Mathematics · 2026
JEE Main · 21 January 2026, Shift 1 · Q19
Let y=y(x) be the solution curve of the differential equation (1+x^2) d y+(y- tan^-1 x) d x=0, y(0)=1. Then the value of y(1) is:
Let $\displaystyle y=y(x)$ be the solution curve of the differential equation $\displaystyle \left(1+x^2\right) \mathrm{d} y+\left(y-\tan ^{-1} x\right) d x=0, y(0)=1$. Then the value of $\displaystyle y(1)$ is :
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle \frac{2}{\mathrm{e}^{\pi / 4}}+\frac{\pi}{4}-1$
More from Differential Equations
- Let y=y(x) be the solution curve of the differential equation (1+ sin x) (d y)/(d x)+(y+1) cos x=0, y(0)=0. If the curve y=y(x) passes through the…2026
- Let y=y(x) be the solution of the differential equation x √(1-x^2) d y+(y √(1-x^2)-x cos^-1 x) d x=0, x ∈(0,1), lim_x → 1^- y(x)=1. Then y(1/2)…2026
- Let f: R → R be such that f(x y)=f(x) f(y), for all x, y ∈ R and f(0) ≠ 0. Let g:[1, ∞) → R be a differentiable function such that x^2 g(x)=∫_1^x(t^2…2026
- Let f be a twice differentiable function such that f(x)=∫_0^x tan (t-x) d t-∫_0^x f(t) tan t d t, x ∈(-(π)/2, (π)/2). Then f^′ ′((π)/6)+12…2026
- Let y=y(x) be the solution of the differential equation x sin (y/x) d y=(y sin (y/x)-x) d x, y(1)=(π)/2 and let α= cos ((y(e^12))/(e^12)). Then the…2026
- If the curve y=f(x) passes through the point ( 1, e ) and satisfies the differential equation d y=y(2+ log_e x) d x, x>0, then f(e) is equal to:2026
- Let y=y(x) be the solution of the differential equation (x^2-x √(x^2-1)) d y+(y(x-√(x^2-1))-x) d x=0, x ≥ 1. If y(1)=1, then the greatest integer…2026
- Let x=x(y) be the solution of the differential equation 2 y^2 (d x)/(d y)-2 x y+x^2=0, y>1, x(e)=e. Then x(e^2) is equal to:2026
JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.