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

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 y², 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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