Consider the differential equation $\frac{d y}{d x} = 1 + y^{2}$ which one of the following can be particular solution of this differential equations?

y = tan (x + 3)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

y = (tan x) +3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = tan (y + 3)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = (tan y) + 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A y = tan (x + 3)
Given $\frac{d y}{d x} = 1 + y^{2} . . . (i)$

$\Rightarrow \frac{d y}{1 + y^{2}} = d x$
$\Rightarrow t a n^{- 1} (y) = x + c$

$\Rightarrow y = t a n (x + c)$ , where c is arbitrary constant

Suggest Corrections

Similar questions

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

y = 1

x = 0.

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

x > 0

\frac{d y}{d x} = - 4 x y^{2}, y (0) = 1

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

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

Join BYJU'S Learning Program