Question

\[\left( 2x - 10 y^3 \right)\frac{dy}{dx} + y = 0\]

Answer 1

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