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Question Number 71570 by TawaTawa last updated on 17/Oct/19

$$\mathrm{Solve}:(2\mathrm{x}+{\mathrm{x}}^2){\mathrm{y}}^{″}-2(1+\mathrm{x}){\mathrm{y}}^{\prime }+2\mathrm{y}=0$$$$\mathrm{Given}\mathrm{y}={\mathrm{x}}^2\mathrm{is}\mathrm{a}\mathrm{solution}.$$

Answered by mind is power last updated on 17/Oct/19

$$\mathrm{y}={\mathrm{zx}}^2$$$${\mathrm{y}}^{\prime }=2\mathrm{xz}+{\mathrm{z}}^{\prime }{\mathrm{x}}^2$$$${\mathrm{y}}^{″}=2\mathrm{z}+{4\mathrm{xz}}^{\prime }+{\mathrm{z}}^{″}{\mathrm{x}}^2$$$$(2\mathrm{x}+{\mathrm{x}}^2){\mathrm{y}}^{″}-2(1+\mathrm{x}){\mathrm{y}}^{\prime }+2\mathrm{y}=(2\mathrm{x}+{\mathrm{x}}^2)(2\mathrm{z}+{4\mathrm{xz}}^{\prime }+{\mathrm{z}}^{″}{\mathrm{x}}^2)-2(1+\mathrm{x})(2\mathrm{xz}+{\mathrm{z}}^{\prime }{\mathrm{x}}^2)$$$$+{2\mathrm{x}}^2\mathrm{z}=0$$$$\Rightarrow ({2\mathrm{x}}^2+{2\mathrm{x}}^2+4\mathrm{x}-4\mathrm{x}-{4\mathrm{x}}^2)\mathrm{z}+(2\mathrm{x}+{\mathrm{x}}^2){\mathrm{x}}^2{\mathrm{z}}^{″}+({8\mathrm{x}}^2+{4\mathrm{x}}^3-{2\mathrm{x}}^2-{2\mathrm{x}}^3){\mathrm{z}}^{\prime }=0$$$$({2\mathrm{x}}^3+{\mathrm{x}}^4){\mathrm{z}}^{″}+({6\mathrm{x}}^2+{2\mathrm{x}}^3){\mathrm{z}}^{\prime }=0$$$$\iff ({\mathrm{x}}^2+2\mathrm{x}){\mathrm{z}}^{″}+2(3+\mathrm{x}){\mathrm{z}}^{\prime }=0..\mathrm{\forall }\mathrm{x}\ne 0$$$$\mathrm{Y}={\mathrm{z}}^{\prime }$$$$({\mathrm{x}}^2+2\mathrm{x}){\mathrm{Y}}^{\prime }+2(3+\mathrm{x})\mathrm{Y}=0$$$$\Rightarrow \frac{{\mathrm{Y}}^{\prime }}{\mathrm{Y}}=\frac{6+2\mathrm{x}}{\mathrm{x}(\mathrm{x}+2)}=\frac{3}{\mathrm{x}}-\frac{1}{\mathrm{x}+2}$$$$\Rightarrow \ln \mid \mathrm{Y}\mid =3\ln \mid \mathrm{x}\mid -\ln \mid \mathrm{x}+2\mid$$$$\ln \mid \mathrm{Y}\mid =\ln \mid \frac{{\mathrm{x}}^3}{\mathrm{x}+2}\mid +\mathrm{c}$$$$\Rightarrow \mid \mathrm{Y}\mid =\mathrm{k}.\mid \frac{{\mathrm{x}}^3}{\mathrm{x}+2}\mid$$$$\mathrm{Y}=\mathrm{c}.\frac{{\mathrm{x}}^3}{\mathrm{x}+2}....\mathrm{c}=\underset{-}{+}\mathrm{k}$$$${\mathrm{z}}^{\prime }=\mathrm{c}\frac{{\mathrm{x}}^3}{\mathrm{x}+2}=\mathrm{c}\frac{(\mathrm{x}+2)({\mathrm{x}}^2-2\mathrm{x}+4)-8}{\mathrm{x}+2}$$$${\mathrm{z}}^{\prime }=\mathrm{c}({\mathrm{x}}^2-2\mathrm{x}+4)-\frac{8\mathrm{c}}{\mathrm{x}+2}$$$$\Rightarrow \mathrm{z}=\mathrm{c}(\frac{{\mathrm{x}}^3}{3}-{\mathrm{x}}^2+4\mathrm{x})-8\mathrm{cln}\mid \mathrm{x}+2\mid +\mathrm{k}$$$$\mathrm{y}={\mathrm{zx}}^2=\mathrm{c}(\frac{{\mathrm{x}}^5}{3}-{\mathrm{x}}^4+{4\mathrm{x}}^3)+{\mathrm{kx}}^2-{8\mathrm{cx}}^2\ln \mid \mathrm{x}+2\mid$$$$\mathrm{c},\mathrm{k}\mathrm{constante}$$$$\mathrm{solution}\mathrm{is}\mathrm{iner}\mathrm{space}\mathrm{of}\mathrm{TWo}\mathrm{Dimensiln}\mathrm{S}=\mathrm{vect}(\frac{{\mathrm{x}}^5}{3}-{\mathrm{x}}^4+{4\mathrm{x}}^3-8\ln \mid \mathrm{x}+2\mid ,{\mathrm{x}}^2)$$$$$$$$$$

Commented by TawaTawa last updated on 17/Oct/19

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by mind is power last updated on 17/Oct/19

$${\mathrm{y}}^{\prime }\mathrm{re}\mathrm{welcom}$$

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