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