datavorous/entrance-exam-dataset huggingface dataset:
The gas released during anaerobic degradation of vegetation may lead to:

Options:
A Corrosion of metals
B Ozone hole
C Global warming and cancer
D Acid rain

The correct answer is:
Global warming and cancer

Solution:
The gas CH4 evolved due to anaerobic degradation of vegetation, which causes global warming and cancer.

Anaerobic degradation means the breakdown in the absence of oxygen. During anaerobic degradation of vegetation, CH4 is produced, which is responsible for global warming as it absorbs sun’s heat. It is also known to cause cancer.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The gas which has similar shape and bond order as that of azide ion is:

Options:
A Sulphur dioxide
B Ozone
C Sulphur trioxide
D Carbon dioxide

The correct answer is:
Carbon dioxide

Solution:
Azide ion N3- has linear shape and bond order = 2, same as CO2.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The gas which is absorbed by ferrous sulphate solution giving blackish brown colour is

Options:
A $\mathrm{NO}$
B $\mathrm{CO}$
C $\mathrm{N}_2$
D $\mathrm{NH}_2$

The correct answer is:
$\mathrm{NO}$

Solution:
$3 \mathrm{FeSO}_4+\mathrm{NO}_2+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+\underset{\text { (BeSlack bromn) }}{\mathrm{FO}_4 \cdot \mathrm{NO}}+\mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Practical Chemistry, JEE Main
The gaseous plant growth regulator is used in plants to:

Options:
A promote root growth and roothair formation to increase the absorption surface
B help overcome apical dominance
C kill dicotyledonous weeds in the fields
D speed up the malting process

The correct answer is:
promote root growth and roothair formation to increase the absorption surface

Solution:
Ethylene is a gaseous plant growth regulator, synthesised by most ripening fruits and ageing tissues. Ethylene is used as both plant growth promoters and plant growth inhibitors. It is mainly used to stimulate the ripening of fruits. It also promotes root growth and root hair formation to increase the absorption surface. Cytokinin can overcome apical dominance caused by auxin. Synthetic auxin kill dicotyledonous weeds in the fields. Gibberellins are used to speed up the malting process in brewing industry.

Topic: Biology, Plant - Growth and Development, NEET, NEET 2022 (Phase 1)
The gaseous reaction $A+B \rightleftharpoons 2 C+D+Q$ is most favoured at

Options:
A low temperature and high pressure
B high temperature and high pressure
C high temperature and low pressure
D low temperature and low pressure

The correct answer is:
low temperature and low pressure

Solution:
For the reaction
$$
A+B \rightleftharpoons 2 C+D+Q
$$
As the reaction is exothermic, low temperature favours formed direction according to Le-Chatelier's principle. Also $\Delta n_{(g)}=(2+1)-(1+1)=1$
$\begin{aligned} \because \quad \Delta n_{\mathrm{g}}=&+\text { ve } \\ & \text { (low pressure favours forward direction) } \end{aligned}$
$\therefore$ Low pressure and temperature favours formed direction for the reaction.

Topic: Chemistry, Chemical Equilibrium, JEE Main
The general displacement of a simple harmonic oscillator is x=A sin ωt. Let T be its time period. The slope of its potential energy (U) – time (t) curve will be maximum when t=Tβ. The value of β is ______.

Options:

The correct answer is:
8

Solution:
Given here, displacement x=A sin ωt

Now, potential energy of a simple harmonic oscillator is U=12mω2x2=12mω2A2sin2ωt,

So, slope of potential energy and time curve is dUdt=12mω3A22sin2ωt

Now, dUdtmax will be maximum when sin2ωt=1

⇒2ωt=π2⇒22πTt=π2⇒t=T8⇒β=8

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (30 Jan Shift 1)
The general formula $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}} \mathrm{O}_2$ could be for open chain

Options:
A
carboxylic acids

B
diols

C
dialdehydes

D
deketones

The correct answer is:
carboxylic acids

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2003
The general formula for alkyl halides is

Options:
A $C_n H_{2 n+1} X$
B $C_n H_{2 n+2} X$
C $\mathrm{C}_n \mathrm{H}_{n+1} \mathrm{X}$
D $C_n H_{2 n} X$

The correct answer is:
$C_n H_{2 n+1} X$

Solution:
General formula of alkanes is $\mathrm{C} \mathrm{H} 2 \mathrm{n}+2$. In case of alkyl halide, one halide $(X)$ is replaced by one hydrogen atom and general formula of alkyl halide becomes $\mathrm{C} \mathrm{H} 2 \mathrm{n}+1 \mathrm{X}$.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The general solution of $\left(\frac{d y}{d x}\right)^2=1-x^2-y^2+x^2 y^2$ is

Options:
A $2 \sin ^{-1} y=x \sqrt{1-x^2}+\sin ^{-1} x+C$
B $\cos ^{-1} y=x \cos ^{-1} x$
C $\sin ^{-1} y=\frac{1}{2} \sin ^{-1} x+C$
D $2 \sin ^{-1} y=x \sqrt{1-y^2}+C$

The correct answer is:
$2 \sin ^{-1} y=x \sqrt{1-x^2}+\sin ^{-1} x+C$

Solution:
$\begin{aligned}
& \text { Given, }\left(\frac{d y}{d x}\right)^2=1-x^2-y^2+x^2 y^2 \\
& \Rightarrow \quad\left(\frac{d y}{d x}\right)^2=\left(1-y^2\right)-x^2\left(1-y^2\right)=\left(1-x^2\right)\left(1-y^2\right) \\
& \therefore \quad \frac{d y}{d x}=\sqrt{\left(1-x^2\right)\left(1-y^2\right)} \\
& \Rightarrow \quad \frac{d y}{\sqrt{1-y^2}}=\sqrt{1-x^2} d x
\end{aligned}$
On integrating both sides, we get
$\begin{aligned}
& \int \frac{d y}{\sqrt{1-y^2}}=\int \sqrt{1-x^2} d x \\
\Rightarrow \quad & \sin ^{-1} y=\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x+\frac{C}{2} \\
\Rightarrow \quad & 2 \sin ^{-1} y=x \sqrt{1-x^2}+\sin ^{-1} x+C
\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
The general solution of $\frac{d y}{d x}+y \tan x=2 x+x^2 \tan x$

Options:
A $y-x^2=c \sec x$
B $y \cos x=x^2 \sec x+c$
C $y \sec x=x^2+c \cos x$
D $y=x^2+c \cos x$

The correct answer is:
$y=x^2+c \cos x$

Solution:
Given differential equation,
$\because$ The differential equation is in linear form, so Integrating factor (I.F.) $=e^{\int \tan x d x}=\sec x$ So, solution of given differential Eq. (i), is
$$
\begin{aligned}
y(\sec x) & =\int\left(2 x+x^2 \tan x\right) \sec x d x \\
& =\int 2 x \sec x d x+\int x^2 \tan x \sec x d x \\
& =\int 2 x \sec x d x+x^2 \sec x-\int 2 x \sec x d x \\
\Rightarrow \quad y \sec x & =x^2 \sec x+c \\
\Rightarrow \quad y & =x^2+c \cos x .
\end{aligned}
$$

Topic: Mathematics, Differential Equations, JEE Main
The general solution of the differential equation $\sqrt{1-x^{2} y^{2}} \cdot d x=y \cdot d x+x \cdot d y$ is

Options:
A $\sin (x y)=x+C$
B $\sin ^{-1}(x y)+x=C$
C $\sin (x+C)=x y$
D $\sin (x y)+x=C$

The correct answer is:
$\sin (x+C)=x y$

Solution:
Given differential equation is
$\begin{array}{ll}
& \sqrt{1-x^{2} y^{2}} \cdot d x=y \cdot d x+x d y \\
\Rightarrow \quad & \sqrt{1-(x y)^{2}} d x=d(x y) \\
\Rightarrow \quad & \int d x=\int \frac{d(x y)}{\sqrt{1-(x y)^{2}}} \quad \text { (on integrating) } \\
\Rightarrow \quad & x=\sin ^{-1}(x y)-C \\
\Rightarrow \quad & \sin (x+C)=x y
\end{array}$

Topic: Mathematics, Differential Equations, JEE Main
The general solution of the differential equation 1+x2+y2+x2y2+xydydx=0 (where C is a constant of integration)

Options:
A 1+y2+1+x2=12loge1+x2-11+x2+1+C
B 1+y2-1+x2=12loge1+x2-11+x2+1+C
C 1+y2+1+x2=12loge1+x2+11+x2-1+C
D 1+y2-1+x2=12loge1+x2+11+x2-1+C

The correct answer is:
1+y2+1+x2=12loge1+x2+11+x2-1+C

Solution:
1+x21+y2+xydydx=0

Integrating,

⇒∫2y21+y2dy=-∫1+x2x1+x2dx
⇒1+y2=-∫x1+x2dx-∫xx21+x2dx

Put 1+x2=t⇒x2=t2-1 to solve RHS 2nd integration,

⇒1+y2=-1+x2-∫tt2-1tdt
⇒1+y2+1+x2=-∫1t2-1dt
⇒1+y2+1+x2=12ln1+x2+11+x2-1+C

Topic: Mathematics, Differential Equations, JEE Main
The general solution of the differential equation 1+x2+y2+x2y2+xydydx=0 (where C is a constant of integration)

Options:
A 1+y2+1+x2=12loge1+x2-11+x2+1+C
B 1+y2-1+x2=12loge1+x2-11+x2+1+C
C 1+y2+1+x2=12loge1+x2+11+x2-1+C
D 1+y2-1+x2=12loge1+x2+11+x2-1+C

The correct answer is:
1+y2+1+x2=12loge1+x2+11+x2-1+C

Solution:
1+x21+y2+xydydx=0

Integrating,

⇒∫2y21+y2dy=-∫1+x2x1+x2dx
⇒1+y2=-∫x1+x2dx-∫xx21+x2dx

Put 1+x2=t⇒x2=t2-1 to solve RHS 2nd integration,

⇒1+y2=-1+x2-∫tt2-1tdt
⇒1+y2+1+x2=-∫1t2-1dt
⇒1+y2+1+x2=12ln1+x2+11+x2-1+C

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (06 Sep Shift 1)
The general solution of the differential equation $(2 x-y+1) d x+(2 y-x+1) d y=0$ is

Options:
A $x^2+y^2+x y-x+y=c$
B $x^2+y^2-x y+x+y=c$
C $x^2-y^2+2 x y-x+y=c$
D $x^2-y^2-2 x y+x-y=0$

The correct answer is:
$x^2+y^2-x y+x+y=c$

Solution:
$\begin{aligned} & (2 x-y+1) d x+(2 y-x+1) d y=0 \\ & \frac{d y}{d x}=\frac{2 x-y+1}{x-2 y-1}, \text { put } x=X+h, y=Y+k \\ & \frac{d Y}{d X}=\frac{2 X-Y+2 h-k+1}{X-2 Y+h-2 k-1} \\ & 2 h-k+1=0 \Rightarrow h-2 k-1=0\end{aligned}$
On solving $h=-1, k=-1 ; \therefore \frac{d Y}{d X}=\frac{2 X-Y}{X-2 Y}$
$\begin{aligned} & \text { Put } Y=v X ; \therefore \frac{d Y}{d X}=v+X \frac{d v}{d X} \\ & v+X \frac{d v}{d X}=\frac{2 X-v X}{X-2 v X}=\frac{2-v}{1-2 v} \\ & X \frac{d v}{d X}=\frac{2-2 v+2 v^2}{1-2 v}=\frac{2\left(v^2-v+1\right)}{1-2 v} \\ & \quad \frac{d X}{X}=\frac{(1-2 v)}{2\left(v^2-v+1\right)} d v \\ & \therefore \quad\end{aligned}$
Put $v^2-v+1=t \Rightarrow(2 v-1) d v=d t$
$\begin{aligned} & \therefore \frac{d X}{X}=-\frac{d t}{2 t} \therefore \log X=\log t^{-1 / 2}+\log c \\ & \therefore X=t^{-1 / 2} c \Rightarrow X=\left(v^2-v+1\right)^{-1 / 2} \cdot c \\ & X^2\left(v^2-v+1\right)=\text { constant } \\ & (x+1)^2\left(\frac{(y+1)^2}{(x+1)^2}-\frac{(y+1)}{x+1}+1\right)= \\ & (y+1)^2-(y+1)(x+1)+(x+1)^2=c \\ & y^2+x^2-x y+x+y=c .\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
The general solution of the differential equation 2x-y+1dx+2y-x+1dy=0 is

Options:
A x2+y2+xy-x+y=c
B x2+y2-xy+x+y=c
C x2-y2+2xy-x+y=c
D x2-y2-2xy+x-y=c

The correct answer is:
x2+y2-xy+x+y=c

Solution:
xdy-2ydy-dy=2xdx-ydx+dx
xdy+ydx-2ydy-dy-2xdx-dx=0
d(xy)-2ydy-dy-2xdx-dx=0

on integrating, we get,
xy-y2-y-x2-x=c

Topic: Mathematics, Differential Equations, JEE Main
The general solution of the differential equation 2xy â xdy+y dx=0 is (Here x, y > 0)

Options:
A log⁡x+yx=c
B log⁡y– xy=c
C log⁡y+xy=c
D None of these

The correct answer is:
log⁡y+xy=c

Solution:
We have, dydx=yx-2xy which is homogeneous.
Put, y=Vx so that dydx=xdVdx+V
â xdVdx=V1-2V- V=2V3/21-2V
â dxx=1-2V2V3/2 dV=12V3/2-1V dV
Integrating, we get,
- c+logâ¡x=- V-1/2- logâ¡V=-xy-logâ¡y+logâ¡x
âlogâ¡y+xy=c.

Topic: Mathematics, Differential Equations, JEE Main
The general solution of the differential equation $(3 y-7 x+7) d x+(7 y-3 x+3) d y=0$ is

Options:
A $(x-y+1)^2(x+y-1)^5=C$
B $(x+y+1)^5(x-y-1)^2=C$
C $(x-y-1)^2(x+y-1)^5=C$
D $(x+y-1)^7=C$

The correct answer is:
$(x-y-1)^2(x+y-1)^5=C$

Solution:
We have,
$(3 y-7 x+7) d x+(7 y-3 x+3) d y=0$
$\Rightarrow(7 x-3 y-7) d x+(3 x-7 y-3) d y=0$ $\ldots(\mathrm{i})$
Given equation is non-homogeneous and
$a_1 b_2-a_2 b_1=-40 \neq 0$
On solving $7 x-3 y-7=0$ and $3 x-7 y-3=0$
we get $x=1, y=0$
Now, substitude $x=1+u$ and $y=0+v=v$
$\therefore d x=d u$ and $d y=d v \ldots$ in Eq. (i),
we get $(7 u-3 v) d u+(3 u-7 v) d v=0$ $\ldots(\mathrm{ii})$
This homogeneous equation in $u$ and $v$.
Put, $u=t v \Rightarrow d u=t d v+v d t$ in Eq. (ii),
we get $(7 t v-3 v)(t d v+v d t)+(3 t v-7 v) d v=0$
$(7 t-3)(t d v+v d t)+(3 t-7) d v=0$
$\Rightarrow \quad\left(7 t^2-7\right) d v+v(7 t-3) d t=0$
$\Rightarrow \quad \int \frac{d v}{v}+\int \frac{7 t-3}{7 t^2-7} d t=0$
$\Rightarrow \log |v|+\int \frac{t-3 / 7}{t^2-1} d t=C_1$
$\Rightarrow \log |v|+\left[\int \frac{2}{7(t-1)} d t+\int \frac{5 / 7}{(t+1)} d t\right]=\log C_1$
$\Rightarrow \log |v|+\frac{2}{7} \log |t-1|+\frac{5}{7} \log |t+1|=\log C_1$
$\Rightarrow \log |v|+\frac{2}{7} \log \left|\frac{u}{v}-1\right|+\frac{5}{7} \log \left|\frac{u}{v}+1\right|=\log C_1$
$\Rightarrow \log |v|+\frac{2}{7} \log \left(\frac{u-v}{v}\right)+\frac{5}{7} \log \left(\frac{u+v}{v}\right)=\log C_1$
$\Rightarrow \log |v|+\log \left(\frac{u-v}{v}\right)^{2 / 7}+\log \left(\frac{u+v}{v}\right)^{5 / 7}=\log C_1$
$\Rightarrow \log \left[|v| \cdot \frac{(u-v)^{2 / 7}}{(v)^{2 / 7}} \cdot \frac{(u+v)^{5 / 7}}{(v)^{5 / 7}}\right]=\log C_1$
$\Rightarrow \quad \log \left[(u-v)^{2 / 7}(u+v)^{5 / 7}\right]=\log C_1$
$\Rightarrow \log \left[(x-y-1)^{2 / 7}(x+y-1)^{5 / 7}\right]=\log C_1$
$\Rightarrow \quad(x-y-1)^{2 / 7}(x+y-1)^{5 / 7}=C_1$
$\Rightarrow \quad(x-y-1)^2(x+y-1)^5=C$

Topic: Mathematics, Differential Equations, JEE Main
The general solution of the differential equation $\frac{d y}{d x}+\frac{1+\cos 2 y}{1-\cos 2 y}=0$ is given by

Options:
A $\tan y+\cot x=c$
B $\tan y-\cot x=c$
C $\tan x-\cot y=c$
D $\tan x+\cot x=c$

The correct answer is:
$\tan y-\cot x=c$

Solution:
We have,
$\quad \frac{d y}{d x}+\frac{1+\cos 2 y}{1-\cos 2 x}=0$ $\Rightarrow \quad \quad \frac{d y}{d x}=-\left(\frac{1+\cos 2 y}{1-\cos 2 x}\right)$ $\Rightarrow \quad \quad \frac{d y}{d x}=-\frac{2 \cos ^{2} y}{2 \sin ^{2} x}$ $\Rightarrow \quad \sec ^{2} y d y=-\operatorname{cosec}^{2} x d x$ $\Rightarrow \quad \int \sec ^{2} y d y+\int \operatorname{cosec}^{2} x d x=0$ $\Rightarrow \quad \tan y-\cot x=c$

Topic: Mathematics, Differential Equations, JEE Main
The general solution of the differential equation $\frac{d y}{d x}=\frac{x+7 y+3}{3 x+5 y+9}$ is

Options:
A $(x-3)^4(y-x+3)^4=c(5 y+x-3)^5$
B $(x+3)^4(y-x-3)^4=c(5 y+x+3)^5$
C $(y-x+3)^4=c|5 y+x-3|$
D $(y-x+3)^4=c|5 y+x+3|$

The correct answer is:
$(y-x+3)^4=c|5 y+x-3|$

Solution:
We have, $\frac{d y}{d x}=\frac{x+7 y+3}{3 x+5 y+9}$
Put, $x=x+h$ and $y=y+k$
$\therefore \quad \frac{d y}{d x}=\frac{(x+7 y)+(h+7 k+3)}{(3 x+5 y)+(3 h+5 k+9)}$
Put $h+7 k+3=0$ and $3 h+5 k+9=0$
$\Rightarrow \quad h=-3, k=0$
$\therefore \quad \frac{d y}{d x}=\frac{x+7 y}{3 x+5 y}$
Now, put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\Rightarrow \quad x \frac{d v}{d x}=\frac{1+7 v-3 v-5 v^2}{3+5 v}$
$\begin{aligned} & \Rightarrow \quad x \frac{d v}{d x}=-\left[\frac{5 v^2-4 v-1}{5 v+3}\right] \\ & \Rightarrow \int \frac{5 v+3}{5 v^2-4 v-1} d v=\int-\frac{1}{x} d x \\ & \Rightarrow \frac{1}{2} \int \frac{10 v+6}{5 v^2-4 v-1} d v=-\int \frac{1}{x} d x \\ & \Rightarrow \frac{1}{2} \int \frac{10 v-4+10}{5 v^2-4 v-1} d v=-\int \frac{1}{x} d x \\ & \Rightarrow \frac{1}{2} \int \frac{10 v-4}{5 v^2-4 v-1} d v+5 \int \frac{1}{5 v^2-4 v-1} d v \\ & =-\log x+\log k \\ & \Rightarrow \frac{1}{2} \log \left[5 v^2-4 v-1\right]+\frac{5}{5} \int \frac{1}{v^2-\frac{4}{5} v-\frac{1}{5}} d v=\log \left(\frac{k}{x}\right) \\ & \Rightarrow \frac{1}{2} \log \left[5 v^2-4 v-1\right]+\int \frac{1}{\left(v-\frac{2}{5}\right)^2-\frac{1}{5}-\frac{4}{25}} d v=\log \frac{k}{x} \\ & \Rightarrow \frac{1}{2} \log \left[5 v^2-4 v-1\right]+\int \frac{1}{\left(v-\frac{2}{5}\right)^2-\left(\frac{3}{5}\right)^2} d v=\log \frac{k}{x} \\ & \Rightarrow \frac{1}{2} \log \left[5 v^2-4 v-1\right]+\frac{1}{2 \times \frac{3}{5}} \log \frac{v-\frac{2}{5}-\frac{3}{5}}{v-\frac{2}{5}+\frac{3}{5}}=\log \frac{k}{x} \\ & \Rightarrow \frac{1}{2} \log \left[5 v^2-4 v-1\right]+\frac{5}{6} \log \frac{5 v-5}{5 v+1}=\log \frac{k}{x} \\ & \end{aligned}$
$\begin{aligned} & \Rightarrow \frac{1}{6}\left[\log \left(5 v^2-4 v-1\right)^3+\log \left(\frac{5 v-5}{5 v+1}\right)^5\right]=\log \frac{k}{x} \\ & \Rightarrow \log \left[\left(5 v^2-4 v-1\right)^3 \times 5^5 \frac{(v-1)^5}{(5 v+1)^5}\right]=\log \frac{k^6}{x^6} \\ & \Rightarrow(5 v+1)^3(v-1)^3 \times \frac{(v-1)^5}{(5 v+1)^5}=\frac{c}{x^6}, \text { where } c=\frac{k^6}{5^5} \\ & \Rightarrow \quad \frac{(v-1)^8}{(5 v+1)^2}=\frac{c}{x^6} \Rightarrow \frac{(v-1)^4}{5 v+1}=\frac{c}{x^3} \\ & \Rightarrow \quad \frac{(y-x)^4}{x^4\left[\frac{5 y+x}{x}\right]}=\frac{c}{x^3} \Rightarrow(y-x)^4=c(5 y+x) \\ & \Rightarrow \quad \frac{(y-x+3)^4}{2}=c(5 y+x-3)\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
The general solution of the differential equation $\frac{d y}{d x}+\sin (x+y)=\sin (x-y)$ is

Options:
A $\log \tan y+\sin x=\mathrm{C}$
B $\quad \log \tan \frac{y}{2}+2 \sin x=\mathrm{C}$
C $\tan \frac{y}{2}+\log \sin x=\mathrm{C}$
D None of these

The correct answer is:
$\quad \log \tan \frac{y}{2}+2 \sin x=\mathrm{C}$

Solution:
The equation is,

$\frac{d y}{d x}=\sin (x-y)-\sin (x+y)=2 \cos x \sin (-y)$

$\Rightarrow \frac{d y}{\sin y}+2 \cos x d x=0$

$\Rightarrow \int \cos e c y d y+2 \int \cos x d x=\mathrm{C}$

$\Rightarrow \log \tan \frac{y}{2}+2 \sin x=\mathrm{C}$

Topic: Mathematics, Differential Equations, JEE Main
The general solution of the differential equation $\frac{d y}{d x}+\frac{2}{x} y=x^2$ is

Options:
A
$y=c x^{-3}-\frac{x^2}{4}$

B
$y=c x^3-\frac{x^2}{4}$

C
$y=c x^2+\frac{x^3}{5}$

D
$y=c x^{-2}+\frac{x^3}{5}$

The correct answer is:
$y=c x^{-2}+\frac{x^3}{5}$

Solution:
Given differential equation is
$$
\frac{d y}{d x}+\frac{2}{x} \cdot y=x^2
$$
This is of the linear form.
$$
\begin{aligned}
& \therefore P=\frac{2}{x}, Q=x^2 \\
& \text { I.F }=e^{\int \frac{2}{x} d x}=e^{\log x^2}=x^2
\end{aligned}
$$
Solution is
$$
\begin{aligned}
& y \cdot x^2=\int x^2 \cdot x^2 d x+c=\frac{x^5}{5}+c \\
& y=\frac{x^3}{5}+c x^{-2}
\end{aligned}
$$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2012 (19 May Online)
The general solution of the differential equation dydx=2ytanâ¡x+tan2â¡x, âxâ0,Ï2 is yfx=x2-sinâ¡2x4+C, (where, C is an arbitrary constant). If fÏ4=12, then the value of fÏ3 is equal to

Options:
A 12
B 14
C 2
D 4

The correct answer is:
14

Solution:
Given equation is
dydx+y-2tan⁡x=tan2⁡x
IF=e2ln⁡cos⁡x=cos2⁡x
Thus, the solution is
y⋅cos2⁡x=∫tan2⁡x·cos2⁡xdx
⇒y⋅cos2⁡x=∫1-cos⁡2x2dx
=x2-sin⁡2x4+C
⇒fx=cos2⁡x
⇒fπ3=122=14

Topic: Mathematics, Differential Equations, JEE Main
The general solution of the differential equation dydx+sin⁡x+y2=sin⁡x-y2 is (where c is an arbitrary constant)

Options:
A lntanâ¡y2=c-2sinâ¡x
B lnâ¡tanâ¡y4=c-2sinâ¡x2
C lnâ¡tanâ¡y2+Ï4 =c-2sinâ¡x
D lnâ¡tanâ¡y4+Ï4 =c-2sinâ¡ x2

The correct answer is:
ln⁡tan⁡y4=c-2sin⁡x2

Solution:
Given equation
dydx+sinâ¡x+y2=sinâ¡x-y2
â dydx=sinâ¡x-y2-sinâ¡x+y2
â dydx=-2sinâ¡y2 cosâ¡x2
âcosec y2dy=-2cosâ¡x2dx
On integrating both sides, we get
â«cosec y2dy=-â«2cosâ¡x2dx
â lnâ¡(tanâ¡y4) 12=-2sinâ¡x212+c
âln(tanâ¡y4)=c-2sinâ¡x2

Topic: Mathematics, Differential Equations, JEE Main
The general solution of the differential equation \( \frac{d y}{d x}+\sin \left(\frac{x+y}{2}\right)=\sin \left(\frac{x-y}{2}\right) \) is (where \( c \) is an arbitrary constant)

Options:
A \( \ln \tan \left(\frac{y}{2}\right)=c-2 \sin x \)
B \( \ln \tan \left(\frac{y}{4}\right)=c-2 \sin \left(\frac{x}{2}\right) \)
C \( \ln \tan \left(\frac{y}{2}+\frac{\pi}{4}\right)=c-2 \sin x \)
D \( \ln \tan \left(\frac{y}{4}+\frac{\pi}{4}\right)=c-2 \sin \left(\frac{x}{2}\right) \)

The correct answer is:
\( \ln \tan \left(\frac{y}{4}\right)=c-2 \sin \left(\frac{x}{2}\right) \)

Solution:
Given equation

dydx+sinx+y2=sinx-y2

⇒ dydx=sinx-y2-sinx+y2

⇒ dydx=-2siny2 cosx2

⇒ cosec y2dy=-2cosx2dx

On integrating both sides, we get

∫cosec y2dy=-∫2cosx2dx

⇒lntany412=−2sinx212+c

⇒lntany4=c−2sinx2

Topic: Mathematics, Differential Equations, JEE Main
The general solution of the differential equation, $\sin 2 x\left(\frac{d y}{d x}-\sqrt{\tan x}\right)-y=0$, is :

Options:
A
$y \sqrt{\tan x}=x+c$

B
$y \sqrt{\cot x}=\tan x+c$

C
$y \sqrt{\tan x}=\cot x+c$

D
$y \sqrt{\cot x}=x+c$

The correct answer is:
$y \sqrt{\cot x}=x+c$

Solution:
Given, $\sin 2 x\left(\frac{d y}{d x}-\sqrt{\tan x}\right)-y=0$ or, $\frac{d y}{d x}=\frac{y}{\sin 2 x}+\sqrt{\tan x}$ or, $\frac{d y}{d x}-y \operatorname{cosec} 2 x=\sqrt{\tan x}$
Now, integrating factor (I.F) $=e^{\int-\operatorname{cosec} 2 x}$ or, I.F $=e^{-\frac{1}{2} \log |\tan x|}=e^{\log (\sqrt{\tan x})^{-1}}$ $=\frac{1}{\sqrt{\tan x}}=\sqrt{\cot x}$
Now, general solution of eq. (1) is written as
$$
\begin{aligned}
&\text { y (I. F.) }=\int \mathrm{Q}(\text { I.F.) } d x+c \\
&\therefore y \sqrt{\cot x}=\int \sqrt{\tan x} \cdot \sqrt{\cot x} d x+c \\
&\therefore y \sqrt{\cot x}=\int 1 . d x+c
\end{aligned}
$$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2014 (12 Apr Online)
The general solution of the differential equation $\left(\tan ^{-1} y-x\right) d y=\left(1+y^{2}\right) d x$ is

Options:
A $x=\left(\tan ^{-1} y+1\right)+\mathrm{C} e^{-\tan ^{-1} y}$
B $x=\left(\tan ^{-1} y-1\right)+\mathrm{C} e^{-\tan ^{-1} y}$
C $x=\left(\tan ^{-1} x-1\right)+\mathrm{C} e^{-\tan ^{-1} x}$
D $x=\left(\tan ^{-1} x+1\right)+\mathrm{Ce}^{-\tan ^{-1} x}$

The correct answer is:
$x=\left(\tan ^{-1} y-1\right)+\mathrm{C} e^{-\tan ^{-1} y}$

Solution:
The given differential equation can be written as $\frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{\tan ^{-1} y}{1+y^{2}} \ldots$ (i)

Now, eq. (i) is a linear differential equation of the form $\frac{d x}{d y}+\mathrm{P}_{1} x=\mathrm{Q}_{1}$

where $\mathrm{P}_{1}=\frac{1}{1+y^{2}}$ and $\mathrm{Q}_{1}=\frac{\tan ^{-1} y}{1+y^{2}}$

Therefore, I.F $=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y}$

Thus, the solution of the given differential equation is given by $x e^{\tan ^{-1} y}=\int\left(\frac{\tan ^{-1} y}{1+y^{2}}\right) e^{\tan ^{-1} y} \mathrm{dy}+\mathrm{C} \ldots$ (ii)

Let $\mathrm{I}=\int\left(\frac{\tan ^{-1} y}{1+y^{2}}\right) e^{\tan ^{-1} y} \mathrm{dy}$

On substituting $\tan ^{-1} y=t$, so that

$\left(\frac{1}{1+y^{2}}\right) \mathrm{dy}=\mathrm{dt}$, we get

$I=\int t e^{t} d t=t e^{t}-\int 1 \cdot e^{t} d t=t e^{t}-e^{t}=e^{t}(t-1)$

or $\mathrm{I}=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)$

On substituting the value of 1 in equation (i1), we get

$x \cdot e^{\tan ^{-1} y}=\cdot e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)+\mathrm{C}$

or $x=\left(\tan ^{-1} y-1\right)+\mathrm{C} e^{\tan ^{-1} y}$

which is the general solution of the given differntial equation.

Topic: Mathematics, Differential Equations, JEE Main
The general solution of the differential equation x2+xyy'=y2 is

Options:
A eyx=cx
B cy=e-yx
C e-yx=cxy
D e-2yx=cy

The correct answer is:
cy=e-yx

Solution:
Given that,

x2+xyy1=y2

x2+xydydx=y2
dydx=y2x2+xy …1

Let y=vx

Differentiating w.r.t. 'x'?

dydx=ddxvx

dydx=v+x·dvdx

y2x2+xy=v+x·dvdx

v2x2x2+xvx=v+xdvdx

v21+v=v+x·dvdx

v21+v-v=xdvdx

v2-v-v21+v=xdvdx

-v1+v=xdvdx

1xdx=-1+vv·dv

Integrating on both sides

∫1xdx=-∫(1+v)vdv

logx=-∫1vdv+∫vvdv

logx=-logv+v+logc

logx=-logyx+yx+logc

logx+logyx+logc=-yx

logx·yx·c=-yx

cy=e-yx

Topic: Mathematics, Differential Equations, JEE Main