datavorous/entrance-exam-dataset huggingface dataset:
The solution of the differential equation $\left(1+x^2\right) \frac{d y}{d x}=x$ is

Options:
A $y=\tan ^{-1} x+c$
B $y=-\tan ^{-1} x+c$
C $y=\frac{1}{2} \log _e \left(1+x^2\right)+c$
D $y=-\frac{1}{2} \log _e\left(1+x^2\right)+c$

The correct answer is:
$y=\frac{1}{2} \log _e \left(1+x^2\right)+c$

Solution:
$\begin{aligned} & \left(1+x^2\right) \frac{d y}{d x}=x \Rightarrow d y=\frac{x}{1+x^2} d x \\ & \Rightarrow \int d y=\int \frac{x}{1+x^2} d x+c \Rightarrow y=\frac{1}{2} \log _e\left(1+x^2\right)+c\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation 1x2dydx2+6=5xdydx is y=λx2+c (where, c is an arbitrary constant). The sum of all the possible value of λ is

Options:
A 32
B 52
C 25
D 2

The correct answer is:
52

Solution:
Given equation is dydx2-5xdydx+6x2=0 or dydx-3xdydx-2x=0
⇒y=32x2+k or ⇒y=x2+Î¼

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation 1-x2dydx-xy=1 is (where, x<1, x∈R and C is an arbitrary constant)

Options:
A y1-x2=tan-1x+C
B y1-x2=tan-1x+C
C y1-x2=sin-1x+C
D y⋅1-x2=sin-1x+C

The correct answer is:
y1-x2=sin-1x+C

Solution:
dydx-x1-x2y=11-x2
Ι.F.=e-∫x1-x2dx
=e12∫-2x1-x2dx=e12ln1-x2=1-x2

Hence, the solution of the differential equation is
y1-x2=∫1-x21-x2dx
y1-x2=∫11-x2dx
y1-x2=sin-1x+C

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation 1−x2⋅dydx+xy=x−x3y12, ∀x<1 is 9y=−fx+c1−x214, where c is an arbitrary constant and f12=34. Then, fx is

Options:
A an odd function
B an even function
C a periodic function
D symmetric about line x=1

The correct answer is:
an even function

Solution:
Given equation is 1ydydx+x1-x2y=x
Let 2y=ν
⇒1ydydx=dνdx
Thus, we have dνdx+x21-x2⋅ν=x
∴I.F.=e∫x21-x2dx
=e-14ln⁡1-x2
=1-x2-14
Thus, the solution is ν1-x2-14=∫x1-x2-14dx
or ν⋅1-x2-14=-231-x234+c"
or 2y=-231-x2+c"1-x214
⇒y=-1-x23+c'1-x214

⇒9y=-1-x2+c1-x214
⇒fx=1-x2
∴fx is an even function

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation $\left(1+y^2\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0$, is

Options:
A $x e^{\tan ^{-1} y}=\tan ^{-1 y}+C$
B $x e^{2 \tan ^{-1} y}=e^{-\tan ^{-1} y}+C$
C $2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+C$
D $x^2 e^{\tan ^{-1} y}=4 e^{2 \tan ^{-1} y}+C$

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

Solution:
Given differential equation,
$\begin{aligned}
& \left(1+y^2\right)\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0 \\
& \frac{d y}{d x}+\frac{x}{1+y^2}=\frac{e^{\tan ^{-1} y}}{1+y^2} \\
& P=\frac{x}{1+y^2}, \mathrm{Q}=\frac{e^{\tan ^{-1}} y}{1+y^2}
\end{aligned}$
Now, integration factor,
$\mathrm{IF}=e^{\int p d y}=e^{l \frac{1}{1+y^2} d y}=e^{\tan ^{-1} y}$
The solution of differential equation,
$\begin{aligned}
& x . I F=\int Q . I F d y+C \\
& \Rightarrow x e^{\tan ^{-1} y}=\int \frac{e^{2 \tan ^{-1} y}}{1+y^2} d y+C \\
& \Rightarrow \quad x e^{\tan ^{-1} y}=\frac{e^{2 \tan ^{-1} y}}{1+y^2}+C \\
& \Rightarrow \quad 2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+C
\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation $\left(1+y^2\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0$, is

Options:
A
$x e^{2 \tan ^{-1} y}=e^{\tan ^{-1} y}+k$

B
$(x-2)=k e^{2 \tan ^{-1} y}$

C
$2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+k$

D
$x e^{\tan ^{-1} y}=\tan ^{-1} y+k$

The correct answer is:
$2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+k$

Solution:
$\left(1+y^2\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0$
$\left(1+y^2\right) \frac{d x}{d y}+x=e^{\tan ^{-1} y} \Rightarrow \frac{d x}{d y}+\frac{x}{\left(1+y^2\right)}=\frac{e^{\tan ^{-1} y}}{\left(1+y^2\right)}$
I.F. $=e^{\int \frac{1}{\left(1+y^2\right)} d y}=e \tan ^{-1} y$
$x\left(e^{\tan ^{-1} y}\right)=\int \frac{e^{\tan ^{-1} y}}{1+y} e^{\tan ^{-1} y d y}$
$x\left(e^{\tan ^{-1} y}\right)=\frac{e^{2 \tan ^{-1} y}}{2}+C \quad \therefore 2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+k$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2003
The solution of the differential equation 2xexdy+exydx=xsinxdx is (where, c is an arbitrary constant)

Options:
A 2yex+sin⁡x=c
B ysin⁡x=ex+c
C yex+sin⁡x=c
D 2yex+cos⁡x=c

The correct answer is:
2yex+cos⁡x=c

Solution:
The given equation is e x ⋅dy+ e x 2 x dx⋅y= sinx 2 dx
or dexy=12sin⁡xdx
On integrating, we get,
ex⋅y=−cosx2+c'

i.e. 2yex+cos⁡x=c

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation 2ydx+xdy=2xydx is (where, C is an arbitrary constant)

Options:
A xy=x+C
B xy=x22+C
C xy=x+C
D xy=C

The correct answer is:
xy=x22+C

Solution:
Given equation can be written as dxy+x12ydy=xdx
or dxy=xdx
∫dxy=∫xdx⇒xy=x22+C

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation 3sin2⁡xcos⁡xy2dx+2ysin3⁡xdy=sin⁡xdx is (where, C is an arbitrary constant)

Options:
A 2y2sin⁡x=cos⁡x+C
B y2sin3⁡x+cos⁡x=C
C y3sin2⁡x+sin⁡x=C
D ysin⁡x=cos2⁡x+C

The correct answer is:
y2sin3⁡x+cos⁡x=C

Solution:
The given equation is dy2sin3⁡x=sin⁡xdx
On integrating, we get,

∫dy2sin3⁡x=∫sin⁡xdx
⇒y2sin3⁡x=-cos⁡x+C
or y2sin3⁡x+cos⁡x=C

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation 3x2sin1x+ydx=xcos1xdx-xdy is

(where, c is an arbitrary constant)

Options:
A sin1x=xy+c
B x3sin1x+xy=c
C x3sin1x=xy+c
D sinx=x3y+c

The correct answer is:
x3sin1x+xy=c

Solution:
Given equation can be written as dx3sin1x+dxy=0
∴ solution is x3sin1x+xy=c

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation $\left(3 x y+y^2\right) d x+\left(x^2+x y\right) d y=0$ is

Options:
A $x^2\left(2 x y+y^2\right)=c^2$
B $x^2\left(2 x y-y^2\right)=c^2$
C $x^2\left(y^2-2 x y\right)=c^2$
D None of these

The correct answer is:
$x^2\left(2 x y+y^2\right)=c^2$

Solution:
It can be written in the form of homogeneous equation $\frac{d y}{d x}=-\frac{3 x y+y^2}{x^2+x y}$
So, now put $y=v x$ and $\frac{d y}{d x}-v+x \frac{d v}{d x}$, we get
$\begin{aligned} & \quad v+x \frac{d v}{d x}=-\frac{3 x^2 v+x^2 v^2}{x^2+x^2 v} \Rightarrow x \frac{d v}{d x}=\frac{-2 v(v+2)}{v+1} \\ & \Rightarrow \quad \frac{1}{x} d x=-\frac{v+1}{2 v(v+2)} d v=-\left[\frac{1}{2(v+2)}+\frac{1}{2 v(v+2)}\right] d v \\ & \Rightarrow \quad-\frac{2}{x} d x=\left[\frac{1}{v+2}+\frac{1}{2 v}-\frac{1}{2 v(v+2)}\right]\end{aligned}$
On integrating, we get
$\begin{aligned} & -2 \log _e x=\frac{1}{2} \log (v+2)+\frac{1}{2} \log v+\log c \\ & \Rightarrow v(v+2) x^4=c^2 \Rightarrow \frac{y}{x}\left(\frac{y}{x}+2\right) x^4=c^2 \quad\left(v v=\frac{y}{x}\right)\end{aligned}$
Hence required solution is $\left(y^2+2 x y\right) x^2=c^2$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation $\left(3 x y+y^{2}\right) d x+\left(x^{2}+x y\right) d y=0$ is

Options:
A $x^{2}\left(2 x y+y^{2}\right)=c^{2}$
B $x^{2}\left(2 x y-y^{2}\right)=c^{2}$
C $x^{2}\left(y^{2}-2 x y\right)=c^{2}$
D None of these

The correct answer is:
$x^{2}\left(2 x y+y^{2}\right)=c^{2}$

Solution:
Homogeneous equation can be written in the form of
Put
$$
\begin{aligned}
\frac{d y}{d x} &=-\frac{3 x y+y^{2}}{x^{2}+x y} \\
y=& v x \text { and } \frac{d y}{d x}=v+x \frac{d v}{d x}, \text { we get }
\end{aligned}
$$
$$
v+x \frac{d v}{d x}=-\frac{3 x^{2} v+x^{2} v^{2}}{x^{2}+x^{2} v}
$$
$$
\Rightarrow \quad x \frac{d v}{d x}=\frac{-2 v(v+2)}{v+1}
$$
$$
\begin{array}{ll}
\Rightarrow \quad & \frac{1}{x} d x=-\frac{(v+1)}{2 v(v+2)} d v \\
\Rightarrow \quad & -\frac{2}{x}=-\left[\frac{1}{2(v+2)}+\frac{1}{2 v}\right] d v
\end{array}
$$
On integrating, we get
$$
\begin{aligned}
&-2 \log _{e} x=\frac{1}{2} \log (v+2)+\frac{1}{2} \log v-\log c \\
\Rightarrow & v(v+2) x^{4}=c^{2} \\
\Rightarrow & \frac{y}{x}\left(\frac{y}{x}+2\right) x^{4}=c^{2} &\left(\because v=\frac{y}{x}\right)
\end{aligned}
$$
Hence, required solution is $\left(y^{2}+2 x y\right) x^{2}=c^{2}$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation cosâ¡xdydx+ysinâ¡x=1 is (where, c is an arbitrary constant)

Options:
A y=csin⁡x+cos⁡x
B y=sin⁡x+ccos⁡x
C y=tan⁡x+c
D ysin⁡x=sin⁡x+c

The correct answer is:
y=sin⁡x+ccos⁡x

Solution:
dydx+ytan⁡x=sec⁡x
Ι.F.=e∫tan⁡xdx=elogsecx=sec⁡x
Hence, the solution is, y⋅sec⁡x=∫sec2xdx
⇒ysec⁡x=tan⁡x+c
⇒y=sin⁡x+ccos⁡x

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation $\frac{d^2 y}{d x^2}=0$ represents

Options:
A all circles in a plane
B all straight lines in a plane
C all parabolas in a plane
D all ellipses in a plane

The correct answer is:
all straight lines in a plane

Solution:
Given, differential equation is $\frac{d^2 y}{d x^2}=0$
On integrating, we get $\frac{d y}{d x}=c$, where $c$ is a constant
$$
\Rightarrow d y=c d x
$$
Again integrating,
$$
\begin{aligned}
& \Rightarrow \quad \int d y=c \int d x \\
& \Rightarrow \quad y=c x+d, \text { where } d \text { is a constant. }
\end{aligned}
$$
$\therefore$ The solution of differential equation represents all straight lines in a plane.

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation $\frac{d y}{d x}=(1+x)\left(1+y^2\right)$ is

Options:
A $y=\tan \left(x^2+x+c\right)$
B $y=\tan \left(2 x^2+x+c\right)$
C $y=\tan \left(x^2-x+c\right)$
D $y=\tan \left(\frac{x^2}{2}+x+c\right)$

The correct answer is:
$y=\tan \left(\frac{x^2}{2}+x+c\right)$

Solution:
$\frac{d y}{d x}=(1+x)\left(1+y^2\right) \Rightarrow \frac{d y}{1+y^2}=(1+x) d x$
On integrating both sides, we get
$\tan ^{-1} y=\frac{x^2}{2}+x+c \Rightarrow y=\tan \left(\frac{x^2}{2}+x+c\right)$.

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

Options:
A $\cos ^{-1} x+\cos ^{-1} y=c \quad$
B $\sin ^{-1} x+\sin ^{-1} y=c$
C $$
\text { } \cosh ^{-1} x+\cosh ^{-1} y=c
$$
D $$
\sinh ^{-1} x+\sinh ^{-1} y=c
$$

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

Solution:
Given, $\frac{d y}{d x}+\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}=0$
$$
\Rightarrow \quad \frac{d y}{\sqrt{1-y^2}}+\frac{d x}{\sqrt{1-x^2}}=0
$$
On integrating, we get
$$
\begin{aligned}
& \Rightarrow \quad \int \frac{d y}{\sqrt{1-y^2}}+\int \frac{d x}{\sqrt{1-x^2}}=0 \\
& \Rightarrow \quad \sin ^{-1} y+\sin ^{-1} x=c
\end{aligned}
$$

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

Options:
A $1+x y+d(y+x)=0$
B $x+y=c(1-x y)$
C $y-x=c(1+x y)$
D $1+x y=c(x+y)$

The correct answer is:
$y-x=c(1+x y)$

Solution:
$\frac{d y}{d x}=\frac{1+y^2}{1+x^2} \Rightarrow \frac{1}{1+y^2} d y=\frac{1}{1+x^2} d x$
Now on integrating both sides, we get
$\begin{aligned} & \tan ^{-1} y=\tan ^{-1} x+\tan ^{-1} c \Rightarrow \tan ^{-1} y=\tan ^{-1}\left(\frac{x+c}{1-c x}\right) \\ & \Rightarrow y=\frac{x+c}{1-c x} \Rightarrow y-x=c(1+x y) .\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation $\frac{d y}{d x}+\frac{3 x^2}{1+x^3} y=\frac{\sin ^2 x}{1+x^3}$ is

Options:
A $y\left(1+x^3\right)=x+\frac{1}{2} \sin 2 x+c$
B $y\left(1+x^3\right)=c x+\frac{1}{2} \sin 2 x$
C $y\left(1+x^3\right)=c x-\frac{1}{2} \sin 2 x$
D $y\left(1+x^3\right)=\frac{x}{2}-\frac{1}{4} \sin 2 x+c$

The correct answer is:
$y\left(1+x^3\right)=\frac{x}{2}-\frac{1}{4} \sin 2 x+c$

Solution:
$\begin{aligned} & \frac{d y}{d x}+\frac{3 x^2}{1+x^3} y=\frac{\sin ^2 x}{1+x^3} \\ & \text { Here, } \quad P=\frac{3 x^2}{1+x^3} \Rightarrow \text { I.F. }=e^{\int P d x}=e^{\log \left(1+x^3\right)}=1+x^3\end{aligned}$
Thus the solution is
$\begin{aligned} & y \cdot\left(1+x^3\right)=\int \frac{\sin ^2 x}{1+x^3}\left(1+x^3\right) d x=\int \frac{1-\cos 2 x}{2} d x \\ & \Rightarrow y\left(1+x^3\right)=\frac{1}{2} x-\frac{\sin 2 x}{4}+c\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation $\frac{d y}{d x}=1+x+y+x y$ is

Options:
A $\log (1+y)=x+\frac{x^2}{2}+c$
B $(1+y)^2=x+\frac{x^2}{2}+c$
C $\log (1+y)=\log (1+x)+c$
D None of these

The correct answer is:
$\log (1+y)=x+\frac{x^2}{2}+c$

Solution:
$\begin{aligned} & \frac{d y}{d x}=1+x+y+x y \\ & \Rightarrow \frac{d y}{d x}=(1+x)(1+y) \Rightarrow \frac{d y}{1+y}=(1+x) d x\end{aligned}$
On integrating, we get $\log (1+y)=\frac{x^2}{2}+x+c$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation $\frac{d y}{d x}=\frac{x-y+3}{2(x-y)+5}$ is

Options:
A $2(x-y)+\log (x-y)=x+c$
B $2(x-y)-\log (x-y+2)=x+c$
C $2(x-y)+\log (x-y+2)=x+c$
D None of the above

The correct answer is:
$2(x-y)+\log (x-y+2)=x+c$

Solution:
Given differential equation is
$$
\frac{d y}{d x}=\frac{x-y+3}{2(x-y)+5}
$$
Let $\quad x-y=v \Rightarrow \frac{d y}{d x}=1-\frac{d v}{d x}$
$\therefore$
$$
\frac{d v}{d x}=\frac{v+2}{2 v+5}
$$
$$
\begin{array}{l}
\Rightarrow \quad \int \frac{2 v+5}{v+2} d v=\int d x \\
\Rightarrow \quad \int\left(2+\frac{1}{v+2}\right) d v=\int d x
\end{array}
$$
$\Rightarrow \quad 2 v+\log (v+2)=x+c$
$\Rightarrow 2(x-y)+\log (x-y+2)=x+c$

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

Options:
A $y=\sec x+\tan x+c$
B $y=\sec x+\cot x+c$
C $y=\sec x-\tan x+c$
D None of these

The correct answer is:
$y=\sec x+\tan x+c$

Solution:
$\frac{d y}{d x}=\sec x(\sec x+\tan x) \Rightarrow \frac{d y}{d x}=\sec ^2 x+\sec x \tan x$
Now integrating both sides, we get $y=\tan x+\sec x+c$.

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

Options:
A $y=-\frac{1}{2} \tan ^{-1} x+c$
B $y+\log x+\frac{x^2}{2}+c=0$
C $y=\frac{1}{2} \tan ^{-1} x+c$
D $y-\log x-\frac{x^2}{2}=c$

The correct answer is:
$y+\log x+\frac{x^2}{2}+c=0$

Solution:
$\frac{d y}{d x}+\frac{1+x^2}{x}=0 \Rightarrow d y+\left(\frac{1}{x}+x\right) d x=0$
On integrating, we get$y+\log x+\frac{x^2}{2}+c=0$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation dxdy=x2ey-x âx>0 is Î»x+2cx2ey=ey (where, c is an arbitrary constant). Then, Î» is equal to

Options:
A 2
B 4
C 12
D 14

The correct answer is:
2

Solution:
The given equation is e −y dy dx + e −y ⋅ 1 x = 1 x 2
Let, -e-y=t⇒e-ydydx=dtdx
Thus we have,
dtdx-t1x=1x2
Integrating factor =e-∫1xdx=e-ln⁡x=1x
Thus, the solution is
t⋅ 1 x =∫ 1 x 2 ⋅ 1 x dx
⇒-eyx=x-2-2+c
⇒-2xe-y=-1+2cx2 or 2xe-y=1-2cx2
Hence, λ=2

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation $\frac{d y}{d x}=x^2+\sin 3 x$ is

Options:
A $y=\frac{x^3}{3}+\frac{\cos 3 x}{3}+c$
B $y=\frac{x^3}{3}-\frac{\cos 3 x}{3}+c$
C $y=\frac{x^3}{3}+\sin 3 x+c$
D None of these

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

Solution:
$\frac{d y}{d x}=x^2+\sin 3 x$. On integrating, $y=\frac{x^3}{3}-\frac{\cos 3 x}{3}+c$

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

Options:
A $a y^2=e^{x^2 / y^2}$
B $a y=e^{x / y}$
C $y=e^{x^2}+e^{y^2}+c$
D $y=e^{x^2}+y^2+c$

The correct answer is:
$a y^2=e^{x^2 / y^2}$

Solution:
Given $\frac{d y}{d x}=\frac{x y}{x^2+y^2}$. Put $y=v x ; \frac{d y}{d x}=v+x \cdot \frac{d v}{d x}$
$\begin{aligned} & \therefore v+x \frac{d v}{d x}=\frac{(x)(v x)}{x^2+v^2 x^2} \\ & \Rightarrow v+x \cdot \frac{d v}{d x}=\frac{v}{1+v^2} \Rightarrow x \frac{d v}{d x}=\frac{-v^3}{1+v^2} \\ & \Rightarrow \frac{\left(1+v^2\right)}{v^3} d v=-\frac{d x}{x} \Rightarrow\left(\frac{1}{v^3}+\frac{1}{v}\right) d v=-\frac{d x}{x}\end{aligned}$
Integrating both sides, $\int \frac{d v}{v^3}+\int \frac{d v}{v}=-\int \frac{d x}{x}$
$\begin{aligned} & \Rightarrow-\frac{1}{2 v^2}+\log v=-\log x-\log c \\ & \Rightarrow-\frac{x^2}{2 y^2}+\log y=-\log c \Rightarrow \log c y=\frac{x^2}{2 y^2} \\ & \Rightarrow c y=e^{x^2 / 2 y^2} \Rightarrow c^2 y^2=e^{x^2 / y^2} \\ & \therefore y^2 a=e^{x^2 / y^2}, \text { where } c^2=a .\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation $\frac{d y}{d x}=\frac{x+y}{x}$ satisfying the condition $y(1)=1$ is

Options:
A
$y=\ln x+x$

B
$y=x \ln x+x^2$

C
$y=x e^{(x-1)}$

D
$y=x \ln x+x$

The correct answer is:
$y=x \ln x+x$

Solution:
$$
y=v x
$$
$$
\begin{aligned}
& \frac{d y}{d x}=v+x \frac{d v}{d x} \\
& v+x \frac{d v}{d x}=1+v \\
& \Rightarrow d v=\frac{d x}{x} \\
& \therefore v=\log x+c \\
& \Rightarrow \frac{y}{x}=\log x+c
\end{aligned}
$$
Since, $y(1)=1$, we have
$$
y=x \log x+x
$$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2008
The solution of the differential equation $\frac{d y}{d x}=(x+y)^2$ is

Options:
A $\tan ^{-1}(x+y)=x+C$
B $\tan ^{-1}(x+y)=0$
C $\cot ^{-1}(x+y)=C$
D $\cot ^{-1}(x+y)=x+C$

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

Solution:
$\frac{d y}{d x}=(x+y)^2$
Let $x+y=t \Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x}$
Eq. (i) can be written as
$$
\begin{gathered}
\frac{d t}{d x}-1=t^2 \\
\Rightarrow \quad \frac{d t}{d x}=t^2+1 \Rightarrow \frac{d t}{t^2+1}=d x
\end{gathered}
$$
$\begin{aligned} & \Rightarrow \int \frac{d t}{t^2+1}=\int d x \Rightarrow \tan ^{-1} t=x+C \\ & \Rightarrow \quad \tan ^{-1}(x+y)=x+C\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation, $\frac{\mathrm{d} y}{\mathrm{~d} x}=(x-y)^{2}$, when $y(1)=1,$ is:

Options:
A $\quad \log _{e}\left|\frac{2-x}{2-y}\right|=x-y$
B $-\log _{e}\left|\frac{1-x+y}{1+x-y}\right|=2(x-1)$
C $-\log _{e}\left|\frac{1+x-y}{1-x+y}\right|=x+y-2$
D $\log _{\mathrm{e}}\left|\frac{2-y}{2-x}\right|=2(y-1)$

The correct answer is:
$-\log _{e}\left|\frac{1-x+y}{1+x-y}\right|=2(x-1)$

Solution:
The given differential equation

$\frac{d y}{d x}=(x-y)^{2}$

Let $x-y=t \Rightarrow 1-\frac{d y}{d x}=\frac{d t}{d x}$

$\Rightarrow \frac{d y}{d x}=1-\frac{d t}{d x}$

Now, from equation (1)

$\left(1-\frac{d t}{d x}\right)=(t)^{2}$

$\Rightarrow \quad 1-t^{2}=\frac{d t}{d x} \Rightarrow \int d x=\int \frac{d t}{1-t^{2}}$

$\Rightarrow-x=\frac{1}{2 \times 1} \ln \left|\frac{t-1}{t+1}\right|+c$

$\Rightarrow-x=\frac{1}{2} \ln \left|\frac{x-y-1}{x-y+1}\right|+c$

$\because$ The given condition $y(1)=1$

$-1=\frac{1}{2} \ln \left|\frac{1-1-1}{1-1+1}\right|+c \Rightarrow c=-1$

Hence, $2(x-1)=-\ln \left|\frac{1-x+y}{1-y+x}\right|$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The solution of the differential equation $\frac{d y}{d x}+\frac{y}{x}=x^2$ is

Options:
A $4 x y=x^4+c$
B $x y=x^4+c$
C $\frac{1}{4} x y=x^4+c$
D $x y=4 x^4+c$

The correct answer is:
$4 x y=x^4+c$

Solution:
The given equation $\frac{d y}{d x}+\frac{y}{x}=x^2$ is of the form
$\frac{d y}{d x}+P y=Q$. So, I.F. $=e^{\int \frac{1}{x} d x}=e^{\log x}=x$
Hence required solution $x y=\int x \cdot x^2 d x+c$
$\Rightarrow x y=\frac{x^4}{4}+c \Rightarrow 4 x y=x^4+c$.

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

Options:
A $y=\frac{1}{2}(\cos x+\sin x)+c e^{-x}$
B $y=\frac{1}{2}(\cos x-\sin x)+c^{-x}$
C $y=\cos x+\sin x+c e^{-x}$
D None of these

The correct answer is:
$y=\frac{1}{2}(\cos x+\sin x)+c e^{-x}$

Solution:
It is linear equation of the form $\frac{d y}{d x}+P y=Q$
So, I.F. $=e^{\int 1 d x}=e^x$
Hence solution is $y \cdot e^x=\int \cos x \cdot e^x d x+c$
$\Rightarrow y=\frac{1}{2}(\cos x+\sin x)+c e^{-x}$

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

Options:
A $y \sin x+\cos 2 x=2 c$
B $2 y \sin x+\cos x=c$
C $y \sin x+\cos x=c$
D $2 y \sin x+\cos 2 x=c$

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

Solution:
$\frac{d y}{d x}+y \cot x=2 \cos x$
It is linear equation of the form $\frac{d y}{d x}+P y=Q$
So, I.F. $=e^{\int P d x}=e^{\int \cot x d x}=e^{\log \sin x}=\sin x$
Hence the solution is$y \sin x=\int 2 \sin x \cos x d x+c$
$\Rightarrow y \sin x=-\frac{1}{2} \cos 2 x+c \Rightarrow 2 y \sin x+\cos 2 x=c$.

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation $\frac{d y}{d x}+\frac{x}{y} \cdot \frac{x^2+y^2-1}{2\left(x^2+y^2\right)+1}=0$ is

Options:
A $x^2+y^2+3 \log \left(x^2+y^2\right)=c$
B $x^2+3 x y-3 \log \left(x^2+y^2+2\right)=c$
C $x^2+2 y^2-3 \log \left(x^2+y^2+2\right)=c$
D $-x^2-2 y^2-3 \log \left(x^2+y^2\right)=c$

The correct answer is:
$x^2+2 y^2-3 \log \left(x^2+y^2+2\right)=c$

Solution:
We have,
$$
\begin{aligned}
& \frac{d y}{d x}+\frac{x}{y}\left(\frac{x^2+y^2-1}{2\left(x^2+y^2\right)+1}\right)=0 \\
\Rightarrow & 2 y\left(x^2+y^2\right) d y+y d y+x\left(x^2+y^2\right) d x-x d x=0 \\
\Rightarrow & 2 y\left(x^2+y^2+2-2\right) d y+y d y+ \\
& x\left(x^2+y^2+2-2\right) d x-x d x=0 \\
\Rightarrow & 2 y\left(x^2+y^2+2\right) d y-4 y d y+y d y \\
\Rightarrow & 2 y\left(x^2+y^2+2\right) d y+x\left(x^2+y^2+2\right) d x-3 x d x=0 \\
\Rightarrow & 2 y d y+x d x=3\left(\frac{x d x+y d y}{x^2+y^2+2}\right) \quad=3 x d x+3 y d y \\
\Rightarrow & 2 y d y+x d x=\frac{3}{2}\left(\frac{2 x d x+2 y d y}{x^2+y^2+2}\right) \\
\Rightarrow & 2 y d y+x d x=\frac{3}{2} d\left(\log \left(x^2+y^2+2\right)\right.
\end{aligned}
$$
On integrating both sides, we get
$$
\begin{aligned}
& 2 y^2+x^2=3 \log \left(x^2+y^2+2\right)+c \\
& x^2+2 y^2-3 \log \left|x^2+y^2+2\right|=c
\end{aligned}
$$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation dydx+1x=eyx2, âx>0 is Î»xe-y=1-2cx2 (where c is an arbitrary constant). Then, the value of Î» is equal to

Options:
A 2
B 4
C 12
D 14

The correct answer is:
2

Solution:
The given equation is e-ydydx+e-y⋅1x=1x2
Let, -e-y=t⇒e-ydydx=dtdx
Thus we have,
dtdx-t1x=1x2
Integrating factor =e-∫1xdx=e-ln⁡x=1x
Thus, the solution is
t⋅1x=∫1x2⋅1xdx
⇒-e-yx=x-2-2+c
⇒-e-yx=-12x2+c
⇒-2xe-y=-1+2cx2 or 2xe-y=1-2cx2
Hence, λ=2

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation dydx=2x-yx-6y is (where c is an arbitrary constant)

Options:
A 4xy=x2-3y+c
B xy=x2-y2+c
C xy=x2+3y2+c
D xy=x2+c

The correct answer is:
xy=x2+3y2+c

Solution:
Given equation is xdy-6ydy=2xdx-ydx
⇒xdy+ydx=2xdx+6ydy

or dxy=2xdx+6ydy

On integrating, we get,

xy=x2+3y2+c

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation dydx4-dydx2-2=0 is y=Â±Î»x+C (where, C is an arbitrary constant). Then, Î»2 is equal to

Options:
A 2
B 4
C 8
D 16

The correct answer is:
4

Solution:
Let dydx2=t
∴t2-t-2=0
⇒t-2t+1=0
⇒dydx2=2 or -1 (rejected)
⇒dydx=±2
or dy=±2dx
⇒∫dy=±2∫1dx
⇒y=±2x+C
∴λ=2⇒λ2=4

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation dydx=ax+b, a≠0 represents

Options:
A a parabola
B an ellipse
C a circle
D a hyperbola

The correct answer is:
a parabola

Solution:
Given equation is
∫dy=∫ax+bdx
⇒y=ax22+bx+C
Which is a parabola as a≠0.

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation dydx=-x2+3y23x2+y2,y1=0 is

Options:
A logex+y-xyx+y2=0
B logex+y+xyx+y2=0
C logex+y+2xyx+y2=0
D logex+y-2xyx+y2=0

The correct answer is:
logex+y+2xyx+y2=0

Solution:
Given,

Differential equation,

dydx=-x2+3y23x2+y2 & y1=0

Now let y=vx

v+xdvdx=dydx

⇒v+xdvdx=-1+3v23+v2

⇒xdvdx=-v+133+v2

⇒3+v2dvv+13+dxx=0

⇒∫4dvv+13+∫dvv+1-∫2dvv+12+∫dxx=0

⇒-2v+12+lnv+1+2v+1+lnx=c

⇒-2x2x+y2+lnx+yx+2xx+y+lnx=c

⇒2xyx+y2+lnx+y=c

∵ y1=0

∴ c=0, as x=1,y=0

⇒2xyx+y2+lnx+y=0

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (30 Jan Shift 2)
The solution of the differential equation dydx+x2x+y=x32x+y3-2 is (C being an arbitrary constant)

Options:
A 12x+xy=x2+1+Cex
B 12x+y2=x2+1+Cex2
C 12x+y=x+1+Ce-x2
D 12x+y2=x2+1+C

The correct answer is:
12x+y2=x2+1+Cex2

Solution:
Let, 2x+y=t⇒dydx+2=dtdx
dtdx+xt=x3t3⇒1t3dtdx+1t2x=x3
Let, 1t2=u⇒-2t3dtdx=dudx
dudx+-2xu=-2x3
I.F.=e-∫2xdx=e-x2⇒u.e-x2=∫e-x2-2x3dx
e-x22x+y2=-2∫e-x2.x3dx
e-x22x+y2=∫e-x2.x2-2xdx

Let, -x2=v
−2xdx=dv⇒e−x2(2x+y)2=−∫evvdv
e−x2(2x+y)2+v⋅ev−ev=C⇒e−x2(2x+y)2−x2e−x2−e−x2=C
1(2x+y)2=x2+1+Cex2

Topic: Mathematics, Differential Equations, JEE Main