Use-the-laplace-tranform-to-solve-d-2-y-dx-2-5-dy-dx-6y-e-x-for-y-0-and-dy-dx-1-when-x-0-

Question Number 98993 by Rio Michael last updated on 17/Jun/20

Use the laplace tranform to solve \frac{d^{2} y}{{d x}^{2}} + 5 \frac{d y}{d x} + 6 y = e^{- x}

for y = 0, and \frac{d y}{d x} = 1 when x = 0

Answered by mathmax by abdo last updated on 18/Jun/20

y^{^{″}} + {5 y}^{^{'}} + 6 y = e^{- x} with y (0) = 0 and y^{^{'}} (0) = 1

(e) \Rightarrow L (y^{^{″}}) + 5 L (y^{^{'}}) + 6 L (y) = L (e^{- x}) \Rightarrow

x^{2} L (y) - x y (0) - y^{^{'}} (0) + 5 (x L (y) - y (0)) + 6 L (y) = L (e^{- x}) \Rightarrow

(x^{2} + 5 x + 6) L (y) - 1 = L (e^{- x}) we have L (e^{- x}) = \int_{0}^{\infty} e^{- t} e^{- xt} dt

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

(e) \Rightarrow (x^{2} + 5 x + 6) L (y) = 1 + \frac{1}{x + 1} = \frac{x + 2}{x + 1} \Rightarrow L (y) = \frac{x + 2}{(x + 1) (x^{2} + 5 x + 6)} \Rightarrow

y = L^{- 1} (\frac{x + 2}{(x + 1) (x^{2} + 5 x + 6)}) let decompose f (x) = \frac{x + 2}{(x + 1) (x^{2} + 5 x + 6)}

x^{2} + 5 x + 6 = 0 \to Δ = 25 - 24 = 1 \Rightarrow x_{1} = \frac{- 5 + 1}{2} = - 2 and x_{2} = \frac{- 5 - 1}{2} = - 3 \Rightarrow

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

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

the solution is y (x) = \frac{e^{- x} - e^{- 3 x}}{2}

Commented by Rio Michael last updated on 18/Jun/20

wow nice solution sir, but i have problems with

L (y^{″}) and L (y^{″}) the derivations

Commented by abdomathmax last updated on 18/Jun/20

you must use a table of formula for laplace