Question

The solution of the differential equation $\frac { d y } { d x } + \frac { 1 + x ^ { 2 } } { x } = 0$ is

• A. $y = - \frac { 1 } { 2 } \tan ^ { - 1 } x + c$
• B. $y + \log x + \frac { x ^ { 2 } } { 2 } + c = 0$
• C. $y = \frac { 1 } { 2 } \tan ^ { - 1 } x + c$
• D. $y - \log x - \frac { x ^ { 2 } } { 2 } = c$

Answer 1

Answer: (B)

$y + \log x + \frac { x ^ { 2 } } { 2 } + c = 0$

Answer 2

$\frac { d y } { d x } + \frac { 1 + x ^ { 2 } } { x } = 0$ ⇒ $d y + \left( \frac { 1 } { x } + x \right) d x = 0$

On integrating, we get $y + \log x + \frac { x ^ { 2 } } { 2 } + c = 0$ .