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 Solve the differential equation  \frac{dy}{dx}=1+x+y^{2}+xy^{2} when y = 0, x = 0.

Answers (1)

Given:

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

To find: solution of the differential equation

Rewriting the given equation as,

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

Integrating on both the sides

\int \frac{dy}{1+y^{2}}=\int \left ( 1+x \right )dx\\ arctan\; y=x+\frac{x^{2}}{2}+c\\ Formula:\; \int \frac{dy}{1+y^{2}}=\tan^{-1}y \\ \int x^{n}dx=\frac{x^{n+1}}{n+1}\\

Substitute(0,0) to find c’s value

0+0=c

c=0

Hence, the solution is

\tan^{-1}y=x+\frac{x^{2}}{2}\\ y=\tan\left ( x+\frac{x^{2}}{2} \right )

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