Question

Find the particular solution of the differential equation \[\frac{dy}{dx} = - 4x y^2\]  given that y = 1, when x = 0.

Answer 1

We have,
\[\frac{dy}{dx} = - 4x y^2 \]
\[ \Rightarrow \frac{1}{y^2}dy = - 4x dx\]
Integrating both sides, we get
\[\int\frac{1}{y^2}dy = - 4\int x dx \]
\[ \Rightarrow - \frac{1}{y} = - 4 \times \frac{x^2}{2} + C\]
\[ \Rightarrow - \frac{1}{y} = - 2 x^2 + C . . . . . (1)\]
\[\text{ It is given that at }x = 0, y = 1 . \]
Substituting the values of x and y in (1), we get
\[C = - 1\]
Therefore, substituting the value of C in (1), we get 
\[ - \frac{1}{y} = - 2 x^2 - 1\]
\[ \Rightarrow y = \frac{1}{2 x^2 + 1}\]
\[\text{ Hence, }y = \frac{1}{2 x^2 + 1}\text{ is the required solution .} \]