Given the differential equation, dydx=1+y^2 or, dy1+y^2=dx Now integrating we get, tan^ 1y=x+c or, y=tan (x+c) . [ Where c is integratin ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Variable Separable Method

Solve: dydx=...

Question

Solve:
$\frac{d y}{d x} = 1 + y^{2}$ .

Open in App

Solution

Given the differential equation,
$\frac{d y}{d x} = 1 + y^{2}$
or, $\frac{d y}{1 + y^{2}} = d x$
Now integrating we get,
${tan}^{- 1} y = x + c$
or, $y = tan (x + c)$ . [ Where $c$ is integrating constant]

Suggest Corrections

Similar questions

\frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})

\frac{d y}{d x} + \sqrt{\frac{1 - y^{2}}{(1 + y)^{2}}} = 0

\frac{d y}{d x} = y^{2}

y (1) = 1

\frac{d y}{d x} = \frac{y^{2} - x}{2 y (x + 1)}

\frac{d y}{d x} + \sqrt{\frac{1 - y^{2}}{1 - x^{2}}} = 0

Join BYJU'S Learning Program