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

$$solveDifferentialequation$$$$x^2\frac{d^{2}y}{{dx}^2}-4x\frac{dy}{dx}+3y=x^3+2x+5$$

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

$${\mathrm{x}}^2\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-4\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+3\mathrm{y}=0..\mathrm{E}$$$$\mathrm{theorie}\mathrm{of}\mathrm{linear}\mathrm{algebra}\mathrm{applied}\mathrm{in}\mathrm{equation}$$$$\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{solution}\mathrm{of}\mathrm{linear}\mathrm{Differtionsl}\mathrm{equation}$$$$\mathrm{forme}\mathrm{vectoriel}\mathrm{subspace}\mathrm{of}\dim 2$$$$\mathrm{let}\mathrm{y}={\mathrm{x}}^{\mathrm{m}}$$$$\Rightarrow ((\mathrm{m}(\mathrm{m}-1)-4\mathrm{m}+3)=0$$$$\Rightarrow {\mathrm{m}}^2-5\mathrm{m}+3=0$$$${\mathrm{m}}_1=\frac{5-\sqrt{13}}{2},{\mathrm{m}}_2=\frac{5+\sqrt{13}}{2}$$$${\mathrm{y}}_1={\mathrm{x}}^{\frac{5-\sqrt{13}}{2}},{\mathrm{y}}_2={\mathrm{x}}^{\frac{5+\sqrt{13}}{2}}\mathrm{are}\mathrm{independente}\Rightarrow$$$$({\mathrm{y}}_1,{\mathrm{y}}_2)\mathrm{Bases}\mathrm{of}\mathrm{solutions}\mathrm{of}\mathrm{E}$$$$\mathrm{General}\mathrm{solution}\mathrm{of}\mathrm{E}\mathrm{is},\mathrm{\forall }\mathrm{y}\in \mathrm{E}\mathrm{\exists }!(\mathrm{a},\mathrm{b})\in {\mathbb{R}}^2\mid \mathrm{y}={\mathrm{ax}}^{\frac{5-\sqrt{13}}{2}}+{\mathrm{bx}}^{\frac{5+\sqrt{13}}{2}}$$$$\mathrm{particulare}\mathrm{solution}$$$$\mathrm{P}\left(\mathrm{x}\right)={\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}$$$$\Rightarrow (6\mathrm{a}-12\mathrm{a}+3\mathrm{a})=1$$$$(2\mathrm{b}-8\mathrm{b}+3\mathrm{b})=0$$$$(-4\mathrm{c}+3\mathrm{c})=2$$$$3\mathrm{d}=5$$$$\mathrm{a}=-\frac{1}{3}$$$$\mathrm{b}=0$$$$\mathrm{c}=-2$$$$\mathrm{d}=\frac{5}{3}$$$$\mathrm{p}\left(\mathrm{x}\right)=\frac{-{\mathrm{x}}^3}{3}-2\mathrm{x}+\frac{5}{3}$$$$\mathrm{Y}={\mathrm{ax}}^{\frac{5-\sqrt{13}}{2}}+{\mathrm{bx}}^{\frac{5+\sqrt{13}}{2}}\frac{-{\mathrm{x}}^3-6\mathrm{x}+5}{3},(\mathrm{a},\mathrm{b})\in {\mathbb{R}}^2$$$$$$

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

$$thankyou$$

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

$$wherethis6a-12a+3acamefrpm$$

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