Given differential equation is
dxdy=2x+2x+yloge22x⋅y+2y⋅2x
⇒dxdy=2x(1+2yloge2)2x(y+2y)
⇒∫y+2y1+2yloge2dy=∫dx
⇒∫y+2yd(y+2y)=∫dx
⇒ln∣y+2y∣=x+C[∵∫f(x)f′(x)dx=ln(f(x))+C]
Now ∵y(0)=0
⇒C=0
∴ln∣y+2y∣=x
Now for y=1 we have
x=ln(1+2)=ln3∈(1,2)
JEE Main 2021 — Mathematics Calculus
If dxdy=2x+2x+yloge22xy+2y⋅2x,y(0)=0, then for y=1, the value of x lies in the interval :
Held on 31 Aug 2021 · Verified 6 Jul 2026.
(1,2)
(21,1]
(2,3)
(0,21]
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
If $\int\left(\frac{1-5 \cos ^{2} x}{\sin ^{5} x \cos ^{2} x}\right) d x=f(x)+\mathrm{C}$, where C is the constant of integration, then $f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)$ is equal to
If the solution curve $y=f(x)$ of the differential equation $\left(x^{2}-4\right) y^{\prime}-2 x y+2 x\left(4-x^{2}\right)^{2}=0, x>2$, passes through the point $(3,15)$, then the local maximum value of $f$ is $\_\_\_\_$.
Let $[\cdot]$ be the greatest integer function. If $\alpha=\int_{0}^{64}\left(x^{1 / 3}-\left[x^{1 / 3}\right]\right) \mathrm{d} x$, then $\frac{1}{\pi} \int_{0}^{\alpha \pi}\left(\frac{\sin ^{2} \theta}{\sin ^{6} \theta+\cos ^{6} \theta}\right) \mathrm{d} \theta$ is equal to $\_\_\_\_$ .
Let $P_{1}: y=4 x^{2}$ and $P_{2}: y=x^{2}+27$ be two parabolas. If the area of the bounded region enclosed between $P_{1}$ and $P_{2}$ is six times the area of the bounded region enclosed between the line $y=\alpha x, \alpha>0$ and $P_{1}$, then $\alpha$ is equal to :
If $y=y(x)$ satisfies the differential equation $16(\sqrt{x+9 \sqrt{x}})(4+\sqrt{9+\sqrt{x}}) \cos y \mathrm{~d} y=(1+2 \sin y) \mathrm{d} x, x>0$ and $y(256)=\frac{\pi}{2}, y(49)=\alpha$, then $2 \sin \alpha$ is equal to :
Work through every JEE Main Calculus PYQ, year by year.