^(@bemath@) (1) (d^2 y/dx^2 ) −6(dy/dx) + 9y = 1+x+x^2 (2) { ((x^3 +3y^3 = 11)),((x^2 y +xy^2 = 6)) :}


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Question Number 106775 by bemath last updated on 07/Aug/20
$^(@bemath@) (1) (d^2 y/dx^2 ) −6(dy/dx) + 9y = 1+x+x^2 (2) { ((x^3 +3y^3 = 11)),((x^2 y +xy^2 = 6)) :}$
$^{@ bemath @}$ $(1) \frac{d^{2} y}{{dx}^{2}} - 6 \frac{dy}{dx} + 9 y = 1 + x + x^{2}$ $(2) {\begin{cases} x^{3} + {3 y}^{3} = 11 \\ x^{2} y + {xy}^{2} = 6 \end{cases}$
	Answered by abdomathmax last updated on 07/Aug/20
	$y^(′′) −6y^′ +9y =1+x +x^2 h→r^2 −6r +9 =0 ⇒(r−3)^2 =0 ⇒r =3 ⇒ y =(ax+b)e^(3x) =ax e^(3x) +be^(3x) =au_(1 ) +bu_2 W(u_1 ,u_2 ) = determinant (((xe^(3x) e^(3x) )),(((3x+1)e^(3x) 3e^(3x) )))=3x e^(6x) −(3x+1)e^(6x) =−e^(6x ) ≠0 W_1 = determinant (((0 e^(3x) )),((1+x+x^2 3e^(3x) )))=−(1+x+x^2 )e^(3x) W_2 = determinant (((xe^(3x) 0)),(((3x+1)e^(3x) 1+x+x^2 )))=(x+x^2 +x^3 )e^(3x) v_1 =∫ (w_1 /w)dx =−∫ (((1+x+x^2 )e^(3x) )/(−e^(6x) ))dx =∫ (1+x+x^2 )e^(−3x) dx =((1+x+x^2 )/(−3)) e^(−3x) +(1/3)∫ (2x+1) e^(−3x) dx =−((1+x+x^2 )/3)e^(−3x) +(1/3){ ((2x+1)/(−3))e^(−3x) +(1/3)∫2 e^(−3x ) dx} =−((1+x+x^2 )/3)e^(−3x) −(1/9)(2x+1)e^(−3x) −((2e^(−3x) )/(27)) v_2 =∫ (w_2 /w)dx =∫ (((x+x^2 +x^3 )e^(3x) )/(−e^(6x) ))dx =−∫ (x+x^2 +x^3 )e^(−3x) dx =....(same way) ⇒y_p =u_1 v_(1 ) +u_2 v_2 and general solution is y =y_h +y_p y$
	$y^{^{″}} - {6 y}^{^{'}} + 9 y = 1 + x + x^{2}$ $h \to r^{2} - 6 r + 9 = 0 \Rightarrow {(r - 3)}^{2} = 0 \Rightarrow r = 3 \Rightarrow$ $y = (ax + b) e^{3 x} = ax e^{3 x} + {be}^{3 x} = {au}_{1} + {bu}_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} {xe}^{3 x} e^{3 x} \\ (3 x + 1) e^{3 x} {3 e}^{3 x} \end{matrix} \| = 3 x e^{6 x} - (3 x + 1) e^{6 x}$ $= - e^{6 x} \neq 0$ $W_{1} = \| \begin{matrix} 0 e^{3 x} \\ 1 + x + x^{2} {3 e}^{3 x} \end{matrix} \| = - (1 + x + x^{2}) e^{3 x}$ $W_{2} = \| \begin{matrix} {xe}^{3 x} 0 \\ (3 x + 1) e^{3 x} 1 + x + x^{2} \end{matrix} \| = (x + x^{2} + x^{3}) e^{3 x}$ $v_{1} = \int \frac{w_{1}}{w} dx = - \int \frac{(1 + x + x^{2}) e^{3 x}}{- e^{6 x}} dx$ $= \int (1 + x + x^{2}) e^{- 3 x} dx$ $= \frac{1 + x + x^{2}}{- 3} e^{- 3 x} + \frac{1}{3} \int (2 x + 1) e^{- 3 x} dx$ $= - \frac{1 + x + x^{2}}{3} e^{- 3 x} + \frac{1}{3} {\frac{2 x + 1}{- 3} e^{- 3 x} + \frac{1}{3} \int 2 e^{- 3 x} dx}$ $= - \frac{1 + x + x^{2}}{3} e^{- 3 x} - \frac{1}{9} (2 x + 1) e^{- 3 x} - \frac{{2 e}^{- 3 x}}{27}$ $v_{2} = \int \frac{w_{2}}{w} dx = \int \frac{(x + x^{2} + x^{3}) e^{3 x}}{- e^{6 x}} dx$ $= - \int (x + x^{2} + x^{3}) e^{- 3 x} dx = . . . . (same way)$ $\Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2} and general solution is$ $y = y_{h} + y_{p}$ $y$
	Answered by bobhans last updated on 07/Aug/20
	$(1) Homogenous Equation λ^2 −6λ+9=0→(λ−3)^2 =0 ; λ=3,3 y_h = Ae^(3x) + Bxe^(3x) Particular solution y_p =Px^2 +Qx+R → { ((y′=2Px+Q)),((y′′=2P)) :} comparing coefficient 2P−6(2Px+Q)+9Px^2 +9Qx+9R=1+x+x^2 2P−6Q+9R = 1 ; 9P = 1→P=(1/9) 9Q−12P =1 →Q=((1+(4/3))/9)=(7/(27)) R=((1+6Q−2P)/9)=((1+((14)/9)−(2/9))/9)=((21)/(81)) = (7/(27)) Particular solution y_p = (1/9)x^2 +(7/(27))x+(7/(27)) General solution y = Ae^(3x) +Bxe^(3x) +(1/9)x^2 +(7/(27))x+(7/(27)) ^(≻bobhans≺ )$
	$(1) H omogenous E quation$ $λ^{2} - 6 λ + 9 = 0 \to {(λ - 3)}^{2} = 0; λ = 3, 3$ $y_{h} = {Ae}^{3 x} + {Bxe}^{3 x}$ $P articular solution$ $y_{p} = {Px}^{2} + Qx + R \to {\begin{cases} y^{'} = 2 Px + Q \\ y^{″} = 2 P \end{cases}$ $comparing coefficient$ $2 P - 6 (2 Px + Q) + {9 Px}^{2} + 9 Qx + 9 R = 1 + x + x^{2}$ $2 P - 6 Q + 9 R = 1; 9 P = 1 \to P = \frac{1}{9}$ $9 Q - 12 P = 1 \to Q = \frac{1 + \frac{4}{3}}{9} = \frac{7}{27}$ $R = \frac{1 + 6 Q - 2 P}{9} = \frac{1 + \frac{14}{9} - \frac{2}{9}}{9} = \frac{21}{81} = \frac{7}{27}$ $P articular solution y_{p} = \frac{1}{9} x^{2} + \frac{7}{27} x + \frac{7}{27}$ $G eneral solution$ $y = {Ae}^{3 x} + {Bxe}^{3 x} + \frac{1}{9} x^{2} + \frac{7}{27} x + \frac{7}{27}$ $^{≻ bobhans ≺}$
	Answered by 1549442205PVT last updated on 07/Aug/20
	$(2) { ((x^3 +3y^3 = 11(1))),((x^2 y +xy^2 = 6(2))) :} Multiplying obliquely two equalities we get 6x^3 +18y^3 =11x^2 y+11xy^2 and setting y=kx we obtain: 6x^3 +18k^3 x^3 =11kx^3 +11k^2 x^3 .Dividing both sides of above eqn. by x^3 ≠0 we get 18k^3 −11k^2 −11k+6=0 ⇔(2k−1)(9k^2 −k−6)=0 i)2k−1=0⇒k=1/2. one root⇒y=(1/2)x⇔x=2y.Replace into the first eqn.of the given we get 8y^3 +3y^3 =11⇔y^3 =1⇔y=1⇒x=2 ii)9k^2 −k−6=0⇒k=((1±(√(2017)))/(18)) a)For k=((1+(√(2017)))/(18))⇒y=((1+(√(2017)))/(18))x. Replace into (1)we get (3k^3 +1)x^3 =11 ⇔x=^3 (√((11)/(3k^3 +1)))=^3 (√(((21384)/(7996+2020(√(2017)))) ))≈0.600 y=kx=((1+(√(2017)))/(18))×^3 (√(((21384)/(7996+2020(√(2017)))) )) ≈1.5318 b)For k=((1−(√(2017)))/(18))⇒y=((1−(√(2017)))/(18))x Similarly,we get x=^3 (√((11)/(3k^3 +1)))=^3 (√(((21384)/(7996−2020(√(2017)))) )) ≈−06370 ⇒y=((1−(√(2017)))/(18))×^3 (√(((64512)/(23548−2020(√(2017)))) )) ≈1.5540 Thus,the given system of equtions has three solution :(x;y)∈{(2;1); (0.600;1.5318);(−0.6370;1.5540)$
	$(2) {\begin{cases} x^{3} + {3 y}^{3} = 11 (1) \\ x^{2} y + {xy}^{2} = 6 (2) \end{cases}$ $Multiplying obliquely two equalities$ $we get {6 x}^{3} + {18 y}^{3} = {11 x}^{2} y + {11 xy}^{2} and$ $setting y = kx we obtain :$ ${6 x}^{3} + {18 k}^{3} x^{3} = {11 kx}^{3} + {11 k}^{2} x^{3} . Dividing$ $both sides of above eqn . by x^{3} \neq 0 we get$ ${18 k}^{3} - {11 k}^{2} - 11 k + 6 = 0$ $\Leftrightarrow (2 k - 1) ({9 k}^{2} - k - 6) = 0$ $i) 2 k - 1 = 0 \Rightarrow k = 1 / 2 .$ $one root \Rightarrow y = \frac{1}{2} x \Leftrightarrow x = 2 y . Replace$ $into the first eqn . of the given we get$ ${8 y}^{3} + {3 y}^{3} = 11 \Leftrightarrow y^{3} = 1 \Leftrightarrow y = 1 \Rightarrow x = 2$ $ii) {9 k}^{2} - k - 6 = 0 \Rightarrow k = \frac{1 \pm \sqrt{2017}}{18}$ $a) For k = \frac{1 + \sqrt{2017}}{18} \Rightarrow y = \frac{1 + \sqrt{2017}}{18} x .$ $Replace into (1) we get ({3 k}^{3} + 1) x^{3} = 11$ $\Leftrightarrow x =^{3} \sqrt{\frac{11}{{3 k}^{3} + 1}} =^{3} \sqrt{\frac{21384}{7996 + 2020 \sqrt{2017}}} \approx 0.600$ $y = kx = \frac{1 + \sqrt{2017}}{18} \times^{3} \sqrt{\frac{21384}{7996 + 2020 \sqrt{2017}}} \approx 1.5318$ $b) For k = \frac{1 - \sqrt{2017}}{18} \Rightarrow y = \frac{1 - \sqrt{2017}}{18} x$ $Similarly, we get x =^{3} \sqrt{\frac{11}{{3 k}^{3} + 1}} =^{3} \sqrt{\frac{21384}{7996 - 2020 \sqrt{2017}}} \approx - 06370$ $\Rightarrow y = \frac{1 - \sqrt{2017}}{18} \times^{3} \sqrt{\frac{64512}{23548 - 2020 \sqrt{2017}}} \approx 1.5540$ $Thus, the given system of equtions$ $has three solution : (x; y) \in {(2; 1);$ $(0.600; 1.5318); (- 0.6370; 1.5540)$
	Commented by bemath last updated on 07/Aug/20

	Commented by bemath last updated on 07/Aug/20

	$no sir . in graph has 3 roots$
	Commented by 1549442205PVT last updated on 07/Aug/20

	$Excuse, me . I mistaked and now i am$ $correcting (since the equation of$ $degree 3 w . r . t .^{'} k^{'} has 3 roots)$
	Commented by bemath last updated on 07/Aug/20

	$great sir$
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