y′′′+2y′−3y= e^x (x+3)


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Question Number 95933 by john santu last updated on 28/May/20

$y^{‴} + {2 y}^{'} - 3 y = e^{x} (x + 3)$
Commented by Sourav mridha last updated on 29/May/20

$are you sure it is 3 rd order ODE, i think$ $it should be 2 nd order .$
Commented by john santu last updated on 29/May/20

$yes .$
	Answered by bobhans last updated on 29/May/20

	$homogenous solution$ $λ^{3} + 2 λ - 3 = 0$ $(λ - 1) (λ^{2} + λ + 3) = 0$ $λ = 1; - \frac{1}{2} \pm \frac{i \sqrt{11}}{2}$ $y_{h} = A_{1} e^{x} + e^{- \frac{1}{2} x} (A_{2} \cos (\frac{x \sqrt{11}}{2}) + A_{3} \sin (\frac{x \sqrt{11}}{2}))$
	Answered by mathmax by abdo last updated on 29/May/20
	$let try with Laplace transform (e) ⇒L(y^((3)) )+2L(y^′ )−3L(y) =L((x+3)e^x ) ⇒ x^3 L(y)−x^2 y(0)−xy^′ (0)−y^(′′) (0)+2(xL(y)−y(0))−3L(y) =L((x+3)e^x )⇒ (x^3 +2x−3)L(y)−x^2 y(0)−xy^′ (0)−2y(0)−y^(′′) (0) =L{(x+3)e^x } ⇒ (x^3 +2x−3)L(y)=L{(x+3)e^x }+(x^2 +2)y(0)+xy^′ (0)+y^(′′) (0) we have L((x+3)e^x ) =∫_0 ^∞ (t+3)e^t e^(−xt) dt =∫_0 ^∞ (t+3)e^((1−x)t) dt =[((t+3)/(1−x))e^((1−x)t) ]_0 ^∞ −∫_0 ^∞ (1/(1−x))e^((1−x)t) dt =−(3/(1−x))−(1/(1−x))[(1/(1−x))e^((1−x)t) ]_0 ^∞ =(3/(x−1))+(1/((1−x)^2 )) (e)⇒(x^3 +2x−3)L(y) =(3/((x−1))) +(1/((x−1)^2 )) +(x^2 +2)y(0)+xy^′ (0)+y^((2)) (0) ⇒L(y) =(3/((x−1)(x^3 +2x−3))) +(1/((x−1)^2 (x^3 +2x−3))) +y(0)((x^2 +2)/(x^3 +2x−3)) +y^′ (0)(x/(x^3 +2x−3)) +((y^((2)) (0))/(x^3 +2x−3)) ⇒ y =3L^(−1) ((1/((x−1)(x^3 +2x−3)))) +L^(−1) ((1/((x−1)^2 (x^3 +2x−3))))+y(o)L^(−1) (((x^2 +2)/(x^3 +2x−3))) +y^′ (0)L^(−1) ((x/(x^3 +2x−3)))+y^((2)) (0)L^(−1) ((1/(x^3 +2x−3))) rest to decompose those fractions ...be continued...$
	$let try with Laplace transform$ $(e) \Rightarrow L (y^{(3)}) + 2 L (y^{^{'}}) - 3 L (y) = L ((x + 3) e^{x}) \Rightarrow$ $x^{3} L (y) - x^{2} y (0) - {xy}^{^{'}} (0) - y^{^{″}} (0) + 2 (xL (y) - y (0)) - 3 L (y) = L ((x + 3) e^{x}) \Rightarrow$ $(x^{3} + 2 x - 3) L (y) - x^{2} y (0) - {xy}^{^{'}} (0) - 2 y (0) - y^{^{″}} (0) = L {(x + 3) e^{x}} \Rightarrow$ $(x^{3} + 2 x - 3) L (y) = L {(x + 3) e^{x}} + (x^{2} + 2) y (0) + {xy}^{^{'}} (0) + y^{^{″}} (0)$ $we have L ((x + 3) e^{x}) = \int_{0}^{\infty} (t + 3) e^{t} e^{- xt} dt = \int_{0}^{\infty} (t + 3) e^{(1 - x) t} dt$ $= {[\frac{t + 3}{1 - x} e^{(1 - x) t}]}_{0}^{\infty} - \int_{0}^{\infty} \frac{1}{1 - x} e^{(1 - x) t} dt = - \frac{3}{1 - x} - \frac{1}{1 - x} {[\frac{1}{1 - x} e^{(1 - x) t}]}_{0}^{\infty}$ $= \frac{3}{x - 1} + \frac{1}{{(1 - x)}^{2}}$ $(e) \Rightarrow (x^{3} + 2 x - 3) L (y) = \frac{3}{(x - 1)} + \frac{1}{{(x - 1)}^{2}} + (x^{2} + 2) y (0) + {xy}^{^{'}} (0) + y^{(2)} (0)$ $\Rightarrow L (y) = \frac{3}{(x - 1) (x^{3} + 2 x - 3)} + \frac{1}{{(x - 1)}^{2} (x^{3} + 2 x - 3)} + y (0) \frac{x^{2} + 2}{x^{3} + 2 x - 3}$ $+ y^{^{'}} (0) \frac{x}{x^{3} + 2 x - 3} + \frac{y^{(2)} (0)}{x^{3} + 2 x - 3} \Rightarrow$ $y = {3 L}^{- 1} (\frac{1}{(x - 1) (x^{3} + 2 x - 3)}) + L^{- 1} (\frac{1}{{(x - 1)}^{2} (x^{3} + 2 x - 3)}) + y (o) L^{- 1} (\frac{x^{2} + 2}{x^{3} + 2 x - 3})$ $+ y^{^{'}} (0) L^{- 1} (\frac{x}{x^{3} + 2 x - 3}) + y^{(2)} (0) L^{- 1} (\frac{1}{x^{3} + 2 x - 3}) rest to decompose those$ $fractions . . . be continued . . .$
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