datavorous/entrance-exam-dataset huggingface dataset:
Let $y=y(x)$ be the solution of the differential equation $\left(1+y^2\right) e^{\tan x} d x+\cos ^2 x\left(1+e^{2 \tan x}\right) d y=0, y(0)=1$. Then $y\left(\frac{\pi}{4}\right)$ is equal to

Options:
A $\frac{2}{e}$
B $\frac{2}{e^2}$
C $\frac{1}{e}$
D $\frac{1}{e^2}$

The correct answer is:
$\frac{1}{e}$

Solution:
$\begin{aligned} & \left(1+y^2\right) e^{\tan x} d x+\cos ^2 x\left(1+e^{2 \tan x}\right) d y=0 \\ & \int \frac{\sec ^2 x e^{\tan x}}{1+e^{2 \tan x}} d x+\int \frac{d y}{1+y^2}=C \\ & \Rightarrow \tan ^{-1}\left(e^{\tan x}\right)+\tan ^{-1} y=C \\ & \text { for } x=0, y=1, \tan ^{-1}(1)+\tan ^{-1} 1=C\end{aligned}$
$\mathrm{C}=\frac{\pi}{2}$
$\tan ^{-1}\left(e^{\tan x}\right)+\tan ^{-1} y=\frac{\pi}{2}$
Put $x=\pi, \tan ^{-1} e+\tan ^{-1} y=\frac{\pi}{2}$
$\tan ^{-1} y=\cot ^{-1} e$
$y=\frac{1}{e}$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Let y=yx be the solution of the differential equation, 2+sinxy+1.dydx=−cosx, y>0, y0=1. If yπ=a and dydx at x=π is b, then the ordered pair a, b is equal to

Options:
A 2, 32
B 1, −1
C 1, 1
D 2, 1

The correct answer is:
1, 1

Solution:
dy1+y=-cosx2+sinxdx

ln1+y=-ln2+sinx+lnc

1+y2+sinx=c

We know y0=1

Soc=4

1+y=42+sinx ⇒ y=42+sinx-1

yπ=2-1=1=a

dydx=-42+sinx2·cosx=1

At x=π⇒b=1

a,b=1,1

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (02 Sep Shift 1)
Let $y=y(x)$ be the solution of the differential equation $\left(2 x \log _e x\right) \frac{d y}{d x}+2 y=\frac{3}{x} \log _e x, x>0$ and $y\left(e^{-1}\right)=0$. Then, $y(e)$ is equal to

Options:
A $-\frac{3}{\mathrm{e}}$
B $-\frac{3}{2 e}$
C $-\frac{2}{3 e}$
D $-\frac{2}{\mathrm{e}}$

The correct answer is:
$-\frac{3}{\mathrm{e}}$

Solution:
$\begin{aligned} & \frac{d y}{d x}+\frac{y}{x \ell \ln x}=\frac{3}{2 x^2} \\ & \therefore \text { I.F. }=e^{\int \frac{1}{x \ln x} d x}=e^{\ln (\ln (x))}=\ln x \\ & \therefore y \ell \ln x=\int \frac{3 \ln x}{2 x^2} d x \\ & =\frac{3 \ln x}{2} \int x^{-2} d x-\int\left(\frac{3}{2 x} \cdot \int x^{-2} d x\right) d x \\ & =\frac{3 \ln x}{2}\left(-\frac{1}{x}\right)-\int \frac{3}{2 x}\left(-\frac{1}{x}\right) d x \\ & \text { y. } \ln x=\frac{-3 \ell n x}{2 x}-\frac{3}{2 x}+C\end{aligned}$
$\begin{aligned} & \because \mathrm{y}\left(\mathrm{e}^{-1}\right)=0 \\ & \therefore 0(-1)=\frac{3 \mathrm{e}}{2}-\frac{3 \mathrm{e}}{2}+\mathrm{C} \Rightarrow \mathrm{C}=0 \\ & \therefore \mathrm{y}=\frac{-3 \ell \mathrm{nx}}{2 \mathrm{x}}-\frac{3}{2 \mathrm{x}} \\ & \therefore \mathrm{y}(\mathrm{e})=\frac{-3}{2 \mathrm{e}}-\frac{3}{2 \mathrm{e}}=\frac{-3}{\mathrm{e}}\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (06 Apr Shift 1)
Let y=y(x) be the solution of the differential equation 3y2-5x2ydx+2xx2-y2dy=0 such that y(1)=1. Then (y(2))3-12y(2) is equal to :

Options:
A 64
B 322
C 32
D 162

The correct answer is:
322

Solution:
Given,

3y2-5x2ydx+2xx2-y2dy=0

⇒dydx=y5x2-3y22xx2-y2

This is homogeneous differential equation

Let y=tx⇒dydx=t+xdtdx

t+xdtdx=(tx)5x2-3t2x22xx2-t2x2=t5-3t221-t2

⇒xdtdx=5t-3t32-2t2-t=3t-t32-2t2

⇒21-t23t-t3dt=1xdx

Taking integration on both sides we get,

∫21-t23t-t3dt=∫1xdx......1

Now let 3t-t3=z⇒31-t2dt=dz

⇒1-t2dt=13dz

Substitute in equation 1 we get,

23∫dzz=∫1xdx

⇒23ln(z)=lnx+lnc

⇒z23=cx⇒z=c32x32

⇒3t-t3=c32x32

Now put t=yx

⇒3yx-y3x3=c32x32

Given y(1)=1 when x=1,y=1

⇒c32=2

So, equation becomes 3yx-y3x3=2x32

⇒3x2y-y3=2x92......2

Put x=2 in equation 2,

12y(2)-y(2)3=2(2)92=322

∴y(2)3-12y(2)=322

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Let y=y(x) be the solution of the differential equation cosxdydx+2ysinx=sin2x,x∈0,π2
If y(π/3)=0, then y(π/4) is equal to :

Options:
A 2-2
B 2+2
C 2-2
D 12-1

The correct answer is:
2-2

Solution:
dydx+2tanx·y=2sinx

I.F. =e∫2tanxdx=sec2x

solution is
y·sec2x=∫2sinx·sec2xdx+C

ysec2x=2secx+C
⇒0=2.2+C⇒C=-4
ysec2x=2secx-4
yπ4=2-2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (05 Sep Shift 2)
Let y=yx be the solution of the differential equation cosec2xdy+2dx=1+ycos2xcosec2xdx, with yπ4=0. Then, the value of y0+12 is equal to:

Options:
A e1/2
B e-1/2
C e-1
D e

The correct answer is:
e-1

Solution:
Given, cosec2xdy+2dx=1+ycos2xcosec2xdx

⇒dysin2x+2dx=1+ycos2xsin2xdx

⇒dydx+2sin2x=1+ycos2x

⇒dydx+2sin2x-1=ycos2x

Using, cos2x=1-2sin2x, we get

dydx+-cos2x=ycos2x

⇒dydx+-cos2xy=cos2x

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

Thus, P=-cos2x & Q=cos2x

Now, we have integrating factor I.F.=e∫Pdx=e∫-cos2xdx=e-sin2x2

And, the solution of the given differential equation is

yI.F.=∫QI.F.dx+c

⇒y·e-sin2x2=∫cos2x·e-sin2x2dx+c

Put sin2x2=t, ⇒2cos2x2dx=dt

⇒y·e-sin2x2=∫e-tdt+c

⇒y·e-sin2x2=-e-tdt+c

⇒y·e-sin2x2=-e-sin2x2+c

Given, yπ4=0

⇒0·e-sinπ22=-e-sinπ22+c

⇒c=-e-12

⇒y·e-sin2x2=-e-sin2x2+e-12

Now, at x=0

⇒y0·e-sin02=-e-sin02+e-12

⇒y0·e0=-e0+e-12

⇒y0=-1+e-12

⇒y0+1=e-12

⇒y0+12=e-1.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Let $y=y(x)$ be the solution of the differential equation
$\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{2 x}{\left(1+x^2\right)^2} y=x \mathrm{e}^{\frac{1}{\left(1+x^2\right)}} ; y(0)=0 .$
Then the area enclosed by the curve $f(x)=y(x) \mathrm{e}^{-\frac{1}{\left(1+x^2\right)}}$ and the line $y-x=4$ is__________

Options:

The correct answer is:
18

Solution:
$\begin{aligned} & I F=e^{\int \frac{2 x}{\left(1+x^2\right)^2} d x}=e^{\frac{-1}{1+x^2}} \\ & y \cdot e^{\frac{-1}{1+x^2}}=\int x \cdot e^{\frac{1}{1+x^2}} \cdot e^{\frac{-1}{1+x^2} d x} \\ & y \cdot e^{\frac{-1}{1+x^2}}=\frac{x^2}{2}+c \\ & (0,0) \Rightarrow C=0 \\ & y(x)=\frac{x^2}{2} e^{\frac{1}{1+x^2}} \\ & f(x)=\frac{x^2}{2}\end{aligned}$
$A=\int_{-2}^4(x+4)-\frac{x^2}{2} d x=18$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}+2 y=f(x)$, where
$$
f(x)=\left\{\begin{array}{lc}
1, & x \in[0,1] \\
0, & \text { otherwise }
\end{array}\right.
$$
If $y(0)=0$, then $y\left(\frac{3}{2}\right)$ is

Options:
A
$\frac{e^2-1}{2 e^3}$

B
$\frac{e^2-1}{e^3}$

C
$\frac{1}{2 e}$

D
$\frac{e^2+1}{2 e^4}$

The correct answer is:
$\frac{e^2-1}{2 e^3}$

Solution:
When $x \in[0,1]$, then $\frac{d y}{d x}+2 y=1$
$$
\begin{aligned}
&\Rightarrow \mathrm{y}=\frac{1}{2}+C_1 e^{-2 x} \\
&\because y(0)=0 \Rightarrow y(x)=\frac{1}{2}-\frac{1}{2} e^{-2 x} \\
&\text { Here, } y(1)=\frac{1}{2}-\frac{1}{2} e^{-2}=\frac{e^2-1}{2 e^2}
\end{aligned}
$$
When $x \notin[0,1]$, than $\frac{d y}{d x}+2 y=0$
$$
\begin{gathered}
\Rightarrow y=c_2 e^{-2 x} \\
\therefore y(1)=\frac{e^2-1}{2 e^2} \Rightarrow \frac{e^2-1}{2}=c^2 e^{-2} \Rightarrow C_2=\frac{e^2-1}{2} \\
\therefore y(x)\left(\frac{e^2-1}{2}\right) e^{-2 x} \Rightarrow y\left(\frac{3}{2}\right)=\frac{e^2-1}{2 e^3}
\end{gathered}
$$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
Let y=y(x) be the solution of the differential equation dydx=1+xey-x, -2<x<2, y0=0, then the minimum value of yx, x∈-2,2 is equal to :

Options:
A 2-3-loge2
B 2+3+loge2
C 1+3-loge3-1
D 1-3-loge3-1

The correct answer is:
1-3-loge3-1

Solution:
Given,

dydx=1+xey-x

⇒dy-dxey-x=xdx

⇒∫dy-xey-x=∫xdx

⇒-ex-y=x22+c

At x=0,y=0⇒c=-1

So, the particular solution is

ex-y=2-x22

⇒y=x-ln2-x22

⇒dydx=1+2x2-x2=2+2x-x22-x2

⇒dydx=x2-2x-2x2-2

⇒dydx=x2-2x-2x+2x-2

If dydx=0⇒x2-2x-2=0

⇒x=2±122

⇒x=1±3

So minimum value occurs at x=1-3

y1-3=1-3-ln2-(4-23)2

=1-3-ln3-1

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Let y=y(x) be the solution of the differential equation dy dx=2y+2sinx-5x-2cosx such that y(0)=7. Then y(π) is equal to

Options:
A 7eπ2+5
B eπ2+5
C 2eπ2+5
D 3eπ2+5

The correct answer is:
2eπ2+5

Solution:
dydx-2xy=22sin x-5x-2cos x

IF=e-x2

So, y e-x2=∫e-x22x2sin x-5-2cos x dx

⇒y·e-x2=e-x25-2sinx+C

y=5-2 sinx+C·ex2

Given at x=0, y=7

⇒7=5+C

⇒C=2

So, y=5-2 sin x+2ex2

Now at x=π

y=5+2eπ2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (27 Aug Shift 1)
Let y=yx be the solution of the differential equation dydx+2y=fx, where fx= 1,x∈0, 10,otherwise. If y(0)=0, then y 32 is

Options:
A e2-1e3
B 12e
C e2+12e4
D e2-12e3

The correct answer is:
e2-12e3

Solution:
dydx+2y=fx is a linear differential equation.

Integrating Factor =e∫2dx=e2x

Solution of the above equation is

y. e2x=∫fx.e2xdx+C

yx=e-2x∫0xfx.e2xdx+Ce-2x

y0=0⇒C=0

⇒yx=e-2x∫0xfx.e2xdx

∴y32=e-3∫01e2xdx+∫13/20.dx=e-32e2-1=e2-12e3

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2018 (15 Apr)
Let y=yx be the solution of the differential equation dydx+5xx5+1y=x5+12x7, x>0. If y1=2, then y2 is equal to

Options:
A 637128
B 679128
C 693128
D 697128

The correct answer is:
693128

Solution:
Given:

dydx+5x1+x5y=1+x52x7

This is linear differential equation.

I.F.=e∫5x1+x5dx

⇒I.F.=e∫5x5x61+x5dx

⇒I.F.=e∫5x6x-5+1dx

⇒I.F.=e-∫-5x6x-5+1dx

⇒I.F.=e-∫dx-5+1x-5+1

⇒I.F.=e-logex-5+1

⇒I.F.=x-5+1-1=x5x5+1

So, solution is given by

yx5x5+1=∫x5x5+1×x5+12x7dx

⇒yx5x5+1=∫x5+1x2dx

⇒yx5x5+1=∫x3+1x2dx

⇒yx5x5+1=x44-1x+C

Since, y1=2, so

22=14-1+C⇒C=74

So,

yx5x5+1=x44-1x+74

Put x=2, then we get

y2×3233=164-12+74

⇒y2=21×33128=693128

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (11 Apr Shift 2)
Let y=y(x) be the solution of the differential equation dydx=tanx+ysinxsecx-sinxtanx, x∈0, π2 satisfying the condition yπ4=2. Then, yπ3 is

Options:
A 32+loge3
B 322+loge3
C 31+2loge3
D 32+loge3

The correct answer is:
32+loge3

Solution:
Given: dydx=tanx+ysinxsecx-sinxtanx

⇒dydx=tanx+ysinx1cosx-sin2xcosx

⇒dydx=tanx+ysinxcos2xcosx

⇒dydx=tanx+ysinxcosx

⇒dydx-2ysin2x=sec2x

⇒IF=e-2∫cosec2xdx

⇒IF=e-logcosec2x-cot2x

⇒IF=1cosec2x-cot2x=cosec2x+cot2x

⇒IF=1sin2x+cos2xsin2x

⇒IF=2cos2x2sinxcosx

⇒IF=cotx

⇒y×cotx=∫cotx×sec2xdx

⇒y×cotx=2∫cosec2xdx

⇒ycotx=logcosec2x-cot2x+c

Now, using yπ4=2 we get,

⇒2cotπ4=logcosecπ2-cotπ2+c

⇒c=2

⇒ycotx=logcosec2x-cot2x+2

Now, finding yπ3 we get,

⇒ycotπ3=logcosec2π3-cot2π3+2

⇒y×13=log23+13+2

⇒y×13=12log3+2

⇒y=32log3+23

⇒y=3log3+2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Let y=y(x) be the solution of the differential equation dydx=y+1y+1ex2/2-x,0<x<2.1, with y(2)=0. Then the value of dydx at x=1 is equal to

Options:
A -e3/2e2+12
B -2e21+e22
C e5/21+e22
D 5e1/2e2+12

The correct answer is:
-e3/2e2+12

Solution:
Let y+1=Y

∴dYdx=Y2ex22-xY

Put -1Y=k

⇒dkdx+k-x=ex22

I.F.=e-x22

∴k=x+cex2/2

Put k=-1y+1

∴y+1=-1(x+c)ex2/2 ...i

when x=2, y=0, then c=-2-1e2

Differentiate equation (i) & put x=1 we get,

dydxx=1=-e3/21+e22

Topic: Mathematics, Differential Equations, JEE Main
Let y=y(x) be the solution of the differential equation dydx=y+1y+1ex2/2-x,0<x<2.1, with y(2)=0. Then the value of dydx at x=1 is equal to

Options:
A -e3/2e2+12
B -2e21+e22
C e5/21+e22
D 5e1/2e2+12

The correct answer is:
-e3/2e2+12

Solution:
Let y+1=Y

∴dYdx=Y2ex22-xY

Put -1Y=k

⇒dkdx+k-x=ex22

I.F.=e-x22

∴k=x+cex2/2

Put k=-1y+1

∴y+1=-1(x+c)ex2/2 ...i

when x=2, y=0, then c=-2-1e2

Differentiate equation (i) & put x=1 we get,

dydxx=1=-e3/21+e22

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let y=yx be the solution of the differential equation, dydx+ytanx=2x+x2tanx, x∈-π2, π2, such that y0= 1. Then

Options:
A y'π4-y'-π4=π- 2
B y'π4+y'-π4=- 2
C yπ4-y-π4= 2
D y'π4+y'-π4=π22+2

The correct answer is:
y'π4-y'-π4=π- 2

Solution:
Given dydx+ytanx=2x+x2tanx

This is a linear differential equation of the type dydx+Py=Q, where P=tanx & Q=2x+x2tanx

Now, the integrating factor I.F.=e∫Pdx=e∫tanx dx

=elnsecx=secx

And, the general solution is yI.F.=∫QI.F.dx+c

y⋅secx=∫2x+x2tanxsecxdx+c

⇒ysecx=∫2xsec xdx+∫x2secx⋅tanxdx+c

Using integration by parts in the second integral, we get

⇒ysecx=∫2xsec xdx+x2∫secx⋅tanxdx-∫ddxx2∫secx⋅tanxdx+c

⇒ysecx=∫2xsec xdx+x2∫secxdx-∫2xsecxdx+c

⇒ysecx=x2secx+c

⇒y=x2+c·cosx

Now, y0=1

⇒0+c=1⇒c=1

So, y=x2+cosx

Hence, yπ4=π216+12 and y-π4=π216+12

Differentiating y(x) with respect to x

⇒y'x=2x-sinx

Hence, y'π4=π2-12 and y'-π4=-π2+12

⇒y'π4-y'-π4=π-2.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Let y=yx be the solution of the differential equation dy=eαx+ydx;α∈N. If yloge2=loge2 and y0=loge12, then the value of α is equal to ___.

Options:

The correct answer is:
2

Solution:
We have, ∫e-ydy=∫eαxdx

⇒-e-y=eαxα+c ...1

Put x,y=ℓn2,ℓn2

-12=2αα+c ...2

Put x,y≡0,-ℓn2

-2=1α+c

From equation 1-2, we get

2α-1α=32

⇒α=2( as α∈N).

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (27 Jul Shift 2)
Let y=yx be the solution of the differential equation ex1-y2 dx+yxdy=0, y1=-1
Then the value of y32 is equal to:

Options:
A 1-4e3
B 1-4e6
C 1+4e3
D 1+4e6

The correct answer is:
1-4e6

Solution:
We have,

ex1-y2 dx+yxdy=0, y1=-1

⇒ ex1-y2 dx=-yxdy

⇒ ∫-y1-y2 dy=∫xexdx

⇒ 12∫d1-y21-y2 dy=∫xexdx

⇒ 1-y2=exx-1+c

Now, at x=1,y=-1, then

⇒ 0=0+c⇒c=0

∴ 1-y2=exx-1

At x=3, then

1-y32=2e3

⇒ 1-y32=4e6

⇒ y32=1-4e6

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Let y=yx be the solution of the differential equation sec2xdx+e2ytan2x+tanxdy=0, 0<x<π2,yπ4=0. If yπ6=α, then e8α is equal to ______.

Options:

The correct answer is:
9

Solution:
Given,

sec2xdx+e2ytan2x+tanxdy=0,

⇒sec2dxdy+e2ytan2x+tanx=0

Now, let tanx=t⇒sec2xdxdy=dtdy

⇒dtdy+e2y×t2+t=0

⇒dtdy+t=−t2.e2y

⇒1t2dtdy+1t=−e2y

Now, taking 1t=u⇒−1t2dtdy=dudy

⇒−dudy+u=−e2y

⇒dudy−u=e2y

Now, finding an integrating factor IF=e−∫dy=e−y

So, the solution is given by,

ue−y=∫e−y×e2ydy

⇒e−yt=ey+c

⇒1tanx×e−y=ey+c

Now, using the given condition at x=π4,y=0 we get,

⇒c=0

⇒1tanx=e2y

Now, putting the value x=π6, y=α we get,

e2α=3

⇒e8α=9

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Let y=yx be the solution of the differential equation sin⁡xdydx+ycosx⁡=4x, x∈0, π. If yπ2=0, then yπ6 is equal to

Options:
A -49π2
B 493π2
C -893π2
D -89π2

The correct answer is:
-89π2

Solution:
dydx+ycot x=4xsinx, x∈0, π
I.F. =e∫cotx=eln⁡sinx=sinx
∴y sinx=∫4xsinx.sinx dx
ysinx=2x2+c
∵yπ2=0
∴0×1=2×π22+c
⇒c=-π22
yπ6=2×π62-π2212=π218-π22×2
=π29-π2
=-8π29

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2018 (08 Apr)
Let y=yx be the solution of the differential equation x1-x2dydx+3x2y-y-4x3=0,x>1 with y2=-2. Then y3 is equal to

Options:
A -18
B -12
C -6
D -3

The correct answer is:
-18

Solution:
Given x1-x2dydx+3x2y-y-4x3=0

x-x3dydx+3x2-1y=4x3

dydx+3x2-1x-x3y=4x3x-x3

This is a linear differential equation of the form dydx+Py=Q

Here IF=e∫Pdx=e∫3x2-1x-x3dx

Let x-x3=t⇒IF=e∫-dtt

=e-lnt=1t

∴ IF=1x-x3

So the general solution of the differential equation will be

y×IF=∫Q×IFdx

y1x-x3=∫4x3x-x3×1x-x3dx

=∫4x3x-x32dx

=∫4x1-x22dx

Now let I=∫4x1-x22dx

Substituting 1-x2=z

I=2∫-dzz2

=-2-1z+c=21-x2+c

Hence the solution of the differential equation becomes

yx-x3=21-x2+c

At x=2, y=-2

-22-8=21-4+c

13=-23+c

∴c=1

Hence the particular solution will be yx-x3=21-x2+1

Put x=3

y3-27=21-9+1

y-24=-14+1

-y24=34

y=34-24=-18

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (28 Jun Shift 1)
Let y=y(x) be the solution of the differential equation, x2+12 dydx+2x(x2+1)y=1 such that y0=0. If a y1=π32, then the value of a is

Options:
A 116
B 12
C 14
D 1

The correct answer is:
116

Solution:
x2+12 dydx+2xx2+1y=1
⇒dydx+2xx2+1y=1(x2+1)2
Which is a linear differential equation with integrating factor
I.F. = e∫2xx2+1dx=eln(x2+1)=x2+1
The solution of the given differential equation will be,
y.x2+1=∫x2+1.1x2+12dx=∫dxx2+1=tan-1x+c
Now, y0= 0 ⇒C=0 and y=1(x2+1)tan-1x
ay(1)=π32⇒a=π/32π/8=14
⇒a=116

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2019 (08 Apr Shift 1)
Let y=y(x) be the solution of the differential equation (x2– 3y2)dx+3xy dy=0, y(1)=1. Then 6y2e is equal to

Options:
A 3e2
B e2
C 2e2
D 3e22

The correct answer is:
2e2

Solution:
Given equation is:

(x2– 3y2)dx+3xydy=0

We can re-write equation as

dydx=-x2-3y23xy

⇒dydx=yx-13xy ....(1)

Put y=vx

dydx=v+xdvdx

Equation (1) can be written as

v+xdvdx=v-131v⇒vdv=-13x

Integrating both sides, we get

v22=-13lnx+c

⇒y22x2=-13lnx+c .....(2)

∵ y1=1 (given)

∴ c=12 (from equation 2)

Equation (2) can be written as

y22x2=-13lnx+12

⇒y2=-23x2lnx+x2

Now y2(e)=-23e2lne+e2=-23e2+e2=e23

⇒6y2(e)=2e2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Let y = y(x) be the solution of the differential equation x3 dy + (xy – 1) dx = 0, x > 0, y12=3-e. Then y1 is equal to

Options:
A 1
B e
C 2-e
D 3

The correct answer is:
1

Solution:
Given differential equation is

x3dy+xy-1dx=0

⇒dydx=1-xyx3

⇒dydx=1x3-yx2

⇒dydx+yx2=1x3

This is a linear differential equation.

If IF=e∫1x2dx=e-1x

Solution of a given linear differential equation is

ye-1x=∫e-1x1x3dx

Put -1x=t⇒dxx2=dt

⇒ye-1x=-∫e-ttdt

⇒ye-1x=te-t+e-t+C

⇒y=t+1+Ce1x

⇒y=1x+1+ Ce1x

Put x=12, then

3-e=2+1+Ce2

C=-1e

So,

y=1x+1-1ee1x

So,

y1=1+1-1=1

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Let y=yx be the solution of the differential equation, xdydx+y=xlogex, x>1. If 2y2=loge4-1, then ye is equal to

Options:
A -e2
B e4
C -e22
D e24

The correct answer is:
e4

Solution:
The given differential equation can be written as
dydx+yx=logex, which is a linear differential equation.

Now, integrating factor I.F.=e∫1xdx=elnx=x

Hence, the solution of the given differential equation is

y·x=∫x·logex dx

⇒y·x=logex∫xdx-∫1x∫xdxdx (using integration by parts)

⇒y·x=logexx22-∫1x·x22dx

⇒y·x=logexx22-∫x2dx

⇒xyx=x22logex-x24+C ...1

Putting x=2 we get,

⇒2y2=2loge2-1+C

⇒2y2=loge4-1+C

Given, 2y2=loge4-1

Then we get, C=0

Now, from the equation 1 we get,

yx=x2logex-x4

Putting x=e in above equation, we get,

ye=e2logee-e4=e2-e4=e4

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Let y=y(x) be the solution of the differential equation xdy-ydx=x2-y2dx, x≥1, with y(1)=0. If the area bounded by the line x=1,x=eπ,y=0 and y=y(x) is αe2π+β, then the value of 10(α+β) is equal to ___ .

Options:

The correct answer is:
4

Solution:
Given xdy-ydx=x2-y2dx

⇒xdy-ydxx2=1x1-y2x2dx

⇒∫dyx1-yx2=∫dxx

⇒sin-1yx=lnx+c

at x=1,y=0⇒c=0

y=xsin(lnx)

A=∫1eπxsinlnxdx

x=et,dx=etdt⇒∫0πe2tsintdt=A

αe2π+β=e2t5(2sint-cost)0π=1+e2π5

α=15, β=15

So, 10(α+β)=4

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let y=yx be the solution of the differential equation xdy=y+x3cosxdx with yπ=0, then yπ2 is equal to:

Options:
A π24+π2
B π22+π4
C π22-π4
D π24-π2

The correct answer is:
π24+π2

Solution:
We have, xdy=y+x3cosxdx

⇒xdy=ydx+x3cosxdx

⇒xdy-ydxx2=x3cosxdxx2

⇒ddxyx=∫xcosxdx

⇒yx=xsinx-∫1.sinxdx

Therefore, yx=xsinx+cosx+C

At x=π, y=0,

0=-1+C

⇒C=1,x=π,y=0

So, yx=xsinx+cosx+1

⇒y=x2sinx+xcosx+x

Hence, yπ2=π24+π2.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (25 Jul Shift 2)
Let y=y(x) be the solution of the differential equation xlogexdydx+y=x2logex,(x>1). If y(2)=2, then y(e) is equal to

Options:
A 4+e24
B 1+e24
C 2+e22
D 1+e22

The correct answer is:
4+e24

Solution:
Given,

xlnxdydx+y=x2lnx

⇒dydx+yxlnx=x

This is a linear differential equation.

Now, finding integrating factor we get

I.F=e∫1xlnxdx=elnlnx=lnx

Now, solution of the differential equation is given by

y×I.F=∫x×I.Fdx

⇒y×lnx=∫xlnx dx

⇒y×lnx=lnx·x22-x24+c

Now given y2=2,

2×ln2=ln2·222-224+c

⇒c=1

So, equation will be y×lnx=lnx·x22-x24+1

Now finding ye we get,

y×lne=lne·e22-e24+1

⇒y×1=1·e22-e24+1

⇒y=4+e24

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Let y=yx be the solution of the differential equation x tanyxdy=y tanyx-xdx, -1≤x≤1,y12=π6. Then the area of the region bounded by the curves x=0,x=12 and y=yx in the upper half plane is:

Options:
A 18π-1
B

112π-3

C 14π-2
D 16π-1

The correct answer is:
18π-1

Solution:
We have,

dydx=xyx·tanyx-1xtanyx
⇒dydx=yx-cotyx

Put y=vx

⇒dydx=v+xdvdx

Now, we get

v+xdvdx=v-cotv

⇒∫tanvdv=-∫dxx

⇒lnsecv=-lnx+c

⇒lnsecyx=-lnx+c

⇒lnsecyx+lnx=c

Now, y12=π6, then

lnsecπ3+ln12=c

⇒ln2+ln12=c

⇒ln2-ln2=c

⇒c=0

Hence,

∴ secyx=1x

⇒cosyx=x

⇒y=xcos-1x

So, required bounded area

=∫012cos-1x⏟IxIIdx

=cos-1x·x22012+12∫012x21-x2dx

=π16-12∫0121-x2-11-x2dx

=π16-12∫0121-x2-11-x2dx

=π16-12x21-x2+12sin-1x-sin-1x012

=π16-12x21-x2-12sin-1x012

=π16-1214-π8

=π-18 sq. units

Topic: Mathematics, Differential Equations, JEE Main
Let y=yx be the solution of the differential equation x tanyxdy=y tanyx-xdx, -1≤x≤1,y12=π6. Then the area of the region bounded by the curves x=0,x=12 and y=yx in the upper half plane is:

Options:
A 18π-1
B

112π-3

C 14π-2
D 16π-1

The correct answer is:
18π-1

Solution:
We have,

dydx=xyx·tanyx-1xtanyx
⇒dydx=yx-cotyx

Put y=vx

⇒dydx=v+xdvdx

Now, we get

v+xdvdx=v-cotv

⇒∫tanvdv=-∫dxx

⇒lnsecv=-lnx+c

⇒lnsecyx=-lnx+c

⇒lnsecyx+lnx=c

Now, y12=π6, then

lnsecπ3+ln12=c

⇒ln2+ln12=c

⇒ln2-ln2=c

⇒c=0

Hence,

∴ secyx=1x

⇒cosyx=x

⇒y=xcos-1x

So, required bounded area

=∫012cos-1x⏟IxIIdx

=cos-1x·x22012+12∫012x21-x2dx

=π16-12∫0121-x2-11-x2dx

=π16-12∫0121-x2-11-x2dx

=π16-12x21-x2+12sin-1x-sin-1x012

=π16-12x21-x2-12sin-1x012

=π16-1214-π8

=π-18 sq. units

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Let y=y(x) be the solution of the differential equation, xy'-y=x2(xcosx+sinx),x>0. If y(π)=π, then y''π2+yπ2 is equal to :

Options:
A 2+π2
B 1+π2+π24
C 2+π2+π24
D 1+π2

The correct answer is:
2+π2

Solution:
Given xdydx−y=x2xcos+sinx

⇒dydx−1xy=xxcosx+sinx

∴ I.F =e-lnx=1x

∴ Solution is y.1x=∫1x.xxcosx+sinxdx

yx=∫xcosx+sinxdx

yx=xsinx+C

∵yπ=π ⇒C=1

y=x2sinx+x

dydx=x2cosx+2xsinx+1

d2ydx2=-x2sinx+2xcosx+2sinx+2xcosx

=−x2sinx+4xcosx+2sinx

∴ y''π2+yπ2=−π24+0+2+π24+π2

=−π24+2+π24+π2

=π2+2.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (04 Sep Shift 1)
Let y=yx, x>1, be the solution of the differential equation x-1dydx+2xy=1x-1, with y2=1+e42e4. If y3=eα+1βeα. then the value of α+β is equal to ______.

Options:

The correct answer is:
14

Solution:
Given, x-1dydx+2xy=1x-1

⇒dydx+2xx-1·y=1x-12

I.F=e∫2xx-1=e2x×x-12

Solution will be,

y×I.F=∫1x-12×I.F

Solving above with y2=1+e42e4 we get,

y=1x-12e2x+12e2x

So, y3=e6+18e6, now on comparing with y3=eα+1βeα we get, α+β=14

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Let y=yx, y>0, be a solution curve of the differential equation 1+x2dy=yx-ydx. If y0=1 and y22=β, then

Options:
A e3β-1=e3+22
B e3β-1=e5+2
C eβ-1=e-23+22
D eβ-1=e-25+2

The correct answer is:
e3β-1=e3+22

Solution:
Given,

1+x2dy=yx-ydx

⇒dydx=x1+x2y-y21+x2

⇒-1y2dydx+x1+x2·1y=11+x2

Now let 1y=t, we get,

-1y2y'=dtdx

So, the equation becomes,

dtdx+x1+x2t=11+x2

Now finding, IF=e∫x1+x2dx=1+x2

So, solution of the differential equation is given by,

t1+x2=∫1+x21+x2dx

⇒1y1+x2=∫11+x2dx

⇒1y1+x2=lnx+x2+1+C

∵ y0=1⇒C=1

⇒1y1+x2=lnx+x2+1+1

Now for x=22, y=β we get,

⇒3β=ln22+3+1

⇒β=31+ln22+3

⇒3β-1=1+ln22+3

⇒e3β-1=e·eln3+22

⇒e3β-1=e3+22

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Let \( y=\frac{1}{u^{2}+u-2} \) where \( u=\frac{1}{x-1} \) then \( y \) is discontinuous only at \( x= \)

Options:
A \( 1,2 \)
B \( 1,-2 \)
C \( 1, \frac{1}{2}, 2 \)
D None of these

The correct answer is:
\( 1, \frac{1}{2}, 2 \)

Solution:
u= ∅x=1x-1 is discontinuous at x=1, y=fu=1u2+u-2= 1u+2u-1 is discontinuous at u=-2, u=1 . If u=-2 then -2=1x-1 ⇒x=12. If u=1 then 1=1x-1 ⇒x=2. Hence the composite function is discontinuous only at x=1, 12, 2.

Topic: Mathematics, Continuity and Differentiability, JEE Main