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

\[\int\frac{1}{y^2}dy = - 4\int x dx\]

\[ \Rightarrow \frac{- 1}{y} = - 2 x^2 + C . . . . . \left( 1 \right)\]

Now,

When `x = 0, y = 1`

\[ \therefore - 1 = 0 + C\]

\[ \Rightarrow C = - 1\]

Putting the value of `C` in (1), we get

\[\frac{- 1}{y} = - 2 x^2 - 1\]

\[ \Rightarrow \frac{1}{y} = 2 x^2 + 1\]

\[ \Rightarrow y = \frac{1}{2 x^2 + 1}\]