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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Differential Equations

NDA 2020 · Mathematics · Differential Equations

The solution of the differential equation d y=(1+y^2) d x is

The solution of the differential equation dy=(1+y2)dxd y=\left(1+y^2\right) d x is

Options

  1. Ay=tan⁡x+cy=\tan x+c
  2. By=tan⁡(x+c)y=\tan (x+c)
  3. Ctan⁡−1(y+c)=x\tan ^{-1}(y+c)=x
  4. Dtan⁡−1(y+c)=2x\tan ^{-1}(y+c)=2 x

Correct answer

B. y=tan⁡(x+c)y=\tan (x+c)

Step-by-step solution

dy=(1+y2)⋅dx1(1+y2)⋅dy=dxd y=\left(1+y^2\right) \cdot d x\frac{1}{\left(1+y^2\right)} \cdot d y=d x Integrating both sides, we get ∫dy(1+y2)=∫dxtan⁡−1(y)=x+C\int \frac{d y}{\left(1+y^2\right)}=\int d x\tan ^{-1}(y)=x+C, where C=C= constant. ∴y=tan⁡(x+C)\therefore y=\tan (x+C). \text{. }
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