solve-Differential-equation-x-2-d-2-y-dx-2-4x-dy-dx-3y-x-3-2x-5-

Question Number 76964 by peter frank last updated on 01/Jan/20

$${solve}\:{Differential}\:\:{equation} \\ $$$${x}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\mathrm{4}{x}\frac{{dy}}{{dx}}+\mathrm{3}{y}={x}^{\mathrm{3}} +\mathrm{2}{x}+\mathrm{5} \\ $$

Answered by mind is power last updated on 02/Jan/20

x^2 (d^2 y/dx^2 )−4x(dy/dx)+3y=0..E theorie of linear algebra applied in equation we know that solution of linear Differtionsl equation forme vectoriel subspace of dim 2 let y=x^m ⇒((m(m−1)−4m+3)=0 ⇒m^2 −5m+3=0 m_1 =((5−(√(13)))/2),m_2 =((5+(√(13)))/2) y_1 =x^((5−(√(13)))/2) ,y_2 =x^((5+(√(13)))/2) are independente ⇒ (y_1 ,y_2 ) Bases of solutions of E General solution of E is ,∀y∈E ∃! (a,b)∈R^2 ∣y=ax^((5−(√(13)))/2) +bx^((5+(√(13)))/2) particulare solution P(x)=ax^3 +bx^2 +cx+d ⇒(6a−12a+3a)=1 (2b−8b+3b)=0 (−4c+3c)=2 3d=5 a=−(1/3) b=0 c=−2 d=(5/3) p(x)=((−x^3 )/3)−2x+(5/3) Y=ax^((5−(√(13)))/2) +bx^((5+(√(13)))/2) ((−x^3 −6x+5)/3),(a,b)∈R^2

$$\mathrm{x}^{\mathrm{2}} \frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }−\mathrm{4x}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{3y}=\mathrm{0}..\mathrm{E} \\ $$$$\mathrm{theorie}\:\mathrm{of}\:\mathrm{linear}\:\mathrm{algebra}\:\mathrm{applied}\:\mathrm{in}\:\mathrm{equation} \\ $$$$\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{linear}\:\mathrm{Differtionsl}\:\mathrm{equation} \\ $$$$\mathrm{forme}\:\mathrm{vectoriel}\:\mathrm{subspace}\:\mathrm{of}\:\mathrm{dim}\:\mathrm{2} \\ $$$$\mathrm{let}\:\mathrm{y}=\mathrm{x}^{\mathrm{m}} \\ $$$$\Rightarrow\left(\left(\mathrm{m}\left(\mathrm{m}−\mathrm{1}\right)−\mathrm{4m}+\mathrm{3}\right)=\mathrm{0}\right. \\ $$$$\Rightarrow\mathrm{m}^{\mathrm{2}} −\mathrm{5m}+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{m}_{\mathrm{1}} =\frac{\mathrm{5}−\sqrt{\mathrm{13}}}{\mathrm{2}},\mathrm{m}_{\mathrm{2}} =\frac{\mathrm{5}+\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$$\mathrm{y}_{\mathrm{1}} =\mathrm{x}^{\frac{\mathrm{5}−\sqrt{\mathrm{13}}}{\mathrm{2}}} ,\mathrm{y}_{\mathrm{2}} =\mathrm{x}^{\frac{\mathrm{5}+\sqrt{\mathrm{13}}}{\mathrm{2}}} \:\:\:\mathrm{are}\:\mathrm{independente}\:\Rightarrow \\ $$$$\left(\mathrm{y}_{\mathrm{1}} ,\mathrm{y}_{\mathrm{2}} \right)\:\mathrm{Bases}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{E} \\ $$$$\mathrm{General}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{E}\:\mathrm{is}\:,\forall\mathrm{y}\in\mathrm{E}\:\exists!\:\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{R}^{\mathrm{2}} \mid\mathrm{y}=\mathrm{ax}^{\frac{\mathrm{5}−\sqrt{\mathrm{13}}}{\mathrm{2}}} +\mathrm{bx}^{\frac{\mathrm{5}+\sqrt{\mathrm{13}}}{\mathrm{2}}} \\ $$$$\mathrm{particulare}\:\mathrm{solution} \\ $$$$\mathrm{P}\left(\mathrm{x}\right)=\mathrm{ax}^{\mathrm{3}} +\mathrm{bx}^{\mathrm{2}} +\mathrm{cx}+\mathrm{d} \\ $$$$\Rightarrow\left(\mathrm{6a}−\mathrm{12a}+\mathrm{3a}\right)=\mathrm{1} \\ $$$$\left(\mathrm{2b}−\mathrm{8b}+\mathrm{3b}\right)=\mathrm{0} \\ $$$$\left(−\mathrm{4c}+\mathrm{3c}\right)=\mathrm{2} \\ $$$$\mathrm{3d}=\mathrm{5} \\ $$$$\mathrm{a}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{b}=\mathrm{0} \\ $$$$\mathrm{c}=−\mathrm{2} \\ $$$$\mathrm{d}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\frac{−\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{2x}+\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\mathrm{Y}=\mathrm{ax}^{\frac{\mathrm{5}−\sqrt{\mathrm{13}}}{\mathrm{2}}} +\mathrm{bx}^{\frac{\mathrm{5}+\sqrt{\mathrm{13}}}{\mathrm{2}}} \:\:\:\frac{−\mathrm{x}^{\mathrm{3}} −\mathrm{6x}+\mathrm{5}}{\mathrm{3}},\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{R}^{\mathrm{2}} \\ $$$$ \\ $$

Commented by peter frank last updated on 03/Jan/20

$${thank}\:{you} \\ $$

Commented by peter frank last updated on 03/Jan/20

$${where}\:{this}\:\mathrm{6}{a}−\mathrm{12}{a}+\mathrm{3}{a}\:\:{came}\:{frpm} \\ $$

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