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Question Number 50430 by Abdo msup. last updated on 16/Dec/18

$$solve(x^2+3)y^{{}^{\prime }}+(x^3-1)y=x^2$$

Commented by Abdo msup. last updated on 17/Dec/18

$$\left(he\right)(x^2+3)y^{{}^{\prime }}+(x^3-1)y=0\Rightarrow$$$$(x^2+3)y^{{}^{\prime }}=-(x^3-1)y\Rightarrow \frac{y^{{}^{\prime }}}{y}=-\frac{x^3-1}{x^2+3}\Rightarrow$$$$ln\mid y\mid =\int \frac{1-x^3}{x^2+3}dx+c=\int \frac{1-x(x^2+3-3)}{x^2+3}dx+c$$$$=\int \frac{1+3x}{x^2+3}dx-\int xdx+c$$$$=\frac{3}{2}\int \frac{2x}{x^2+3}dx+\int \frac{dx}{x^2+3}-\frac{x^2}{2}+c$$$$=\frac{3}{2}ln(x^2+3)-\frac{x^2}{2}+\int \frac{dx}{x^2+3}but$$$$\int \frac{dx}{x^2+3}{=}_{x=\sqrt{3}u}\int \frac{\sqrt{3}du}{3(1+u^2)}$$$$=\frac{1}{\sqrt{3}}arctan\left(\frac{x}{\sqrt{3}}\right)\Rightarrow$$$$ln\mid y\mid =\frac{3}{2}ln(x^2+3)-\frac{x^2}{2}+\frac{1}{\sqrt{3}}arctan\left(\frac{x}{\sqrt{3}}\right)+c\Rightarrow$$$$y=K{e}^{-\frac{x^2}{2}}{(x^2+3)}^{\frac{3}{2}}.e^{\frac{1}{\sqrt{3}}arctan\left(\frac{x}{\sqrt{3}}\right)}...becontinued...$$$$$$

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