Differential Equations
Linear ODE — arctan Integrating Factor
nta_pyq_2026_jan
Grade 12
Question:
Let $y=y(x)$ be the solution curve of the differential equation $\left(1+x^2\right)dy+\left(y-\tan^{-1}x\right)dx=0$, $y(0)=1$. Then the value of $y(1)$ is:
\dfrac{4}{e^{\pi/4}}+\dfrac{\pi}{2}-1
\dfrac{2}{e^{\pi/4}}+\dfrac{\pi}{4}-1
\dfrac{2}{e^{\pi/4}}-\dfrac{\pi}{4}-1
\dfrac{4}{e^{\pi/4}}-\dfrac{\pi}{2}-1
Step-by-Step Solution
Key Concept: Rewrite: $\dfrac{dy}{dx}+\dfrac{y}{1+x^2}=\dfrac{\tan^{-1}x}{1+x^2}$. IF $=e^{\tan^{-1}x}$. Let $t=\tan^{-1}x$: $\int te^t\,dt=e^t(t-1)+C$.
$y(1)=\dfrac{\pi}{4}-1+\dfrac{2}{e^{\pi/4}}$.
Correct Answer: 2