Let v=x+y, so dxdv=1+dxdy=1+v2λ2=v2v2+λ2.
Separate: v2+λ2v2dv=dx.
Write as (1−v2+λ2λ2)dv=dx.
Integrate: v−λtan−1(v/λ)=x+C1.
Substitute back: λtan−1(λx+y)=y+C.
CUET UG 2022 — Mathematics Calculus
The solution of the differential equation dxdy=(x+y)2λ2 (λ is constant) is:
Held on 4 Aug 2022 · Verified 13 Jul 2026.
y=x+y−λ2+C
3(x+y)3=λ2x+C
λtan−1(λx+y)=y+C
y=(x+y)3−2λ2+C
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
The derivative of x³ + 2x² - 5x + 1 is:
If the function $f(x) = 2x^3 + 9x^2 + 12x-1$ is given,then $f(x)$ have
If $\int \frac{dx}{(x-1)^3/^4. (x+2)^5/^4} = a[1 - g(x)]^b + c$, where $c$ is a constant of integration, then which of the following are true? (A) $a = \frac{2}{3}$ (B) $\beta = \frac{3}{4}$ (C) $3\alpha + 4\beta = 5$ (D) $g(x) = \frac{3}{(x+2)}$ Choose the **correct** answer from the options given below:
Match **List-I** with **List-II** | List-I | List-II | |---|---| | **Function** | **Increasing on the interval** | | (A) $f(x) = -x^2 - 2x + 1$ | (I) $(-\infty, -1)$ | | (B) $f(x) = x^2 + 1$ | (II) $(1, \infty)$ | | (C) $f(x) = x^2 - 2x + 3$ | (III) $(-\infty, 0)$ | | (D) $f(x) = -x^2$ | (IV) $(0, \infty)$ | Choose the correct answer from the options given below:
The integral $\int e^x\left(\frac{x-1}{2x^2}\right)dx$ is equal to
Work through every CUET UG Calculus PYQ, year by year.