datavorous/entrance-exam-dataset huggingface dataset:
The solution of the differential equation dydx+xy1âx2=xy, x<1 is y=-fx3+C1-x214, where f12=34 and C is an arbitrary constant. Then, the value of f-12 is

Options:
A -34
B 34
C 14
D 32

The correct answer is:
34

Solution:
Given equation is 1ydydx+x1âx2y=x
Let, 2y=Î½
â1ydydx=dÎ½dx
Thus, we have
dÎ½dx+x21-x2Î½=x
â´ I.F. =eâ«x21-x2dx
=eâ«-d1-x241-x2
=e-14ln1-x2
=1-x2-14
Thus, the solution is Î½1-x2-14=â«x1-x2-14dx
or Î½â1âx2â14=â231âx234+C'
or 2y=-231-x2+C'1-x214
ây=-1-x23+C1-x214
âfx=1-x2
âf-12=1-14=34

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation dydx+xylnâ¡y=x3y is equal to (where, C is the constant of integration)

Options:
A ln⁡y=x2+Ce-x2
B ln⁡y=x2-2+Ce-x2
C ln⁡y=x2-2+Ce-x22
D ln⁡y=x2+Ce-x22

The correct answer is:
ln⁡y=x2-2+Ce-x22

Solution:
1ydydx+xlnâ¡y=x3
Let lnâ¡y=t
1ydydx=dtdx
dtdx+tx=x3
I.F. =eâ«xdx=ex22
tex22=â«ex22x3dx
lnâ¡yex22=â«ex22⋅xx2dx
lnâ¡yex22=ex22⋅x2-â«2xex22dx
lnâ¡ex22=ex22x2-2ex22+C
lnâ¡y⋅ex22=ex22x2-2+C
lnâ¡y=x2-2+Ce-x22

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation dydx=x-yx-3y is (where, c is an arbitrary constant)

Options:
A 2xy=x2+3y+c
B xy=x2+y2+c
C 2xy=x2+3y2+c
D xy=x2+x

The correct answer is:
2xy=x2+3y2+c

Solution:
Given equation is xdy-3ydy=xdx-ydx
âxdy+ydx=xdx+3ydy
or dxy=xdx+3ydy
On integrating, we get,
xy=x22+3y22+k
or 2xy=x2+3y2+c

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation dydx=x-yx+4y is (where C is the constant of integration)

Options:
A xy+y2=x+C
B xy-y2=x2+C
C xy+2y2=x2+C
D 2xy+4y2=x2+C

The correct answer is:
2xy+4y2=x2+C

Solution:
Given equation is xdy+4ydy=xdx-ydx
or xdy+ydx+4ydy=xdx
or dxy+4ydy=xdx
Integrating, we get
xy+2y2=x22+C

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation dydx=xy+yxy+x is y-λx=ln⁡xy+C (where, C is an arbitrary constant and x,y>0). Then, the value of λ is equal to

Options:
A 1
B 12
C 2
D 4

The correct answer is:
1

Solution:
The given equation is 1+1ydy=1+1xdx
On integrating, we get,

y+ln⁡y=x+ln⁡x+C or y-x=ln⁡xy+C
∴λ=1.

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation dydx+y2sec⁡x=tan⁡x2y, where 0≤x<π2 and y0=1, is given by

Options:
A y2=1+xsec⁡x+tan⁡x
B y=1+xsec⁡x+tan⁡x
C y=1-xsec⁡x+tan⁡x
D y2=1-xsec⁡x+tan⁡x

The correct answer is:
y2=1-xsec⁡x+tan⁡x

Solution:
dydx+y2sec⁡x=tan⁡x2y
2ydydx+y2sec⁡x=tan⁡x
Put y2=t ⇒2ydydx=dtdx
dtdx+tsec⁡x=tan⁡x
I.F=e∫sec⁡xdx=eln(sec⁡x+tanx)=sec⁡x+tan⁡x
⇒tsec⁡x+tan⁡x=∫sec⁡x+tan⁡xtan⁡xdx
⇒tsec⁡x+tan⁡x=∫sec⁡xtan⁡xdx+∫tan2⁡xdx
⇒y2sec⁡x+tan⁡x=sec⁡x+tan⁡x-x+c
y0=1 ⇒c=0
⇒y2=1-xsec⁡x+tan⁡x

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2016 (10 Apr Online)
The solution of the differential equation dydx=y2+xlnâ¡x2xy is (where, c is the constant of integration)

Options:
A 2x2=ylnâ¡x2+2cy
B 2y2=xlnâ¡x2+2cx
C x2=ylnâ¡x2+c
D 2y2=xylnâ¡x2+cx

The correct answer is:
2y2=xln⁡x2+2cx

Solution:
2xydydx=y2+xln⁡x
2ydydx-y2x=ln⁡x
Put, y2=t⇒2ydydx=dtdx
dtdx-tx=ln⁡x
I.F. =e∫-1xdx=e-lnx=1x

The solution is
tx=∫ln⁡xxdx
tx=ln⁡x22+c
y2x=ln⁡x22+c
2y2=xln⁡x2+2cx

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation dydx-y+3xloge(y+3x)+3=0 is
(where C is a constant of integration)

Options:
A x-12loge(y+3x)2=C
B x-loge(y+3x)=C
C y+3x-12logex2=C
D x-2loge(y+3x)=C

The correct answer is:
x-12loge(y+3x)2=C

Solution:
dydx-y+3xln(y+3x)+3=0
dydx+3=y+3xln(y+3x)
ddx(y+3x)=y+3xln(y+3x)
∫ln(y+3x)(y+3x)d(y+3x)=∫dx
Let ln(y+3x)=t
1(y+3x)d(y+3x)=dt
∫tdt=∫dx
t22=x+c
(ln(y+3x))22=x+c

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The solution of the differential equation dydx = ycos x-y2sinâ¡x is equal to (where c is an arbitrary constant)

Options:
A sin⁡x=x-y+c
B sin⁡x=x+y+c
C sin⁡x=xy+cy
D sin⁡xx=y+c

The correct answer is:
sin⁡x=xy+cy

Solution:
dydx=ycosx-y2sin⁡x
ycos⁡xdx-sin⁡xdy=y2dx
ycos x.dx-sin⁡x.dyy2=dx
dsin⁡xy=dx

On integrating, we get,
sin⁡xy=x+c
sin⁡x=xy+cy

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation dydx+yx=1(1+lnx+lny)2 is (where, c is an arbitrary constant)

Options:
A xy 1+lnâ¡(xy 2]=x22+c
B 1+lnâ¡xy2=x22+y+c
C xy 1+lnâ¡xy=x22+c
D xy 1+lnâ¡xy=x2+c

The correct answer is:
xy 1+lnâ¡(xy 2]=x22+c

Solution:
dydx+yx=1(1+lnxy)2
Let xy=u so that dudx=x(1+ln u)2
∴ ∫(1+lnu)2du=∫x dx+c
⇒u(1+lnu)2-∫21+ln uu·udu=x22+c
⇒u1+2 ln⁡u+2u(ln⁡u)2-2uln⁡u=x22+c
∴xy(1+lnxy)2=x22+c

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation dy-ydx2x=xy dy is

(where, c is an arbitrary constant)

Options:
A yx=y+c
B yx=y22+c
C y=yx+c
D yx=-y2+c

The correct answer is:
yx=y22+c

Solution:
dyx-ydx2x32=ydy⇒dyx=ydy⇒yx=y22+c

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation sin⁡x+ydy=dx is

Options:
A y+tanâ¡x+yÂ âsecâ¡(x+y)=c
B yÂ âtanâ¡x+yâsecâ¡xÂ +Â y=c
C y+tanâ¡x+y+secâ¡x+y=c
D y âtanâ¡x+y+secâ¡x+y=c

The correct answer is:
y âtanâ¡x+y+secâ¡x+y=c

Solution:
Let x+y=v
dvdx-1=1sinv
dx=1-11+sinvdv
⇒x=v-∫(sec2v-tanv⋅secv)dv
⇒x=v-tan⁡v+sec⁡v+c
⇒x=x+y-tan⁡x+y+sec⁡x+y+c

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation sin⁡yexdx-excos⁡ydy=sin2ydx is (where, c is an arbitrary constant)

Options:
A exsiny=x+c
B ex=x+csin⁡y
C ex⋅x=sin⁡y+c
D ex⋅sin⁡y=x2+c

The correct answer is:
ex=x+csin⁡y

Solution:
Given equation is
sin⁡y·exdx-excos⁡ydysin2y=dx
⇒dexsin⁡y=dx
On integrating, we get
exsiny=x+c
or ex=x+csin⁡y

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation

$(x+1) \frac{d y}{d x}-y=e^{3 x}(x+1)^{2}$ is

Options:
A $y=(x+1) e^{3 x}+c$
B $3 y=(x+1)+e^{3 x}+c$
C $\frac{3 y}{x+1}=e^{3 x}+c$
D $y e^{-3 x}=3(x+1)+c$

The correct answer is:
$\frac{3 y}{x+1}=e^{3 x}+c$

Solution:
The given equation is $\frac{d y}{d x}-\frac{y}{x+1}=e^{3 x}(x+1)$

I.F. $=e^{\int-\frac{1}{x+1} d x}=e^{-\log (x+1)}=\frac{1}{x+1}$

The solution is

$y\left(\frac{1}{x+1}\right)=\int e^{3 x}(x+1) \cdot \frac{1}{x+1} d x+a$

$\Rightarrow \frac{y}{x+1}=\int e^{3 x} d x+a=\frac{e^{3 x}}{3}+a$

$\Rightarrow \frac{3 y}{x+1}=e^{3 x}+c, c=3 a$

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

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

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

Solution:
It is homogeneous equation which can be written in the form $\frac{d y}{d x}=\frac{x^2+x y+y^2}{x^2}$
Now put $\boldsymbol{y}=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$
Therefore, $\quad v+x \frac{d v}{d x}=\frac{x^2+v x^2+v^2 x^2}{x^2}=1+v+v^2$
$\begin{aligned} & \Rightarrow x \frac{d v}{d x}=1+v^2 \\ & \Rightarrow \frac{d v}{1+v^2}=\frac{d x}{x}\end{aligned}$
Now integrating both sides, we get $\tan ^{-1} v=\log x+c$
$\Rightarrow \quad \tan ^{-1}\left(\frac{y}{x}\right)=\log x+c$ {$\left.y=v x \Rightarrow v=y / x\right\}$

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

Options:
A $x=c\left(x^2+y^2\right)$
B $x=c\left(x^2-y^2\right)$
C $x+c\left(x^2-y^2\right)=0$
D None of these

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

Solution:
It can be written in the form of homogeneous equation $\frac{d y}{d x}=\frac{x^2+y^2}{2 x y}$
Now solve it by putting $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation $\left(x^2+y^2\right) \mathrm{d} x-5 x y \mathrm{~d} y=0, y(1)=0$, is :

Options:
A $\left|x^2-2 y^2\right|^6=x$
B $\left|x^2-4 y^2\right|^6=x$
C $\left|x^2-4 y^2\right|^5=x^2$
D $\left|x^2-2 y^2\right|^5=x^2$

The correct answer is:
$\left|x^2-4 y^2\right|^5=x^2$

Solution:
$\begin{aligned}
& \left(x^2+y^2\right) d x=5 x y d y \\
& \Rightarrow \frac{d y}{d x}=\frac{x^2+y^2}{5 x y}
\end{aligned}$
Put $y=V x$
$\begin{aligned}
& \Rightarrow \mathrm{V}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1+\mathrm{V}^2}{5 \mathrm{~V}} \\
& \Rightarrow \frac{\mathrm{xdv}}{\mathrm{dx}}=\frac{1-4 \mathrm{~V}^2}{5 \mathrm{~V}} \\
& \Rightarrow \int \frac{\mathrm{V}}{1-4 \mathrm{~V}^2} \mathrm{dV}=\int \frac{\mathrm{dx}}{5 \mathrm{x}}
\end{aligned}$
Let $1-4 \mathrm{~V}^2=\mathrm{t}$
$\Rightarrow-8 \mathrm{~V} \mathrm{dV}=\mathrm{dt}$
$\begin{aligned} & \Rightarrow \int \frac{\mathrm{dt}}{(-8)(\mathrm{t})}=\int \frac{\mathrm{dx}}{5 \mathrm{x}} \\ & \Rightarrow \frac{-1}{8} \ln |\mathrm{t}|=\frac{1}{5} \ln |\mathrm{x}|+\ln \mathrm{C} \\ & \Rightarrow-5 \ln |\mathrm{t}|=8 \ln |\mathrm{x}|+\ln \mathrm{K} \\ & \Rightarrow \ln \mathrm{x}^8+\ln \left|\mathrm{t}^5\right|+\ln \mathrm{K}=0 \\ & \Rightarrow \mathrm{x}^8\left|\mathrm{t}^5\right|=\mathrm{C} \\ & \Rightarrow \mathrm{x}^8\left|1-4 \mathrm{~V}^2\right|^5=\mathrm{C} \\ & \Rightarrow \mathrm{x}^8\left|\frac{\mathrm{x}^2-4 \mathrm{y}^2}{\mathrm{x}^2}\right|^5=\mathrm{C} \\ & \Rightarrow\left|\mathrm{x}^2-4 \mathrm{y}^2\right|^5=\mathrm{Cx} \mathrm{x}^2 \\ & \text { given } \mathrm{y}(1)=0 \\ & \Rightarrow|1|^5=\mathrm{C} \Rightarrow \mathrm{C}=1 \\ & \Rightarrow\left|\mathrm{x}^2-4 \mathrm{y}^2\right|^5=\mathrm{x}^2\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (09 Apr Shift 1)
The solution of the differential equation xcosydydx+sinâ¡y=1 is (Here, x>0 and Î» is an arbitrary constant)

Options:
A x-xcosx=λ
B x+xcosx=λ
C x-xsiny=λ
D x+xcosy=λ

The correct answer is:
x-xsiny=λ

Solution:
Let, sinâ¡y=tâcosâ¡ydydx=dtdx
â´ the equation becomes
xdtdx+t=1 or xdtdx=1-t
âdt1-t=dxx
On integrating, we get,
-lnâ¡1-t=lnâ¡x+lnâ¡C
or 11-t=Cx
i.e. 1-sinâ¡yx=1C=Î»say

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

Options:
A $y=x \log x+x+2$
B $y=x \log x-x+2$
C $y=x \log x+x$
D $y=x \log x-x$

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

Solution:
$\begin{aligned} & x \frac{d^2 y}{d x^2}=1 \Rightarrow \frac{d^2 y}{d x^2}=\frac{1}{x} \Rightarrow \frac{d y}{d x}=\log x+c_1 \\ & \Rightarrow y=x \log x-x+c_1 x+c_2 \quad \text { (on integrating twice) }\end{aligned}$
Given $y=1$ and $\frac{d y}{d x}=0$ at $x=1 \Rightarrow c_1=0$ and $c_2=2$
Therefore, the required solution is $y=x \log x-x+2$

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

Options:
A $x y=\frac{x^3}{3}+\frac{3}{2} x^2+2 x+c$
B $x y=\frac{x^4}{4}+x^3+x^2+c$
C $x y=\frac{x^4}{4}+\frac{x^3}{3}+x^2+c$
D $x y=\frac{x^4}{4}+x^3+x^2+c x$

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

Solution:
$x \frac{d y}{d x}+y=x^2+3 x+2 \Rightarrow \frac{d y}{d x}+\frac{y}{x}=x+3+\frac{2}{x}$
Here $P=\frac{1}{x}, Q=x+3+\frac{2}{x}$, therefore I.F. $=e^{\int \frac{1}{x} d x}=x$ Now solve it.

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation xdx+ydy=xdy-ydxx2+y2 is tan⁡fx,y-C=yx (where, C is an arbitrary constant). If f1,1=1, then fπ,π is equal to

Options:
A 2
B π2
C -1
D π

The correct answer is:
π2

Solution:
Given equation is xdx+ydy=dtan-1⁡yx
On integrating, we get,
x2+y22=tan-1⁡yx+C or tan⁡x2+y22-C=yx
⇒fx,y=x2+y22
∴fπ,π=π2

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation xdx+ysin2⁡xdy=ydy+xsin2⁡ydx is (where, c is an arbitrary constant)

Options:
A xtan⁡x=sec⁡y+c
B xtan⁡y=sec⁡x+c
C xtan⁡x-ln⁡sec⁡x=ytan⁡y-ln⁡sec⁡y+c
D xtan⁡x=ln⁡sec⁡y+c

The correct answer is:
xtan⁡x-ln⁡sec⁡x=ytan⁡y-ln⁡sec⁡y+c

Solution:
The given equation is xcos2⁡ydx=ycos2⁡xdy

∫xsec2⁡xdx=∫ysec2⁡ydy

Using integration by parts, we get,
xtan⁡x-ln⁡sec⁡x=ytan⁡y-ln⁡sec⁡y+c

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation xdydx+2y=x2, (x≠0) with y1=1, is

Options:
A y=x35+15x2
B y=34x2+14x2
C y=x24+34x2
D y=45x3+15x2

The correct answer is:
y=x24+34x2

Solution:
Given, differential equation is dydx+2xy=x

This is a linear differential equation of type dydx+Py=Q, where P & Q are the functions of x or constants.

Thus, P=2x & Q=x

The integrating factor I.F.=e∫Pdx

=e∫2x dx=e2lnx

=elnx2=x2.

The solution of the linear differential equation is y×I.F.=∫Q×I.F.dx+C

So, the solution of the given differential equation is

yx2=∫x·x2dx+C

⇒x2y=∫x3dx+C

Using ∫xndx=xn+1n+1, we get

⇒x2y=x44+C

Since y1=1

⇒1·1=14+C

⇒C=34

⇒x2y=x44+34

⇒y=x24+34x2.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The solution of the differential equation xdydx=yln⁡y2x2 is (where, c is an arbitrary constant)

Options:
A y=x.ecx+1
B y=x.ecx-1
C y=x2.ecx+1
D y=x.ecx2+ 12

The correct answer is:
y=x.ecx2+ 12

Solution:
Putting y=xv and dydx=v+xdvdx
So, v+xdvdx=2vln⁡v
⇒ ∫dxx=∫dvv(2ln⁡v-1)

On integrating, we get,
y=x⋅ecx2+ 12

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

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

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

Solution:
$x d y+y d x=\sqrt{1-x^2 y^2} d x \Rightarrow \frac{x d y+y d x}{\sqrt{1-x^2 y^2}}=d x$
$\frac{d x y}{\sqrt{1-(x y)^2}}=d x$. Integrating both side, we get
$\sin ^{-1} x y=x+c \Rightarrow x y=\sin (x+c)$.

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation xdy-ydx+3x2y2ex3dx=0 is (where, c is an arbitrary constant)

Options:
A x=2yex+c
B x=yex3+cy
C x=y2ex3+c
D xy=ex3+c

The correct answer is:
x=yex3+cy

Solution:
Given equation is ydx-xdyy2=3x2ex3dx

or dxy=dex3
⇒∫dxy=∫dex3
⇒xy=ex3+c

or x=yex3+cy

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation xdy+yxdx=dxx is
(where, c is an arbitrary constant)

Options:
A y=1+ce1/x
B y=ce1/x
C y=ce1/x-1
D xy=1-ce1/x

The correct answer is:
y=1+ce1/x

Solution:
The given differential equation can be written as,

dydx+1x2⋅y=1x2
Integrating factor is e-1/x

⇒y.e-1/x=∫1x2.e-1/x dx
Hence, the solution is y=1+ce1/x

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

Options:
A $x \sec y \tan y=c$
B $\mathrm{cx}=\sec y+\tan y$
C $\quad c y=\sec x \tan x$
D $\operatorname{cy}=\sec x+\tan x$

The correct answer is:
$\mathrm{cx}=\sec y+\tan y$

Solution:
$x \sec y \frac{d y}{d x}=1 \Rightarrow \sec y d y=\frac{d x}{x}$
On integrating both sides, we get
$\log (\sec y+\tan y)=\log x+\log c \Rightarrow \sec y+\tan y=c x$

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation xdy=tan⁡y+e1/x2xsec⁡ydx is (where C is the constant of integration)

Options:
A sinâ¡y=e1x2+C
B 2sinâ¡yx+e1x2=C
C sinâ¡yx-e1x2=C
D sin⁡y-xe1x2=C

The correct answer is:
2sin⁡yx+e1x2=C

Solution:
dydx=tanâ¡yx+e1x2x2secâ¡y
cosâ¡ydydx-sinâ¡yx=e1x2x2
Let, sinâ¡y=t
cosâ¡ydydx=dtdx
âdtdx-tx=e1x2x2
I.F. =e-â«1xdx=elnâ¡1x=1x
âtx=â«e1x2x3dx
âtx=-12â«-2x3e1x2dx
âsinâ¡yx=-12e1x2+C

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

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

The correct answer is:
$x \log x=y c$

Solution:
Given differential equation is
$$
y(1+\log x) \frac{d x}{d y}-x \log x=0
$$
$$
\begin{array}{ll}
\Rightarrow & \int \frac{1+\log x}{x \log x} d x=\int \frac{d y}{y} \\
\Rightarrow & \int \frac{1}{x \log x} d x+\int \frac{1}{x} d x=\int \frac{1}{y} d y \\
\Rightarrow & \log (\log x)+\log x=\log y+\log c \\
\Rightarrow & x \log x=y c
\end{array}
$$

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

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

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

Solution:
Given differential equation is
$\left(y^{2}+2 x\right) \frac{d y}{d x}=y \Rightarrow \frac{d x}{d y}=\left(\frac{y^{2}+2 x}{y}\right)$
$\Rightarrow \quad \frac{d x}{d y}=y+\frac{2 x}{y}$
$\Rightarrow \quad \frac{d x}{d y}-\frac{2}{y} \cdot x=y$
$\mathrm{IF}=e^{\int \frac{2}{y} d y}=e^{-2 \mathrm{bg} \gamma}=\mathrm{y}^{-2}=\frac{1}{\mathrm{y}^{2}}$
$\therefore$ Complete solution is $x \cdot \frac{1}{y^{2}}=\int y \cdot \frac{1}{y^{2}} d y+C$
$\Rightarrow \quad \frac{x}{y^{2}}=\int \frac{d y}{y}+C=\log _{e} y+c$
$\Rightarrow \quad x=y^{2} \log _{e} y+C y^{2}$
$A t+x=1 y=1$ then from Eq $(i),$ we get
$$
1=0+C \Rightarrow C=1
$$
$\therefore$ From Eq. (i), we get
$$
x=y^{2}\left(\log _{e} y+1\right)
$$
Which is the required solution.

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation y2x4+ydy+4xy2-1x2dx=0 is (where C is an arbitrary constant)

Options:
A 3x2y+x3-y3=C
B 3x4y2+y3-x3=C
C 3x2y4+x3-y3=C
D 3x2y4+y3-x3=C

The correct answer is:
3x4y2+y3-x3=C

Solution:
2x4ydy+y2dy+4x3y2dx-x2dx=0
2x3yxdy+2ydx+y2dy-x2dx=0
2x2yx2dy+2xydx+y2dy-x2dx=0
2x2ydx2y+y2dy-x2dx=0

On integrating, we get,
x2y2+y33-x33=C1
3x4y2+y3-x3=C

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation ycos⁡x.dx=sin⁡x.dy+xy2dx is (where, c is an arbitrary constant)

Options:
A sin⁡x=xy2+c
B 2sin⁡x=x2y+cy
C 2sin⁡x=xy2+c
D sin⁡x=x2y+cy

The correct answer is:
2sin⁡x=x2y+cy

Solution:
Given equation is
ycos⁡xdx-sin⁡xdyy2=xdx

or dsin⁡xy=xdx
⇒∫dsin⁡xy=∫xdx
⇒sin⁡xy=x22+c'
or 2 sin⁡x=x2y+cy

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation ydx-x+2y2dy=0 is x=fy. If f-1=1, then f(1) is equal to

Options:
A 2
B 3
C 4
D 1

The correct answer is:
3

Solution:
ydx-xdy=2y2dy
ydx-xdyy2=2dy
dxy=2dy
∫dxy=2∫dy
xy=2y+c
x=2y2+cy
If y= -1, x=1
1=2-c
c=1
∴ Curve would be
x=2y2+y
If y=1 then x=3 .

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2015 (11 Apr Online)
The solution of the differential equation ydx-xdy=xydx is ….

Options:
A x2=exy2
B x=yex
C xy=ex
D x2y2=logx

The correct answer is:
x=yex

Solution:
We have differential equation ydx-xdy=xydx
⇒ydx-xdyxy=dx
⇒dlogxy=dx
On integrating both sides, we get
logâ¡xy=x⇒xy=ex
⇒x=yex

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation ydx-xdyxy=xdx+ydy is (where, C is an arbitrary constant)

Options:
A xy=x+y+C
B xy=x2+y22+C
C ln⁡xy=x2+y2+C
D 2ln⁡xy=x2+y2+C

The correct answer is:
2ln⁡xy=x2+y2+C

Solution:
The given equation is 1xy⋅ydx-xdyy2=xdx+ydy

or dln⁡xy=xdx+ydy
On integrating, we get,

∫dln⁡xy=∫xdx+∫ydy
⇒ln⁡xy=x22+y22+k
⇒2ln⁡xy=x2+y2+C

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

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

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

Solution:
$y d x+x d y=-x^2 y d y \Rightarrow \frac{1}{(x y)^2} d x y=-\frac{d y}{y}$
Integrating both side we get
$-\frac{1}{x y}=-\log y+c \Rightarrow-\frac{1}{x y}+\log y=c$.

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

Options:
A
$-\frac{1}{x y}=C$

B
$-\frac{1}{x y}+\log y=C$

C
$\frac{1}{x y}+\log y=C$

D
$\log y=C x$

The correct answer is:
$-\frac{1}{x y}+\log y=C$

Solution:
$y d x+x d y+x^2 y d y=0$
$\frac{d(x y)}{x^2 y^2}+\frac{1}{y} d y=0 \Rightarrow-\frac{1}{x y}+\log y=C$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2004
The solution of the differential equation ysin2xdy+sin⁡xcos⁡xy2dx=xdx is (where C is the constant of integration)

Options:
A sin2x⋅y=x2+C
B sin2x⋅y2=x2+C
C sinx⋅y2=x2+C
D sin2x⋅y2=x+C

The correct answer is:
sin2x⋅y2=x2+C

Solution:
The given equation is sin2x2ydy+2sinâ¡xcosâ¡xdxy2=2xdx
or dsin2xây2=2xdx
On integrating, we get
sin2xây2=x2+C

Topic: Mathematics, Differential Equations, JEE Main
The solution of the differential equation \( y\left(\sin ^{2} x\right) d y+(\sin x \cos x) y^{2} d x=x d x \) is (where \( C \) is the constant of integration)

Options:
A \( \sin ^{2} x \cdot y=x^{2}+C \)
B \( \sin ^{2} x \cdot y^{2}=x^{2}+C \)
C \( \sin x \cdot y^{2}=x^{2}+C \)
D \( \sin ^{2} x \cdot y^{2}=x+C \)

The correct answer is:
\( \sin ^{2} x \cdot y^{2}=x^{2}+C \)

Solution:
The given equation is sin2x2ydy+2sinâ¡xcosâ¡xdxy2=2xdx
or dsin2xây2=2xdx
On integrating, we get
sin2xây2=x2+C

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

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

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

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

Topic: Mathematics, Differential Equations, JEE Main
The solution of the equation cos-1⁡x+cos-1⁡2x=2π3 is

Options:
A x=327
B x=-327
C x=±12
D None of these

The correct answer is:
x=327

Solution:
cos-1â¡2x=2Ï3-cos-1â¡xâ¦i
Let, cos-1â¡2x=Î±âcosâ¡Î±=2x
cos-1â¡x=Î²âcosâ¡Î²=x
From i Î±=2Ï3-Î²
cosâ¡Î±=cosâ¡2Ï3-Î²
=cosâ¡2Ï3cosâ¡Î²+sinâ¡2Ï3sinâ¡Î²
2x=-x2+321-x2
25x24=341-x2â28x2=3
x=Â±327
But x=-327 does not satisfy the given equation
Hence, x=327

Topic: Mathematics, Functions, JEE Main
The solution of the equation $\frac{d^2 y}{d x^2}=e^{-2 x}$

Options:
A
$\frac{e^{-2 x}}{4}$

B
$\frac{e^{-2 x}}{4}+c x+d$

C
$\frac{1}{4} e^{-2 x}+c x^2+d$

D
$\frac{1}{4} e^{-4 x}+c x+d$

The correct answer is:
$\frac{e^{-2 x}}{4}+c x+d$

Solution:
$\frac{d^2 y}{d x^2}=e^{-2 x} ; \frac{d y}{d x}=\frac{e^{-2 x}}{-2}+c ; y=\frac{e^{-2 x}}{4}+c x+d$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2002
The solution of the equation $\frac{d y}{d x}=\frac{x}{2 y-x}$ is

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

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

Solution:
$\begin{aligned} & \frac{d y}{d x}=\frac{x}{2 y-x} . \text { Put } y=v x \Rightarrow v+x \frac{d v}{d x}=\frac{d y}{d x} \\ & v+x \frac{d v}{d x}=\frac{x}{2 v-x}=\frac{1}{2 v-1} \\ & x \frac{d v}{d x}=\frac{1}{2 v-1}-v=\frac{1-2 v^2+v}{2 v-1}=-\frac{(v-1)(2 v+1)}{2 v-1} \\ & \frac{(2 v-1)}{(2 v+1)(v-1)}=\frac{-d x}{x} ; \frac{1}{3(v-1)}+\frac{4}{3(2 v+1)}=\frac{-d x}{x} \\ & \frac{1}{3} \log (v-1)+\frac{4}{3} \cdot \frac{1}{2} \log (2 v+1)=\log \frac{1}{x}+\log c \\ & \log (v-1)^{1 / 3}+\log (2 v+1)^{2 / 3}=\log \frac{c}{x}\end{aligned}$
$=(v-1)^{1 / 3}(2 v+1)^{2 / 3}=\frac{c}{x}$
$\left(\frac{y-x}{x}\right)\left(\frac{2 y+x}{x}\right)^2=\frac{c^3}{x^3} \Rightarrow(x-y)(x+2 y)^2=c$

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

Options:
A $c\left(x^2+y^2\right)^{1 / 2}+e^{\tan ^{-1}(y / x)}=0$
B $c\left(x^2+y^2\right)^{1 / 2}=e^{\tan ^{-1}(y / x)}$
C $c\left(x^2-y^2\right)=e^{\tan ^{-1}(y / x)}$
D None of these

The correct answer is:
$c\left(x^2+y^2\right)^{1 / 2}=e^{\tan ^{-1}(y / x)}$

Solution:
Given equation, $\frac{d y}{d x}=\frac{x+y}{x-y}$
It is a homogeneous equation so putting $y=v x$
$\begin{aligned} & \text { and } \frac{d y}{d x}-v+x \frac{d v}{d x} \text {, we get } \\ & v+x \frac{d v}{d x}=\frac{x+v x}{x-v x}=\frac{1+v}{1-v} \\ & \Rightarrow x \frac{d v}{d x}=\frac{1+v^2}{1-v} \\ & \Rightarrow \frac{1}{x} d x=\left(\frac{1}{1+v^2}-\frac{v}{1+v^2}\right) d v \\ & \Rightarrow \log _e x=\tan ^{-1} v-\frac{1}{2} \log \left(1+v^2\right)+\log _e c\end{aligned}$
Substituting $v=\frac{y}{x}{ }^{\prime}$ we get
$\begin{aligned} & \log _6 x=\tan ^{-1} \frac{y}{x}-\frac{1}{2} \log \left[1+\left(\frac{y}{x}\right)^2\right]+\log _6 c \\ & \Rightarrow c\left(x^2+y^2\right)^{1 / 2}=e^{\tan ^{-1}(y / x)} .\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main