datavorous/entrance-exam-dataset huggingface dataset:
The difference in isotopes of an element is____

Options:
A Mass number
B Atomic number
C Number of electrons
D Number of protons

The correct answer is:
Mass number

Solution:
- Isotopes are atoms of the same element that have different numbers of neutrons but the same number of protons and electrons.
- The difference in the number of neutrons between the various isotopes of an element means that the various isotopes have different masses.

Topic: Physics, Atomic Physics, JEE Main
The difference in length between two rods $\mathrm{A}$ and $\mathrm{B}$ is $60 \mathrm{~cm}$ at all temperatures. If $\alpha_{\mathrm{A}}=18 \times 10^{-6} /{ }^{\circ} \mathrm{C}$ and $\beta_{\mathrm{B}}=27 \times 10^{-6} /{ }^{\circ} \mathrm{C}$, the lengths of the two rods are

Options:
A $l_{\mathrm{A}}=200 \mathrm{~cm}, l_{\mathrm{B}}=140 \mathrm{~cm}$
B $l_{\mathrm{A}}=180 \mathrm{~cm}, l_{\mathrm{B}}=120 \mathrm{~cm}$
C $l_{\mathrm{A}}=160 \mathrm{~cm}, l_{\mathrm{B}}=100 \mathrm{~cm}$
D $l_{\mathrm{A}}=120 \mathrm{~cm}, l_{\mathrm{B}}=60 \mathrm{~cm}$

The correct answer is:
$l_{\mathrm{A}}=180 \mathrm{~cm}, l_{\mathrm{B}}=120 \mathrm{~cm}$

Solution:
Given: $\Delta l=60 \mathrm{~cm}, \alpha_{\mathrm{A}}=18 \times 10^{-6} /{ }^{\circ} \mathrm{C}$,
$\alpha_{\mathrm{B}}=27 \times 10^{-6} /{ }^{\circ} \mathrm{C}$
$\Delta l$ is constant at all temperatures.
We know $\Delta l=l \alpha \Delta \mathrm{t}$
Let the length of the rods at a temperature $0^{\circ} \mathrm{C}$ be $l_{\mathrm{A}}$ and, $l_{\mathrm{B}}$
$\therefore \quad$ At temperature $\mathrm{t}^{\circ} \mathrm{C}$
$\begin{array}{ll}
& l_{\mathrm{A}} a_{\mathrm{A}} \mathrm{t}_{\mathrm{A}}=l_{\mathrm{B}} a_{\mathrm{B}} \mathrm{t}_{\mathrm{B}} \\
& l_{\mathrm{A}}(18) \times 10^{-6}=l_{\mathrm{B}}(27) \times 10^{-6}.. (i) \\
& \Delta l=l_{\mathrm{A}}-l_{\mathrm{B}} \\
\Delta l & =\frac{3}{2} l_{\mathrm{B}}-l_{\mathrm{B}} .. from (i)\\
& \Delta l=\frac{1}{2} l_{\mathrm{B}} \\
\therefore \quad & l_{\mathrm{B}}=2 \Delta l \\
\therefore \quad & l_{\mathrm{B}}=2 \times 60=120 \mathrm{~cm} \\
\therefore \quad & l_{\mathrm{A}}=\frac{3}{2} \times 120=180 \mathrm{~cm}
\end{array}$

Topic: Physics, Thermal Properties of Matter, JEE Main
The difference in oxidation state of chromium in chromate and dichromate salts is___

Options:

The correct answer is:
0

Solution:
Cr2O72-+OH-⟶2CrO42-

Let the oxidation number of chromium be x

The oxidation state of chromium in Cr2O72- =2x+(-2)7=-2x=+6

Oxidation state of Cr in CrO42-

=x+(-2)4=-2x=+6.

So, the oxidation number of Cr in CrO42- is also +6.

Hence, the difference in oxidation number between ions is 0

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (24 Jun Shift 1)
The difference in the number of unpaired electrons in Co2+ ion in its high-spin and low-spin octahedral complexes is -----.

Options:

The correct answer is:
2

Solution:
Co2+⇒[Ar]3d7
High-spin complex:

3 unpaired electrons
Low-spin complex:

1 unpaired electron
â´ Difference =3-1=2

Topic: Chemistry, Coordination Compounds, JEE Main
The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is:

Options:
A Fe2+
B Co2+
C Ni2+
D Mn2+

The correct answer is:
Co2+

Solution:
The electronic configuration for Co2+z=27=4s03d7
When it combines with weak ligands (high spin complex),t2g2,2,1 eg1,1, it has 3 unpaired electrons, whereas when it bonds with strong ligands (low spin complex), t2g2,2,2 eg1,0, then, it has 1 unpaired electron.

The difference in number of unpaired electrons is equal to two.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The difference in the number of wavelengths, when yellow light propagates through air and vacuum columns of the same thickness, is one. The thickness of the air column is Refractive index of air $\mu_c=1.0003$, Wavelength of yellow light in vacuum $=6000 Å$

Options:
A $1.8 \mathrm{~mm}$
B $2 \mathrm{~mm}$
C $2 \mathrm{~cm}$
D $2.2 \mathrm{~cm}$

The correct answer is:
$2 \mathrm{~mm}$

Solution:
Difference in the number of wavelength $=1$. If ' $L$ ' is the thickness of air and vacuum columns.
$\begin{aligned} & \frac{L}{\lambda_{\text {st }}}-\frac{L}{\lambda_{\text {man }}}=1 \\ & L\left(\frac{1}{\lambda_m}-\frac{1}{\lambda_{n m}}\right)=1 \\ & L\left(\frac{1}{\frac{\lambda_{\mathrm{m}=}}{\mu_{\mathrm{m}}}}-\frac{1}{\lambda_{\mathrm{van}}}\right)=1 \\ & L\left(\frac{\mu_{-}}{\lambda_{\operatorname{man}}}-\frac{1}{\lambda_{\operatorname{man}}}\right)=1 \\ & \frac{L}{\lambda_{\text {eme }}}\left(\mu_{\mathrm{N}}-1\right)=1 \\ & \frac{L}{6000 \times 10^{-10}}(1.0003-1)=1 \\ & L=\frac{6000 \times 10^{-20}}{0.0003}=\frac{6 \times 10^{-7}}{3 \times 10^{-4}} \\ & =2 \times 10^{-2} \mathrm{~m}=2 \mathrm{~mm} \\ & \end{aligned}$

Topic: Physics, Ray Optics, JEE Main
The difference in the number of waves when yellow light propagates through air and vacuum columns of the same thickness is one. The thickness of the air column is _____ mm. [Refractive index of air =1.0003, the wavelength of yellow light in vacuum =6000 A∘]

Options:

The correct answer is:
2

Solution:
Thickness t=nλ

So, n λvac =(n+1)λair

nλ=(n+1)λμair

n=1μair-1=1043

t=nλ

=1043×6000 A0

=2 mm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2021 (27 Jul Shift 2)
The difference in the oxidation numbers of the two types of sulphur atoms in Na2S4O6 is

Options:

The correct answer is:
5

Solution:
S will have oxidation number =+5,0
Difference in oxidation number =5

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The difference in the reaction of phenol with bromine in chloroform and bromine in water medium is due to

Options:
A Hyperconjugation in substrate
B Polarity of solvent
C Free radical formation
D Electromeric effect of the substrate

The correct answer is:
Polarity of solvent

Solution:
Difference in reactions is observed due to solvent polarity, which

(i) Ionizes phenol to make more reactive phenoxide ion

(ii) Increases electrophilicity of bromine.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2022 (26 Jul Shift 1)
The difference in the 'spin-only' magnetic moment values of \(\mathrm{KMnO}_4\) and the manganese product formed during titration of \(\mathrm{KMnO}_4\) against oxalic acid in acidic medium is _______ BM. (nearest integer)

Options:

The correct answer is:
6

Solution:
Spin only magnetic moment of $\mathrm{Mn}$ in $\mathrm{KMnO}_4=0$ Spin only value of manganese product fromed during titration of $\mathrm{KMnO}_4$ aganist oxalic acid in acidic medium is $=6$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (06 Apr Shift 1)
The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the

Options:
A
crystal structure

B
variation of the number of charge carriers with temperature

C
type of bonding

D
variation of scattering mechanism with temperature

The correct answer is:
variation of the number of charge carriers with temperature

Solution:
No Solution Available

Topic: Physics, Semiconductors, JEE Main, JEE Main 2003
The difference in the variation of resistance with temperature in a metal semiconductor arises essentially due to the difference in the

Options:
A crystal structure
B variation of the number of charge carriers with temperature
C type of bonding
D variation of scattering mechanism with temperature

The correct answer is:
variation of the number of charge carriers with temperature

Solution:
The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the variation of the number of charge carriers with temperature.

Topic: Physics, Semiconductors, JEE Main
The difference of speed of light in the two media A and BvA-vB is 2.6×107 m s-1. If the refractive index of medium B is 1.47, then the ratio of refractive index of medium B to medium A is: (Given : speed of light in vacuum c=3×108 m s-1)

Options:
A 1.303
B 1.318
C 1.13
D 0.12

The correct answer is:
1.13

Solution:
The speed of light in the medium B will be, vB=cμB

The speed of light in the medium A will be, vA=cμA.

Now,

vA-vB=c1μA-1μB⇒2.6×107=3×108×1μA-11.47⇒μA=1.13

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (25 Jun Shift 1)
The differential coefficient of log10⁡x with respect to logx10 is

Options:
A 1
B -log10⁡x2
C logx102
D x2100

The correct answer is:
-log10⁡x2

Solution:
Let, u=log10â¡x and v=logxâ¡10
â u=logeâ¡xlogeâ¡10 and v=logeâ¡10logeâ¡x
Now, dudx=1xlogeâ¡10
and dvdx=logeâ¡10-1xlogeâ¡x2
â´ dudv=du/dxdv/dx=1xlogeâ¡10Ã·-logeâ¡10xlogeâ¡x2
=-logeâ¡x2logeâ¡102=-logeâ¡xlogeâ¡102
=-log10â¡x2

Topic: Mathematics, Differentiation, JEE Main
The differential equation corresponding to primitive $y=e^{c x}$ is

Options:
A $\frac{d y}{d x}=\left(\frac{y}{x}\right) \log x$
B $\frac{d y}{d x}=\left(\frac{x}{y}\right) \log y$
C $\frac{d y}{d x}=\left(\frac{y}{x}\right) \log y$
D $\frac{d y}{d x}=\left(\frac{x}{y}\right) \log x$

The correct answer is:
$\frac{d y}{d x}=\left(\frac{y}{x}\right) \log y$

Solution:
$\begin{aligned} & y=e^{m x} \Rightarrow \log y=m x \Rightarrow m=\frac{\log y}{x} \\ & \text { Now } y=e^{m x} \Rightarrow \frac{d y}{d x}=m e^{m x}=\frac{\log y}{x} \cdot y=\left(\frac{y}{x}\right) \log y\end{aligned}$

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

Options:
A Linear
B Homogeneous
C Order two
D Degree one

The correct answers are:
Order two, Degree one

Solution:
Given $\frac{d^2 y}{d x^2}+x \cdot \frac{d y}{d x}+\sin y+x^2=0$
The order of highest derivative $=2$ and degree $=1$.

Topic: Mathematics, Differential Equations, JEE Main
The differential equation dydx=1-y2y represents the arc of a circle in the second and the third quadrant and passing through -12,12. Then, the radius (in units) of the circle is

Options:
A 12
B 14
C 2
D 1

The correct answer is:
1

Solution:
∫y1-y2dy=∫dx
⇒12∫2ydy1-y2=∫dx
⇒-12∫d1-y21-y2=∫dx
⇒-1-y2=x+C
As the curves passes through -12,12;C=0
Thus, we have
-1-y2=x
⇒1-y2=x2
⇒x2+y2=1, which is a circle with radius ‘1’ unit

Topic: Mathematics, Differential Equations, JEE Main
The differential equation for the family of curves $x^2+y^2-2 a y=0$, where $a$ is an arbitrary constant is

Options:
A
$2\left(x^2-y^2\right) y^{\prime}=x y$

B
$2\left(x^2+y^2\right) y^{\prime}=x y$

C
$\left(x^2-y^2\right) y^{\prime}=2 x y$

D
$\left(x^2+y^2\right) y^{\prime}=2 x y$

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

Solution:
$2 x+2 y y^{\prime}-2 a y^{\prime}=0$
$a=\frac{x+y y^{\prime}}{y^{\prime}} \quad$ (eliminating a)
$\Rightarrow\left(x^2-y^2\right) y^{\prime}=2 x y$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2004
The differential equation obtained by eliminating the arbitrary constants a and b from xy=aex+be-x is

Options:
A x d2ydx2+2dydx-xy=0
B d2ydx2+2yÂ dydx-xy=0
C xÂ d2ydx2+2Â dydx+xy=0
D d2ydx2+dydx-xy=0

The correct answer is:
x d2ydx2+2dydx-xy=0

Solution:
Given, xy=aex+be-x ..... (I)
⇒ x dydx+y=aex-be-x
⇒ x d2ydx2+dydx+dydx=aex+be-x
⇒ x d2ydx2+2 dydx-xy=0 [from eq. (I)]

Topic: Mathematics, Differential Equations, JEE Main
The differential equation of all circles passing through the origin and having their centres on the $x$-axis is

Options:
A
$x^2=y^2+x y \frac{d y}{d x}$

B
$x^2=y^2+3 x y \frac{d y}{d x}$

C
$y^2=x^2+2 x y \frac{d y}{d x}$

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

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

Solution:
General equation of all such circles is
$\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}=0$.
Differentiating, we get
$2 \mathrm{x}+2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{g}=0$
$\therefore$ Desired equation is
$\begin{aligned}
& x^2+y^2+\left(-2 x-2 y \frac{d y}{d x}\right) x=0 \\
& \Rightarrow y^2=x^2+2 x y \frac{d y}{d x}.
\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2007
The differential equation of all parabolas whose axes are parallel to $y$-axis is

Options:
A $\frac{d^3 y}{d x^3}=0$
B $\frac{d^2 x}{d y^2}=c$
C $\frac{d^3 y}{d x^3}+\frac{d^2 x}{d y^2}=0$
D $\frac{d^2 y}{d x^2}+2 \frac{d y}{d x}=c$

The correct answer is:
$\frac{d^3 y}{d x^3}=0$

Solution:
The equation of a member of the family of parabolas having axis parallel to $y$-axis is $y=A x^2+B x+C$ ....(i)
where $A, B, C$ are arbitrary constants.
Differentiating (i) w.r.t. $x$, we get $\frac{d y}{d x}=2 A x+B$ ....(ii)
Which on differentiating w.r.t. $x$ gives $\frac{d^2 y}{d x^2}=2 A$ ....(iii)
Differentiating w.r.t. $x$ again, we get $\frac{d^3 y}{d x^3}=0$

Topic: Mathematics, Differential Equations, JEE Main
The differential equation of all straight lines passing through the point $(1,-1)$ is

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

The correct answer is:
$y=(x-1) \frac{d y}{d x}-1$

Solution:
Since the equation of line passing through $(1,-1)$ is $y+1=m(x-1)$
$\Rightarrow \quad y+1=\frac{d y}{d x}(x-1) \Rightarrow y=(x-1) \frac{d y}{d x}-1$.

Topic: Mathematics, Differential Equations, JEE Main
The differential equation of all the lines in the $x y$-plane is

Options:
A $\frac{d y}{d x}-x=0$
B $\frac{d^2 y}{d x^2}-x \frac{d y}{d x}=0$
C $\frac{d^2 y}{d x^2}=0$
D $\frac{d^2 y}{d x^2}+x=0$

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

Solution:
The equation of all the lines in $x y$-plane is given by $y=m x+c$
Differentiate it twice w.r.t. $x$, we get $\frac{d^2 y}{d x^2}=0$

Topic: Mathematics, Differential Equations, JEE Main
The differential equation of the curve for which the point of tangency (closer to the x-axis) divides the segment of the tangent between the coordinate axes in the ratio 1:2, is

Options:
A xdy=2ydx
B xdy=ydx
C xdy+2ydx=0
D xdy+ydx=0

The correct answer is:
xdy+2ydx=0

Solution:
Equation of the tangent is
Y-y=mX-x
âAâ¡x-ym,0, Bâ¡0,y-mx
Using section formula, we get,
y=1ây-mx+201+2
or 3y=y-mx
âm=dydx=-2yx
i.e. xdy+2ydx=0

Topic: Mathematics, Differential Equations, JEE Main
The differential equation of the family of circles passing through the origin and having centre at the line $y=x$ is :

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

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

Solution:
$\begin{aligned}
& C \equiv x^2+y^2+g x+g y=0....(1)\\
& 2 x+2 y y^{\prime}+g+g y^{\prime}=0 \\
& g=-\left(\frac{2 x+2 y y^{\prime}}{1+y^{\prime}}\right)
\end{aligned}$
Put in (1)
$\begin{aligned}
& x^2+y^2-\left(\frac{2 x+2 y y^{\prime}}{1+y^{\prime}}\right)(x+y)=0 \\
& \left(x^2-y^2-2 x y\right) y^{\prime}=x^2-y^2+2 x y
\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The differential equation of the family of circles passing through the points 0,2 and 0,-2 is

Options:
A 2xydydx+x2-y2+4=0
B 2xydydx+x2+y2-4=0
C 2xydydx+y2-x2+4=0
D 2xydydx-x2-y2+4=0

The correct answer is:
2xydydx+x2-y2+4=0

Solution:
Given a circle passing through 0,-2 and 0,2, so its equation will be,

x-x1x-x2+y-y1y-y2+λ1xy1x1y11x2y21=0,λ1∈R

On putting the given values we get,

⇒x-0x-0+y-2y+2+λ1xy10210-21=0,λ1∈R

⇒x2+y2-4+λ14x=0

⇒x2+y2-4+λx=0,λ∈R

Now dividing both side by x we get,

⇒x2+y2-4x+λ=0

Now differentiating with respect to x

⇒x2x+2y·dydx-x2+y2-4·1x2=0

⇒2xy·dydx+x2-y2+4=0

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (28 Jul Shift 2)
The differential equation of the family of circles with fixed radius 5 units and centre on the line $y=2$ is

Options:
A
$(x-2) y^{\prime 2}=25-(y-2)^2$

B
$(y-2) y^{\prime 2}=25-(y-2)^2$

C
$(y-2)^2 y^{\prime 2}=25-(y-2)^2$

D
$(x-2)^2 y^{\prime 2}=25-(y-2)^2$

The correct answer is:
$(y-2)^2 y^{\prime 2}=25-(y-2)^2$

Solution:
$$
\begin{aligned}
& (x-h)^2+(y-2)^2=25 \\
& \Rightarrow 2(x-h)+2(y-2) \frac{d y}{d x}=0 \\
& \Rightarrow(x-h)=-(y-2) \frac{d y}{d x}
\end{aligned}
$$
substituting in (1), we have
$$
\begin{aligned}
& (y-2)^2\left(\frac{d y}{d x}\right)^2+(y-2)^2=25 \\
& (y-2)^2 y^{\prime 2}=25-(y-2)^2
\end{aligned}
$$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2008
The differential equation of the family of curves py2=3x-p is (where p is an arbitrary constant) is

Options:
A ydydx=y+x
B ydydx=1
C y2=dydx
D y2=2xydydx-1

The correct answer is:
y2=2xydydx-1

Solution:
Given family of curves is p3=xy2+1
On differentiating with respect to x, we get,
0= y 2 +1 â1âx 2yâ y â² y 2 +1 2
â2xy y â² = y 2 +1
ây2=2xydydx-1

Topic: Mathematics, Differential Equations, JEE Main
The differential equation of the family of curves represented by the equation $x^2+y^2=a^2$ is

Options:
A $x+y \frac{d y}{d x}=0$
B $y \frac{d y}{d x}=x$
C $y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0$
D None of these

The correct answer is:
$x+y \frac{d y}{d x}=0$

Solution:
Given equation $x^2+y^2=a^2$. Differentiate it w.r.t. $x$, we get $2 x+2 y \frac{d y}{d x}=0 \Rightarrow x+y \frac{d y}{d x}=0$

Topic: Mathematics, Differential Equations, JEE Main
The differential equation of the family of curves represented by the equation $x^2 y=a$, is

Options:
A $\frac{d y}{d x}+\frac{2 y}{x}=0$
B $\frac{d y}{d x}+\frac{2 x}{y}=0$
C $\frac{d y}{d x}-\frac{2 y}{x}=0$
D $\frac{d y}{d x}-\frac{2 x}{y}=0$

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

Solution:
$x^2 y=a$ (On differentiating)
$\begin{aligned} & x^2 \frac{d y}{d x}+y \frac{d}{d x}\left(x^2\right)-0 \Rightarrow x^2 \frac{d y}{d x}+2 x y=0 \\ & \Rightarrow \frac{d y}{d x}+\frac{2 y}{x}=0\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
The differential equation of the family of curves whose tangent at any point makes an angle of π4 with the ellipse x24+y2=1 is

Options:
A dydx=x+yx-y
B dydx=x+4yx-4y
C dydx=x4y
D dydx=4yx

The correct answer is:
dydx=x+4yx-4y

Solution:
Given, x24+y2=1

On differentiating w.r.t. x,
x2+2ydydx=0
dydx=-x4y=m1 let
Let, the slope of the tangent at a point on ellipse where the tangents meet is m2
Given, tanπ4=m1-m21+m1m2
1=-x4y-m21+x4ym2⇒4y-xm2=x+4ym2
4y-xm2=x+4ym2 or 4y-xm2=-x-4ym2
m2=4y-xx+4y or m2=4y+xx-4y
dydx=4y-xx+4y or dydx=4y+xx-4y

Topic: Mathematics, Differential Equations, JEE Main
The differential equation of the family of curves, x2=4by+b, b∈R, is.

Options:
A xy'2=x+2yy'
B xy'2=2yy'-x
C xy''=y'
D xy'2=x-2yy'

The correct answer is:
xy'2=x+2yy'

Solution:
x2=4by+b

Differentiate both sides,

2x=4by'

⇒b=x2y'

So, the differential equation is

x2=2xy'y+xy'2

⟹xdydx2=2ydydx+x

⇒xy'2=x+2yy'

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (08 Jan Shift 2)
The differential equation of the family of curves y=k1x2+k2 is given by (where, k1 and k2 are arbitrary constants and y1=dydx,y2=d2ydx2)

Options:
A y1=x2y2
B y12=xy2
C xy2=y1
D y1y2=x

The correct answer is:
xy2=y1

Solution:
Differentiating with respect to x , we get, dydx=2k1x
ây1x=2k1
Differentiating again, we get, xây2-y1x2=0 or xy2=y1

Topic: Mathematics, Differential Equations, JEE Main
The differential equation of the family of ellipses having centres along the line y=4 and major and minor axes parallel to the coordinate axes is of the order

Options:
A 4
B 3
C 2
D 1

The correct answer is:
3

Solution:
The general equation of the family is x-α2a2+y-42b2=1
which has three arbitrary constants. Hence, the order of its differential equation will be ‘3’

Topic: Mathematics, Differential Equations, JEE Main
The differential equation of the family of ellipses with centre at the origin and the ends of the major axis being ±1,0 is of the degree

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
The general equation of the family is x2+y2a2=1……..i
Differentiating w.r.t. x, we get 2x+2yy'a2=0
⇒x=− y ′ y a 2 ⇒ a 2 =− y ′ x
Putting in (i), we have x 2 + y 2 − x y y ′ =1
or x 2 − xy y ′ =1
Which is of the degree ‘1’

Topic: Mathematics, Differential Equations, JEE Main
The differential equation representing the family of curves $y^2=2 c(x+\sqrt{c})$, where $c$ $>0$, is a parameter, is of order and degree as follows:

Options:
A
order 1 , degree 2

B
order 1, degree 1

C
order 1, degree 3

D
order 2, degree 2

The correct answer is:
order 1, degree 3

Solution:
$$
\begin{aligned}
& y^2=2 c(x+\sqrt{c} ) .....(i)\\
& 2 y y^{\prime}=2 c \cdot 1 \text { or } y y^{\prime}=c ....(ii)\\
& \Rightarrow y^2=2 y y^{\prime}\left(x+\sqrt{y y^{\prime}}\right)
\end{aligned}
$$
[on putting value of c from (ii) in (i)]
On simplifying, we get
$$
\left(y-2 x y^{\prime}\right)^2=4 y y^{\prime 3} .....(iii)
$$
Hence equation (iii) is of order 1 and degree 3.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2005
The differential equation representing the family of ellipses having foci either on the x-axis or on the y-axis, center at the origin and passing through the point 0,3 is

Options:
A xyy'-y2+9=0
B xyy′′+xy′2-yy'=0
C xyy'+y2-9=0
D x+yy′′=0

The correct answer is:
xyy'-y2+9=0

Solution:
Let the equation of ellipse be x2a2+y2b2=1

It passes through 0,3⇒x2a2+y29=1 ...i

Differentiating w.r.t. x,

⇒2xa2+2y9dydx=0

1a2=-y9xy' ...ii

From i and ii we get,

-xy9y'+y29=1

⇒xyy'-y2+9=0

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2018 (16 Apr Online)
The differential equation satisfied by the system of parabolas y2=4a(x+a) is

Options:
A ydydx2-2xdydx-y=0
B ydydx2-2xdydx+y=0
C ydydx2+2xdydx-y=0
D ydydx+2xdydx-y=0

The correct answer is:
ydydx2+2xdydx-y=0

Solution:
y2=4ax+4a2

differentiate with respect to x

⇒2ydydx=4a

⇒a=y2dydx

so, required differential equation is

y2=4×y2dydxx+4y2dydx2

⇒y2dydx2+2xydydx-y2=0

⇒ydydx2+2xdydx-y=0

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The differential equation which represents the family of curves $y=c_1 e^{c_2 x}$, where $c_1$ and $c_2$ are arbitrary constants is

Options:
A
$y^{\prime}=y^2$

B
$y^{\prime \prime}=y^{\prime} y$

C
$y y^{\prime \prime}=y^{\prime}$

D
$y y^{\prime \prime}=\left(y^{\prime}\right)^2$

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

Solution:
$$
\begin{aligned}
& y=c_1 e^{c_2 x} \\
& y^{\prime}=c_2 c_1 e^{c_2 x} \\
& y^{\prime}=c_2 y \\
& y^{\prime \prime}=c_2 y^{\prime}
\end{aligned}
$$
From (2)
$$
c_2=\frac{y^{\prime}}{y}
$$
So, $y^{\prime \prime}=\frac{\left(y^{\prime}\right)^2}{y} \Rightarrow y y^{\prime \prime}=\left(y^{\prime}\right)^2$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2009
The differential equation whose solution is $A x^2+B y^2=1$, where $A$ and $B$ are arbitrary constants is of

Options:
A
second order and second degree

B
first order and second degree

C
first order and first degree

D
second order and first degree

The correct answer is:
second order and first degree

Solution:

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2006
The differential equation whose solution is $y=A \sin x+B \cos x$, is

Options:
A $\frac{d^2 y}{d x^2}+y=0$
B $\frac{d^2 y}{d x^2}-y=0$
C $\frac{d y}{d x}+y=0$
D None of these

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

Solution:
$\begin{aligned} & y=A \sin x+B \cos x \Rightarrow \frac{d y}{d x}=A \cos x-B \sin x \\ & \Rightarrow \frac{d^2 y}{d x^2}=-A \sin x-B \cos x=-(A \sin x+B \cos x)=-y \\ & \Rightarrow \frac{d^2 y}{d x^2}+y=0 \text { is the required differential equation. }\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main