + 1) (x^{2} + 3 x + 2)})

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

= (x + 1) (x + 2) \Rightarrow \frac{x + 5}{x^{2} + 3 x + 2} = \frac{x + 5}{(x + 1) (x + 2)} = \frac{a}{x + 1} + \frac{b}{x + 2}

a = 4 and b = \frac{3}{- 1} = - 3 \Rightarrow \frac{x + 5}{x^{2} + 3 x + 2} = \frac{4}{x + 1} - \frac{3}{x + 2} \Rightarrow

L^{- 1} (\frac{x + 5}{x^{2} + 3 x + 2}) = {4 L}^{- 1} (\frac{1}{x + 1}) - {3 L}^{- 1} (\frac{1}{x + 2}) = {4 e}^{- x} - {3 e}^{- 2 x}

let decompose g (x) = \frac{1}{(x + 1) (x^{2} + 3 x + 2)} \Rightarrow g (x) = \frac{1}{{(x + 1)}^{2} (x + 2)}

= \frac{a}{x + 1} + \frac{b}{{(x + 1)}^{2}} + \frac{c}{x + 2}

b = 1 and c = 1 \Rightarrow g (x) = \frac{a}{x + 1} + \frac{1}{{(x + 1)}^{2}} + \frac{1}{x + 2}

{l i m}_{x \to + \infty} xg (x) = 0 = a + 1 \Rightarrow a = - 1 \Rightarrow g (x) = - \frac{1}{x + 1} + \frac{1}{{(x + 1)}^{2}} + \frac{1}{x + 2} \Rightarrow

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

y (x) = {4 e}^{- x} - {3 e}^{- 2 x} - e^{- x} + {xe}^{- x} + e^{- 2 x} = {3 e}^{- x} - {2 e}^{- 2 x} + {xe}^{- x}

\Rightarrow y (x) = (x + 3) e^{- x} - {2 e}^{- 2 x}