Solve the following differential equation: (2x-10y^3) dx dy +y=0

Question

Solve the following differential equation:
$(2\text{x}-10\text{y}^3)\frac{\text{dx}}{\text{dy}}+\text{y}=0$

✓

Answer

We have, $(2\text{x}-10\text{y}^3)\frac{\text{dx}}{\text{dy}}+\text{y}=0$
$\Rightarrow\ (2\text{x}-10\text{y}^3)\frac{\text{dx}}{\text{dy}}=-\text{y}$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}=-\frac{1}{\text{y}}(2\text{x}-10\text{y}^3)$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}+\frac{2}{\text{y}}\text{x}=10\text{y}^2\ \dots(1)$ Clearly, it is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ where $\text{P}=\frac{2}{\text{y}}$
$\text{Q}=10\text{y}^2$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdy}}$
$=\text{e}^{\int\frac{2}{\text{y}}\text{dy}}$
$=\text{e}^{2\log{\text{y}}}=\text{y}^2$ Multiplying both sides of (1) by $y^2$, we get $\text{y}^2\Big(\frac{\text{dx}}{\text{dy}}+\frac{2}{\text{y}}\text{x}\Big)=\text{y}^2\times10\text{y}^2$
$\Rightarrow\ \text{y}^2\frac{\text{dx}}{\text{dy}}+\frac{2}{\text{y}}\text{xy}^2=10\text{y}^2$ Integrating both sides with respect to y, we get $\text{xy}^2=\int10\text{y}^4\text{dy + C}$
$\Rightarrow\ \text{xy}^2=2\text{y}^5+\text{C}$
$\Rightarrow\ \text{x}=2\text{y}^3+\text{Cy}^{-2}$Hence, $\text{x}=2\text{y}^3+\text{Cy}^{-2}$ is the required solution.

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