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List of top Mathematics Questions on General and Particular Solutions of a Differential Equation asked in AP EAMCET
If
y
=
a
e
b
x
+
c
e
d
x
+
x
e
b
x
y = a e^{bx} + c e^{dx} + x e^{bx}
y
=
a
e
b
x
+
c
e
d
x
+
x
e
b
x
is the general solution of a differential equation, where
a
a
a
and
c
c
c
are arbitrary constants and
b
b
b
is a fixed constant, then the order of the differential equation is:
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
The differential equation formed by eliminating
a
a
a
and
b
b
b
from the equation
y
=
a
e
2
x
+
b
x
e
2
x
y = ae^{2x} + bxe^{2x}
y
=
a
e
2
x
+
b
x
e
2
x
is:
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
Number of solutions of the trigonometric equation
2
tan
2
θ
−
cot
2
θ
+
1
=
0
lying in the interval
[
0
,
π
]
2 \tan 2\theta - \cot 2\theta + 1 = 0 \quad \text{lying in the interval} \quad [0, \pi]
2
tan
2
θ
−
cot
2
θ
+
1
=
0
lying in the interval
[
0
,
π
]
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
If
a
a
a
is a common root of
x
2
−
5
x
+
λ
=
0
x^2 - 5x + \lambda = 0
x
2
−
5
x
+
λ
=
0
and
x
2
−
8
x
−
2
λ
=
0
x^2 - 8x - 2\lambda = 0
x
2
−
8
x
−
2
λ
=
0
(
λ
≠
0
\lambda \neq 0
λ
=
0
) and
β
,
γ
\beta, \gamma
β
,
γ
are the other roots of them, then
a
+
β
+
γ
+
λ
=
a + \beta + \gamma + \lambda =
a
+
β
+
γ
+
λ
=
:
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
The differential equation formed by eliminating arbitrary constants
A
A
A
and
B
B
B
from the equation
y
=
A
cos
3
x
+
B
sin
3
x
y = A \cos 3x + B \sin 3x
y
=
A
cos
3
x
+
B
sin
3
x
is:
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
The general solution of the differential equation
(
x
sin
y
x
)
d
y
=
(
y
sin
y
x
−
x
)
d
x
(x \sin \frac{y}{x}) dy = (y \sin \frac{y}{x} - x) dx
(
x
sin
x
y
)
d
y
=
(
y
sin
x
y
−
x
)
d
x
is:
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
Find the integrating factor of the differential equation
sin
x
d
y
d
x
−
y
cos
x
=
1
\sin x \frac{dy}{dx} - y \cos x = 1
sin
x
d
x
d
y
−
y
cos
x
=
1
:
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
Determine the order and degree of the differential equation
d
3
y
d
x
3
=
[
1
+
(
d
y
d
x
)
2
]
5
/
2
\frac{d^3y}{dx^3} = \left[1 + \left(\frac{dy}{dx}\right)^2\right]^{5/2}
d
x
3
d
3
y
=
[
1
+
(
d
x
d
y
)
2
]
5/2
:
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
The acute angle between the curves
x
2
+
y
2
=
x
+
y
x^2 + y^2 = x + y
x
2
+
y
2
=
x
+
y
and
x
2
+
y
2
=
2
y
x^2 + y^2 = 2y
x
2
+
y
2
=
2
y
is:
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
If
p
1
p_1
p
1
and
p
2
p_2
p
2
are the perpendicular distances from the origin to the tangent and normal drawn at any point on the curve
x
2
/
3
+
y
2
/
3
=
a
2
/
3
x^{2/3} + y^{2/3} = a^{2/3}
x
2/3
+
y
2/3
=
a
2/3
respectively. If
k
1
p
1
2
+
k
2
p
2
2
=
a
2
k_1 p_1^2 + k_2 p_2^2 = a^2
k
1
p
1
2
+
k
2
p
2
2
=
a
2
, then
k
1
+
k
2
=
k_1 + k_2 =
k
1
+
k
2
=
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
If
y
=
t
2
+
t
3
y = t^2 + t^3
y
=
t
2
+
t
3
and
x
=
t
−
t
4
x = t - t^4
x
=
t
−
t
4
, then
d
2
y
d
x
2
\frac{d^2y}{dx^2}
d
x
2
d
2
y
at
t
=
1
t = 1
t
=
1
is:
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
The equation of the circle whose diameter is the common chord of the circles
x
2
+
y
2
−
6
x
−
7
=
0
x^2 + y^2 - 6x - 7 = 0
x
2
+
y
2
−
6
x
−
7
=
0
and
x
2
+
y
2
−
10
x
+
16
=
0
x^2 + y^2 - 10x + 16 = 0
x
2
+
y
2
−
10
x
+
16
=
0
is:
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
If five-digit numbers are formed from the digits 0, 1, 2, 3, 4 using every digit exactly only once, then the probability that a randomly chosen number from those numbers is divisible by 4 is
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
If
∣
f
ˉ
∣
=
10
|\mathbf{\bar{f}}| = 10
∣
f
ˉ
∣
=
10
,
∣
g
ˉ
∣
=
14
|\mathbf{\bar{g}}| = 14
∣
g
ˉ
∣
=
14
and
∣
f
ˉ
−
g
ˉ
∣
=
15
|\mathbf{\bar{f}} - \mathbf{\bar{g}}| = 15
∣
f
ˉ
−
g
ˉ
∣
=
15
, then
∣
f
ˉ
+
g
ˉ
∣
=
|\mathbf{\bar{f}} + \mathbf{\bar{g}}| =
∣
f
ˉ
+
g
ˉ
∣
=
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
If
cos
−
1
(
2
x
)
+
cos
−
1
(
3
x
)
=
π
3
\cos^{-1}(2x) + \cos^{-1}(3x) = \frac{\pi}{3}
cos
−
1
(
2
x
)
+
cos
−
1
(
3
x
)
=
3
π
and
4
x
2
=
a
b
4x^2 = \frac{a}{b}
4
x
2
=
b
a
, then
a
+
b
=
a + b =
a
+
b
=
:
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
If
54
0
∘
<
A
<
63
0
∘
540^\circ<A<630^\circ
54
0
∘
<
A
<
63
0
∘
and
∣
cos
A
∣
=
5
13
|\cos A| = \frac{5}{13}
∣
cos
A
∣
=
13
5
, then
tan
A
2
tan
A
=
\tan\frac{A}{2} \tan A =
tan
2
A
tan
A
=
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
If the roots of
1
−
y
y
+
y
1
−
y
=
5
2
\sqrt{\frac{1-y}{y}} + \sqrt{\frac{y}{1-y}} = \frac{5}{2}
y
1
−
y
+
1
−
y
y
=
2
5
are
α
\alpha
α
and
β
\beta
β
(
β
>
α
\beta>\alpha
β
>
α
) and the equation
(
α
+
β
)
x
4
−
25
α
β
x
2
+
(
γ
+
β
−
α
)
=
0
(\alpha + \beta)x^4 - 25\alpha \beta x^2 + (\gamma + \beta - \alpha) = 0
(
α
+
β
)
x
4
−
25
α
β
x
2
+
(
γ
+
β
−
α
)
=
0
has real roots, then a possible value of
y
y
y
is:
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
Let
[
r
]
[r]
[
r
]
denote the largest integer not exceeding
r
r
r
and the roots of the equation
3
z
2
+
6
z
+
5
+
α
(
x
2
+
2
x
+
2
)
=
0
3z^2 + 6z + 5 + \alpha(x^2 + 2x + 2) = 0
3
z
2
+
6
z
+
5
+
α
(
x
2
+
2
x
+
2
)
=
0
are complex numbers whenever
α
>
L
\alpha>L
α
>
L
and
α
<
M
\alpha<M
α
<
M
. If
(
L
−
M
)
(L-M)
(
L
−
M
)
is minimum, then the greatest value of
[
r
]
[r]
[
r
]
such that
L
y
2
+
M
y
+
r
<
0
Ly^2 + My + r<0
L
y
2
+
M
y
+
r
<
0
for all
y
∈
R
y \in \mathbb{R}
y
∈
R
is:
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
If
z
=
x
+
i
y
z = x+iy
z
=
x
+
i
y
,
x
2
+
y
2
=
1
x^2+y^2 = 1
x
2
+
y
2
=
1
and
z
1
=
e
i
θ
z_1 = e^{i\theta}
z
1
=
e
i
θ
, then the expression
z
1
2
n
−
1
−
1
z
1
2
n
−
1
+
1
\frac{z_1^{2n-1} - 1}{z_1^{2^n-1} + 1}
z
1
2
n
−
1
+
1
z
1
2
n
−
1
−
1
simplifies to:
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
Let
A
,
B
,
C
,
D
,
A, B, C, D,
A
,
B
,
C
,
D
,
and
E
E
E
be
n
×
n
n \times n
n
×
n
matrices, each with non-zero determinant. If
A
B
C
D
E
=
I
ABCDE = I
A
BC
D
E
=
I
, then
C
−
1
C^{-1}
C
−
1
is:
AP EAMCET - 2024
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation
If
1
−
x
6
+
1
−
y
6
=
a
(
x
3
−
y
3
)
,
\sqrt{1-x^{6} }+ \sqrt{1-y^{6}} =a\left(x^{3} -y^{3}\right) ,
1
−
x
6
+
1
−
y
6
=
a
(
x
3
−
y
3
)
,
then
y
2
d
y
d
x
=
y^{2} \frac{dy}{dx} =
y
2
d
x
d
y
=
AP EAMCET - 2019
AP EAMCET
Mathematics
General and Particular Solutions of a Differential Equation