Differential Equations
Differential Equations
nta_pyq_2025_apr
Grade 12
Question:
If $x = f(y)$ is the solution of the differential equation $(1+y^2)+\left(x-2e^{\tan^{-1}y}\right)\dfrac{dy}{dx} = 0$, $y\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ with $f(0) = 1$, then $f\!\left(\dfrac{1}{\sqrt{3}}\right)$ is equal to:
$e^{\pi/12}$
$e^{\pi/4}$
$e^{\pi/3}$
$e^{\pi/6}$
Step-by-Step Solution
Key Concept: Rewrite as $\dfrac{dx}{dy}+\dfrac{x}{1+y^2} = \dfrac{2e^{\tan^{-1}y}}{1+y^2}$; I.F. $= e^{\tan^{-1}y}$; substitute $t = \tan^{-1}y$ to evaluate the right-hand side integral.
$\dfrac{dx}{dy}+\dfrac{x}{1+y^2} = \dfrac{2e^{\tan^{-1}y}}{1+y^2}$. I.F. $= e^{\tan^{-1}y}$.
$xe^{\tan^{-1}y} = \int\dfrac{2e^{2\tan^{-1}y}}{1+y^2}dy$. Let $t=\tan^{-1}y$: $= 2\int e^{2t}dt = e^{2t}+C = e^{2\tan^{-1}y}+C$.
$x = e^{\tan^{-1}y}+Ce^{-\tan^{-1}y}$.
At $y=0$, $x=1$: $1 = 1+C \Rightarrow C=0$.
So $x = e^{\tan^{-1}y}$.
$$f\!\left(\frac{1}{\sqrt{3}}\right) = e^{\tan^{-1}(1/\sqrt{3})} = e^{\pi/6}.$$
Correct Answer: 4