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Question Number 60322 by Sardor2211 last updated on 19/May/19

Commented by Mr X pcx last updated on 19/May/19

$$\left(e\right)\iff {xy}^{{}^{″}}+y^{{}^{\prime }}=x^{3}let{y}^{{}^{\prime }}=z\Rightarrow$$$${xz}^{{}^{\prime }}+z=x^3$$$$\left(he\right)\Rightarrow {xz}^{{}^{\prime }}+z=0\Rightarrow \frac{z^{{}^{\prime }}}{z}=-\frac{1}{x}\Rightarrow$$$$ln\mid z\mid =-ln\mid x\mid +k\Rightarrow z=\frac{c}{\mid x\mid }$$$$on]0,+\mathrm{\infty }[\Rightarrow z=\frac{C}{x}mvcmethodgive$$$$z^{{}^{\prime }}=\frac{C^{{}^{\prime }}}{x}-\frac{C}{x^2}$$$${xz}^{{}^{\prime }}+z=x^3\Rightarrow C^{{}^{\prime }}-\frac{C}{x}+\frac{C}{x}=x^3\Rightarrow$$$$C^{{}^{\prime }}=x^3\Rightarrow C=\frac{x^4}{4}+k\Rightarrow z\left(x\right)=\frac{x^3}{4}+\frac{k}{x}$$$$wehave{y}^{{}^{\prime }}=z\Rightarrow y=\int zdx$$$$=\int (\frac{x^3}{4}+\frac{k}{x})dx+\lambda$$$$y=\frac{1}{16}{x}^4+kln\mid x\mid +\lambda x.$$$$$$

Commented by Mr X pcx last updated on 19/May/19

$$y=\frac{1}{16}{x}^4+kln\mid x\mid +\lambda .$$

Answered by tanmay last updated on 19/May/19

$$\frac{dy}{dx}=p$$$$\frac{d^{2}y}{{dx}^2}=\frac{dp}{dx}$$$$\frac{dp}{dx}+\frac{p}{x}=x^2$$$$intregatingfactor{e}^{\int \frac{dx}{x}}=e^{lnx}=x$$$$x\frac{dp}{dx}+p=x^3$$$$\frac{xdp+pdx}{dx}=x^3$$$$d\left(xp\right)=x^{3}dx$$$$\int d\left(xp\right)=\int x^{3}dx$$$$xp=\frac{x^4}{4}+c$$$$p=\frac{x^3}{4}+\frac{c}{x}$$$$\frac{dy}{dx}=\frac{x^3}{4}+\frac{c}{x}$$$$\int dy=\int \frac{x^3}{4}dx+\int \frac{c}{x}dx$$$$y=\frac{x^4}{16}+clnx+c_1$$

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