Mathematics · 2024
JEE Main · 9 April 2024, Shift 1 · Q10
The solution of the differential equation (x^2+y^2) d x-5 x y d y=0, y(1)=0, is:
The solution of the differential equation $\displaystyle \left(x^2+y^2\right) \mathrm{d} x-5 x y \mathrm{~d} y=0, y(1)=0$, is :
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle \left|x^2-4 y^2\right|^5=x^2$
More from Differential Equations
- Let y=y(x) be the solution of the differential equation (d y)/(d x)=2 x(x+y)^3-x(x+y)-1, y(0)=1. Then, (1/(√2)+y(1/(√2)))^2 equals:2024
- Let y=y(x) be the solution of the differential equation (x+y+2)^2 d x=d y, y(0)=-2. Let the maximum and minimum values of the function y=y(x) in [0,…2024
- Let f(x)=x-1 and g(x)=e^x for x ∈ R. If (d y)/(d x)=(e^-2 √x g(f(f(x)))-y/(√x)), y(0)=0, then y(1) is2025
- Let f(x)=√(lim_r → x{(2 r^2[(f(r))^2-f(x) f(r)])/(r^2-x^2)-r^3 e^((f(r))/r)}) be differentiable in (-∞, 0) ∪(0, ∞) and f(1)=1. Then the value of e a,…2024
- Let f: R → R be a thrice differentiable odd function satisfying f^′(x) ≥ 0, f^′ ′(x)=f(x), f(0)=0, f^′(0)=3. Then 9 f( log_e 3) is equal to ____.2025
- Let y=y(x) be the solution of the differential equation (x^2+4)^2 d y+(2 x^3 y+8 x y-2) d x=0. If y(0)=0, then y(2) is equal to2024
- Let for some function y=f(x), ∫_0^x t f(t) d t=x^2 f(x), x>0 and f(2)=3. Then f(6) is equal to2025
- Let f:[1, ∞) →[2, ∞) be a differentiable function. If 10 ∫_1^x f(t) dt=5 x f(x)-x^5-9 for all x ≥ 1, then the value of f(3) is:2025
JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.