solve y^(′′) +5y^′ −3y =x^2 sin(3x) with y(0) =0 and y^′ (0)=−1


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Question Number 99113 by mathmax by abdo last updated on 18/Jun/20

$solve y^{^{″}} + {5 y}^{^{'}} - 3 y = x^{2} \sin (3 x) with y (0) = 0 and y^{^{'}} (0) = - 1$
	Answered by mathmax by abdo last updated on 19/Jun/20

	$(he) \to y^{^{″}} + {5 y}^{^{'}} - 3 y = 0 \to r^{2} + 5 y - 3 = 0$ $Δ = 25 + 12 = 37 \Rightarrow r_{1} = \frac{- 5 + \sqrt{37}}{2} and r_{2} = \frac{- 5 - \sqrt{37}}{2} \Rightarrow y_{h} = α e^{r_{1} x} + β e^{r_{2} x}$ $= α u_{1} + β u_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} e^{r_{1} x} e^{r_{2} x} \\ r_{1} e^{r_{1} x} r_{2} e^{r_{2} x} \end{matrix} \| = (r_{2} - r_{1}) e^{(r_{1} + r_{2}) x} = - \sqrt{37} e^{- 5 x}$ $W_{1} = \| \begin{matrix} 0 e^{r_{2} x} \\ x^{2} \sin (3 x) r_{2} e^{r_{2} x} \end{matrix} \| = - x^{2} \sin (3 x) e^{r_{2} x}$ $W_{2} = \| \begin{matrix} e^{r_{1} x} 0 \\ r_{1} e^{r_{1} x} x^{2} \sin (3 x) \end{matrix} \| = x^{2} \sin (3 x) e^{r_{1} x}$ $v_{1} = \int \frac{W 1}{W} dx = \int \frac{- x^{2} \sin (3 x) e^{r_{2} x}}{- \sqrt{37} e^{- 5 x}} dx = \frac{1}{\sqrt{37}} \int x^{2} \sin (3 x) e^{(5 + r_{2}) x} dx$ $= \frac{1}{\sqrt{37}} Im (\int x^{2} e^{(3 i + 5 + r_{2}) x} dx) = . . . .$ $v_{2} = \int \frac{W_{2}}{W} dx = \int \frac{x^{2} \sin (3 x) e^{r_{1} x}}{- \sqrt{37} e^{- 5 x}} dx = - \frac{1}{\sqrt{37}} \int x^{2} \sin (3 x) e^{(5 + r_{1}) x} dx$ $= - \frac{1}{\sqrt{37}} Im (\int x^{2} e^{(3 i + 5 + r_{1}) x} dx) = . . . . \Rightarrow$ $y_{p} = u_{1} v_{1} + u_{2} v_{2} and general solution is y = α e^{r_{1} x} + β e^{r_{2} x} + y_{p}$
	Answered by mathmax by abdo last updated on 20/Jun/20
	$let take a try with laplace transform e ⇒L(y^(′′) )+5L(y^′ )−3L(y) =L(x^2 sin(3x)) ⇒ x^2 L(y)−xy(0)−y^′ (0)+5(x L(y)−y(0))−3L(y) =L(x^2 sin(3x)) ⇒ (x^2 +5x−3)L(y) +1 =L(x^2 sin(3x) we have L(x^2 sin(3x)) =∫_0 ^∞ t^2 sin(3t) e^(−xt) dt =Im(∫_0 ^∞ t^2 e^((3i−x)t) dt) ∫_0 ^∞ t^2 e^((−x+3i)t) dt =[(t^2 /(−x+3i)) e^((−x+3i)t) ]_0 ^(+∞) −∫_0 ^∞ ((2t)/(−x+3i)) e^((−x+3i)t) dt =(2/(x−3i)) ∫_0 ^∞ t e^((−x+3i)t) dt =(2/(x−3i)){ [(t/(−x+3i)) e^((−x+3i)t) ]_0 ^(+∞) −∫_0 ^∞ (1/(−x+3i)) e^((−x+3i)t) dt} =(2/(x−3i))×(1/(x−3i)) ∫_0 ^∞ e^((−x+3i)t) dt =(2/((x−3i)^2 ))×[(1/(−x+3i)) e^((−x+3i)t) ]_0 ^(+∞) =(2/((x−3i)^3 )) =((2(x+3i)^3 )/((x^2 +9)^3 )) =((2(x^3 +3x^2 (3i)+3x(3i)^2 +(3i)^3 ))/((x^2 +9)^3 )) =((2(x^3 +9ix^2 −27x −27i))/((x^2 +9)^3 )) ⇒Im(...) =((2(9x^2 −27))/((x^2 +9)^3 )) e ⇒(x^2 +5x−3)L(y) =−1 +((18x^3 −54)/((x^2 +9)^3 )) ⇒ L(y) =−(1/(x^2 +5x−3)) +((18x^3 −54)/((x^2 +5x−3)(x^2 +9)^3 )) ⇒ y(x) =−L^(−1) ((1/(x^2 +5x −3)))+L^(−1) (((18x^3 −54)/((x^2 +5x−3)(x^2 +9)^3 )))...be continued....$
	$let take a try with laplace transform$ $e \Rightarrow L (y^{^{″}}) + 5 L (y^{^{'}}) - 3 L (y) = L (x^{2} \sin (3 x)) \Rightarrow$ $x^{2} L (y) - xy (0) - y^{^{'}} (0) + 5 (x L (y) - y (0)) - 3 L (y) = L (x^{2} \sin (3 x)) \Rightarrow$ $(x^{2} + 5 x - 3) L (y) + 1 = L (x^{2} \sin (3 x) we have$ $L (x^{2} \sin (3 x)) = \int_{0}^{\infty} t^{2} \sin (3 t) e^{- xt} dt = Im (\int_{0}^{\infty} t^{2} e^{(3 i - x) t} dt)$ $\int_{0}^{\infty} t^{2} e^{(- x + 3 i) t} dt = {[\frac{t^{2}}{- x + 3 i} e^{(- x + 3 i) t}]}_{0}^{+ \infty} - \int_{0}^{\infty} \frac{2 t}{- x + 3 i} e^{(- x + 3 i) t} dt$ $= \frac{2}{x - 3 i} \int_{0}^{\infty} t e^{(- x + 3 i) t} dt = \frac{2}{x - 3 i} {{[\frac{t}{- x + 3 i} e^{(- x + 3 i) t}]}_{0}^{+ \infty} - \int_{0}^{\infty} \frac{1}{- x + 3 i} e^{(- x + 3 i) t} dt}$ $= \frac{2}{x - 3 i} \times \frac{1}{x - 3 i} \int_{0}^{\infty} e^{(- x + 3 i) t} dt = \frac{2}{{(x - 3 i)}^{2}} \times {[\frac{1}{- x + 3 i} e^{(- x + 3 i) t}]}_{0}^{+ \infty}$ $= \frac{2}{{(x - 3 i)}^{3}} = \frac{2 {(x + 3 i)}^{3}}{{(x^{2} + 9)}^{3}} = \frac{2 (x^{3} + {3 x}^{2} (3 i) + 3 x {(3 i)}^{2} + {(3 i)}^{3})}{{(x^{2} + 9)}^{3}}$ $= \frac{2 (x^{3} + {9 ix}^{2} - 27 x - 27 i)}{{(x^{2} + 9)}^{3}} \Rightarrow Im (. . .) = \frac{2 ({9 x}^{2} - 27)}{{(x^{2} + 9)}^{3}}$ $e \Rightarrow (x^{2} + 5 x - 3) L (y) = - 1 + \frac{{18 x}^{3} - 54}{{(x^{2} + 9)}^{3}} \Rightarrow$ $L (y) = - \frac{1}{x^{2} + 5 x - 3} + \frac{{18 x}^{3} - 54}{(x^{2} + 5 x - 3) {(x^{2} + 9)}^{3}} \Rightarrow$ $y (x) = - L^{- 1} (\frac{1}{x^{2} + 5 x - 3}) + L^{- 1} (\frac{{18 x}^{3} - 54}{(x^{2} + 5 x - 3) {(x^{2} + 9)}^{3}}) . . . be continued . . . .$
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