Question

The solution of the differential equation $\frac{dy}{dx}-y^2=1$ satisfying the conditions y(0) = 1 is

• A. y = $e^{x^2}$
• B. y = $\sqrt{x}$
• C. y = $cot(x+\frac{\pi }{4})$
• D. y = tan $(x+\frac{\pi }{4})$

Answer 1

The correct option is D y = tan $(x+\frac{\pi }{4})$

Given $\frac{dy}{dx}-y^2=1$

$\Rightarrow \frac{dy}{dx}=1+y^2$

$\Rightarrow \frac{dy}{1+y^2}=dx$

$\int \frac{dy}{1+y^2}=\int dx+c$

$\Rightarrow ta{n}^{-1}\left(y\right)=x+c$ ...(ii)

Using y(0) = 1, c = $\frac{\pi }{4}$

$\therefore$ solution is y = tan$(x+\frac{\pi }{4})$