The general solution of frac{dy}{dx} = frac{x | vedclass

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(A) Given, $\frac{dy}{dx} = \frac{x^3(y^4+1)}{\left[2y^{-2/3} + 3x^2y^{-2/3}\right]^{3/2}} = \frac{x^3(y^4+1)}{\left[y^{-2/3}(2+3x^2)\right]^{3/2}} =

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$\log \left(\frac{y^4}{1+y^4}\right) = \frac{4}{9}\left(\frac{4+3x^2}{\sqrt{2+3x^2}}\right) + C$

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(A) Given, $\frac{dy}{dx} = \frac{x^3(y^4+1)}{\left[2y^{-2/3} + 3x^2y^{-2/3}\right]^{3/2}} = \frac{x^3(y^4+1)}{\left[y^{-2/3}(2+3x^2)\right]^{3/2}} = \frac{x^3(y^4+1)}{y^{-1}(2+3x^2)^{3/2}} = \frac{x^3 y(y^4+1)}{(2+3x^2)^{3/2}}$.Separating the variables, we get $\int \frac{1}{y(y^4+1)} dy = \int \frac{x^3}{(2+3x^2)^{3/2}} dx$.For the left side, $I_1 = \int \frac{1}{y(y^4+1)} dy = \int \frac{y^3}{y^4(y^4+1)} dy$. Let $u = y^4$, then $du = 4y^3 dy$, so $I_1 = \frac{1}{4} \int \frac{du}{u(u+1)} = \frac{1}{4} \int \left(\frac{1}{u} - \frac{1}{u+1}\right) du = \frac{1}{4} \log \left|\frac{u}{u+1}\right| = \frac{1}{4} \log \left(\frac{y^4}{1+y^4}\right)$.For the right side, $I_2 = \int \frac{x^3}{(2+3x^2)^{3/2}} dx$. Let $t^2 = 2+3x^2$, then $2t dt = 6x dx$, so $x dx = \frac{1}{3} t dt$ and $x^2 = \frac{t^2-2}{3}$.$I_2 = \int \frac{x^2 \cdot x dx}{(t^2)^{3/2}} = \int \frac{(\frac{t^2-2}{3}) \cdot \frac{1}{3} t dt}{t^3} = \frac{1}{9} \int \frac{t^2-2}{t^2} dt = \frac{1}{9} \int (1 - 2t^{-2}) dt = \frac{1}{9} (t + \frac{2}{t}) = \frac{1}{9} \left(\frac{t^2+2}{t}\right) = \frac{1}{9} \left(\frac{2+3x^2+2}{\sqrt{2+3x^2}}\right) = \frac{1}{9} \left(\frac{4+3x^2}{\sqrt{2+3x^2}}\right)$.Equating $I_1 = I_2 + C$, we get $\frac{1}{4} \log \left(\frac{y^4}{1+y^4}\right) = \frac{1}{9} \left(\frac{4+3x^2}{\sqrt{2+3x^2}}\right) + C'$.Multiplying by $4$, $\log \left(\frac{y^4}{1+y^4}\right) = \frac{4}{9} \left(\frac{4+3x^2}{\sqrt{2+3x^2}}\right) + C$.

The general solution of $\frac{dy}{dx} = \frac{x^3(y^4+1)}{\left[2y^{-2/3} + 3\left(\frac{x}{y^{1/3}}\right)^2\right]^{3/2}}$ is

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