Question

The solution of the differential equation \[\frac{dy}{dx} = 1 + x + y^2 + x y^2 , y\left( 0 \right) = 0\] is

पर्याय

• \[y^2 = \exp\left( x + \frac{x^2}{2} \right) - 1\]

• \[y^2 = 1 + C \exp\left( x + \frac{x^2}{2} \right)\]

• y = tan (C + x + x²)

• \[y = \tan\left( x + \frac{x^2}{2} \right)\]

Answer 1

\[y = \tan\left( x + \frac{x^2}{2} \right)\]

 

We have,

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

\[ \Rightarrow \frac{dy}{dx} = \left( x + 1 \right) + y^2 \left( x + 1 \right)\]

\[ \Rightarrow \frac{dy}{dx} = \left( x + 1 \right)\left( 1 + y^2 \right)\]

\[ \Rightarrow \frac{dy}{\left( 1 + y^2 \right)} = \left( x + 1 \right)dx\]

Integrating both sides, we get

\[\int\frac{dy}{\left( 1 + y^2 \right)} = \int\left( x + 1 \right)dx\]

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

Now,

\[y\left( 0 \right) = 0\]

\[ \therefore \tan^{- 1} 0 = \frac{0}{2} + 0 + C\]

\[ \Rightarrow C = 0\]

\[\text{Putting the value of C in }\left( 1 \right),\text{ we get }\]

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

\[ \Rightarrow y = \tan\left( \frac{x^2}{2} + x \right)\]