MCQ
The solution of the differential equation $dy = (1 + y^2) dx$ is:
- A$\text{y}=\tan\text{x}+\text{c}$
- ✓$\text{y}=\tan(\text{x}+\text{c})$
- C$\tan^{-1}(\text{y}+\text{c})=\text{x}$
- D$(\tan^{-1}(\text{y}+\text{c})=2\text{x}$
Concept:
$\int\limits\frac{\text{dx}}{1+\text{x}^2}\tan^{-1}\text{x}=\text{c}$
Calculation:
Given: $dy = (1 + y^2) dx$
$\Rightarrow\frac{\text{dy}}{1+\text{y}^2}\text{dx}$
Integrating both sides, we get
$\Rightarrow\int\frac{\text{dy}}{1+\text{y}^2}=\int\text{dx}$
$\Rightarrow\tan^{-1}\text{y}=\text{x}+\text{c}$
$\Rightarrow\text{y}=\tan(\text{x}+\text{c})$
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