Mathematics: CUET Mock Test - 4 - CUET MCQ

# Mathematics: CUET Mock Test - 4 - CUET MCQ

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## 30 Questions MCQ Test CUET Mock Test Series - Mathematics: CUET Mock Test - 4

Mathematics: CUET Mock Test - 4 for CUET 2024 is part of CUET Mock Test Series preparation. The Mathematics: CUET Mock Test - 4 questions and answers have been prepared according to the CUET exam syllabus.The Mathematics: CUET Mock Test - 4 MCQs are made for CUET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Mathematics: CUET Mock Test - 4 below.
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Mathematics: CUET Mock Test - 4 - Question 1

### Which one of the following is correct if we differentiate the equation xy = aex + be-x two times?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 1

We have xy = aex + be-x ……(1)
Differentiating (1) with respect to x, we get
x(dy/dx) + y = aex + be-x …..(2)
Differentiating (2) now, with respect to x, we get
x(d2y/ dx) + dy/dx + dy/dx = aex + be-x
From (1),
aex + be-x = xy, so that we get
x(d2y/ dx) + 2(dy/dx) = xy
which is the required differential equation.

Mathematics: CUET Mock Test - 4 - Question 2

### What is the differential equation of all parabolas whose directrices are parallel to the x-axis?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 2

The equation of family of parabolas is Ax2 + Bx + C = 0 where, A, B, C are arbitrary constant.
By differentiating the equation with respect to x till all the constants get eliminated,
Hence, d3y/dx3 = 0

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Mathematics: CUET Mock Test - 4 - Question 3

### What will be the required solution of d2y/dx2 – 3dy/dx + 4y = 0?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 3

d2y/dx2 – 3dy/dx + 4y = 0 …..(1)
Let, y = emx be a trial solution of (1), then,
⇒ dy/dx = memx and d2y/dx2 = m2emx
Clearly, y = emx will satisfy equation (1). Hence, we have,
m2emx – 3m * emx – 4emx = 0
⇒ m2 – 3m – 4 = 0 (as emx ≠ 0) …….(2)
⇒ m2 – 4m + m – 4 = 0
⇒ m(m – 4) + 1(m – 4) = 0
Or, (m – 4)(m + 1) = 0
Thus, m = 4 or m = -1
Clearly, the roots of the auxiliary equation (2) are real and unequal.
Therefore, the required general solution of (1) is
y = Ae4x + Be-x where A and B are constants.

Mathematics: CUET Mock Test - 4 - Question 4

What is the solution of the given equation (D + 1)2y = 0 given y = 2 loge 2 when x = loge 2 and y = (4/3) loge3 when x = loge3?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 4

(D + 1)2y = 0
Or, (D2 + 2D+ 1)y = 0
⇒ d2y/dx2 + 2dy/dx + y = 0 ……….(1)
Let y = emx be a trial solution of equation (1). Then,
⇒ dy/dx = memx and d2y/dx2 = m2emx
Clearly, y = emx will satisfy equation (1). Hence, we have
⇒ m2.emx + 2m.emx + emx = 0
Or, m2 + 2m +1 = 0 (as, emx ≠ 0) ………..(2)
Or, (m + 1)2 = 0
⇒ m = -1, -1
So, the roots of the auxiliary equation (2) are real and equal. Therefore, the general solution of equation (1) is
y = (A + Bx)e-x where A and B are two independent arbitrary constants ……….(3)
Given, y = 2 loge 2 when x = loge 2
Therefore, from (3) we get,
2 loge 2 = (A + B loge2)e-x
Or, 1/2(A + B loge2) = 2 log e2
Or, A + B loge2 = 4 loge2 ……….(4)
Again y = (4/3) loge3 when x = loge3
So, from (3) we get,
4/3 loge3 = (A + Bloge3)
Or, A + Bloge3 = 4loge3 ……….(5)
Now, (5) – (4) gives,
B(loge3 – loge2) = 4(loge3 – loge2)
⇒ B = 4
Putting B = 4 in (4) we get, A = 0
Thus the required solution of (1) is y = 4xe-x

Mathematics: CUET Mock Test - 4 - Question 5

If, A and B are arbitrary constants then what will be the differential equation of y = Ax + B/x?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 5

Given, y = Ax + B/x
⇒ xy = Ax2 + B ……….(1)
Differentiating (1) with respect to x, we get,
d(xy)/dx = d/dx(Ax2 + B)
or, xdy/dx + y = A * 2x ……….(2)
Differentiating again with respect to x, we get,
x*d2y/dx2 + dy/dx + dy/dx = A*2 ……….(3)
Eliminating A from (2) and (3) we get,
x2 d2 y/dx2 + 2xdy/dx = 2Ax  [multiplying (3) by x]
or, x2 d2 y/dx2 + 2xdy/dx = xdy/dx + y [using (2)]
or, x2 d2 y/dx2 + xdy/dx – y = 0

Mathematics: CUET Mock Test - 4 - Question 6

What will be the value of C if C the constant of the coefficient of the solution of the given equation (D + 1)2y = 0 given y = 2 loge 2 when x = loge 2 and y = (4/3) loge3 when x = loge3?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 6

(D + 1)2y = 0
Or, (D2 + 2D+ 1)y = 0
⇒ d2y/dx2 + 2dy/dx + y = 0 ……….(1)
Let y = emx be a trial solution of equation (1). Then,
⇒ dy/dx = memx and d2y/dx2 = m2emx
Clearly, y = emx will satisfy equation (1). Hence, we have
⇒ m2.emx + 2m.emx + emx = 0
Or, m2 + 2m + 1 = 0 (as, emx ≠ 0) ………..(2)
Or, (m + 1)2 = 0
⇒ m = -1, -1
So, the roots of the auxiliary equation (2) are real and equal. Therefore, the general solution of equation (1) is
y = (A + Bx)e-x where A and B are two independent arbitrary constants ……….(3)
Given, y = 2 loge 2 when x = loge 2
Therefore, from (3) we get,
2 loge 2 = (A + B loge2)e-x
Or, 1/2(A + B loge2) = 2 log e2
Or, A + B loge2 = 4 loge2 ……….(4)
Again y = (4/3) loge3 when x = loge3
So, from (3) we get,
4/3 loge3 = (A + Bloge3)
Or, A + Bloge3 = 4loge3 ……….(5)
Now, (5) – (4) gives,
B(loge3 – loge2) = 4(loge3 – loge2)
⇒ B = 4
Putting B = 4 in (4) we get, A = 0
Thus the required solution of (1) is y = 4xe-x
So, C = 4

Mathematics: CUET Mock Test - 4 - Question 7

What is thedifferential equation whose solution represents the family y = ae3x + bex?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 7

The original given equation is,
y = ae3x + bex ……….(1)
Differentiating the above equation, we get
dy/dx = 3ae3x + bex ……….(2)
d2y/dx2 = 9ae3x + bex ……….(3)
Now, to obtain the final equation we have to make the RHS = 0,
So, to get RHS = 0, we multiply, (1) with 3 and (2) with -4,
Thus , 3y = 3ae3x + 3bex ……….(4)
And -4(dy/dx) = -12ae3x – 4 bex ……….(5)
Now, adding the above three equations,(i.e. (3) + (4) + (5)) we get,
d2y/dx2 – 4dy/dx + 3y = 0

Mathematics: CUET Mock Test - 4 - Question 8

If y = t(x) be a differentiable function ᵾ x € R, then which of the following is always true?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 8

(dy/dx)= (dx/dy)-1
So, d2y/dx2 = – (dx/dy)-2 d/dx(dx/dy)
= – (dy/dx)2(d2x/dy2)(dy/dx)
⇒ d2y/dx2 + (dy/dx)3 d2x/dy2 = 0

Mathematics: CUET Mock Test - 4 - Question 9

What will be the general solution of the differential equation d2y/dx2 = e2x(12 cos3x – 5 sin3x)? (here, A and B are integration constant)

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 9

Given, d2y/dx2 = e2x(12 cos3x – 5 sin3x) ………(1)
Integrating (1) we get,
dy/dx = 12∫ e2xcos3x dx – 5∫ e2x sin3x dx
= 12 * (e2x/22 + 32)[2cos3x + 3sin3x] – 5 * (e2x/22 + 32)[2sin3x – 3cos3x] + A (A is integrationconstant)
So dy/dx = e2x/13 [24 cos3x + 36 sin3x – 10 sin3x + 15 cos3x] + A
= e2x/13(39 cos3x + 26 sin3x) + A
⇒ dy/dx = e2x(3 cos3x + 2 sin3x) + A ……….(2)
Again integrating (2) we get,
y = 3*∫ e2x cos3xdx + 2∫ e2x sin3xdx + A ∫dx
y = 3*(e2x/22 + 32)[2cos3x + 3sin3x] + 2*(e2x/22 + 32)[2sin3x – 3cos3x] + Ax + B (B is integration constant)
y = e2x/13(6 cos3x + 9 sin3x + 4 sin3x – 6 cos3x) + Ax + B
or, y = e2x sin3x + Ax + B

Mathematics: CUET Mock Test - 4 - Question 10

Which of the following is the valid differential equation x = a cos(αt + β)?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 10

Since, x = a cos(αt + β)
Therefore, dx/dt = a cos(αt + β)
And, d2x/dt2 = -a α2 cos(αt + β)
= -α2 a cos(αt + β)
Or, d2x/dt2 = -α2 [as a cos(αt + β) = x]
So, d2x/dt2 + α2x = 0

Mathematics: CUET Mock Test - 4 - Question 11

If, A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant length k. What kind of curve is passing through (0, k)?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 11

Equation of the normal at a point P(x, y) is given by
Y – y = -1/(dy/dx)(X – x) ….(1)
Let the point Q at the x-axis be (x1 , 0).
From (1), we get
y(dy/dx) = x1 – x ….(2)
Now, giving that PQ2 = k2
Or, x1 – x + y2 = k2
=>y(dy/dx) = ± √(k2 – y2) ….(3)
(3) is the required differential equation for such curves,
Now solving (3) we get,
∫-dy/√(k2 – y2) = ∫-dx
Or, x2 + y2 = k2 passes through (0, k)
Thus, it is a circle.

Mathematics: CUET Mock Test - 4 - Question 12

What is the solution of (y(dy/dx) + 2x)2 = (y2 + 2x2)[1 + (dy/dx)2]?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 12

Here, y2(dy/dx)2 + 4x2 + 4xy(dy/dx) = (y2 + 2x2)[1 + (dy/dx)2]
⇒ dy/dx = y/x ± √(1/2(y/x)2) + 1
Let, y = vx
⇒ v + x dv/dx = v ± √(1/2(v)2) + 1
Integrating both sides,
±√dv/(√(1/2(v)2) + 1) = ∫dx/x
cx±1/√2 = y/x + √(y2 + 2x2)/x2

Mathematics: CUET Mock Test - 4 - Question 13

A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t2 – 5t)cm/sec2. What will be the velocity of the particle?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 13

Let x cm be the distance of the moving particle from the origin and v cm/sec be its velocity at the end of t seconds. Then the acceleration of the particle at time t seconds = dv/dt and its velocity at that time = v = dx/dt.
By question, dv/dt = 3t2 – 5t
Or, dv = 3t2 dt – 5tdt
Or, ∫dv = 3∫t2 dt – 5∫t dt
Or, v = t3 – (5/2)t2 + c ……….(1)
Given, v = 5, when t = 0; hence putting these values in equation (1) we get, c = 5
Thus v = t3 – (5/2)t2 + 5
Or, dx/dt = t3 – (5/2)t2 + 5 ………..(2)
Thus, the velocity of the particle at the end of 4 seconds,
= [v]t = 4 = (43 – (5/2)42 + 5 ) cm/sec [putting t = 4 in (2)]
= 29 cm/sec

Mathematics: CUET Mock Test - 4 - Question 14

What will be the differential equation form of √(a2 + x2)dy/dx + y = √(a2 + x2) – x?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 14

The given form of equation can be written as,
dy/dx + 1/√(a2 + x2) * y = (√(a2 + x2) – x)/√(a2 + x2) ……(1)
We have, ∫1/√(a2 + x2)dx = log(x + √(a2 + x2))
Therefore, integrating factor is,
e∫1/√(a2 + x2) = elog(x + √(a2 + x2))
= x + √(a2 + x2)
Therefore, multiplying both sides of (1) by x + √(a2 + x2) we get,
x + √(a2 + x2dy/dx + (x + √(a2 + x2))/ √(a2 + x2)*y = (x + √(a2 + x2))(√(a2 + x2) – x)/√(a2 + x2)
or, d/dx[x + √(a2 + x2)*y] = (a2 + x2) ………..(2)
Integrating both sides of (2) we get,
(x + √(a2 + x2) * y = a2∫dx/√(a2 + x2)
= a2 log (x + √(a2 + x2)) + c

Mathematics: CUET Mock Test - 4 - Question 15

A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What will be the equation of the curve?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 15

The equation of tangent to the curve y = f(x), at point (x, y), is
Y – y = dy/dx * (X – x) …..(1)
Where it meets the x axis, Y = 0 and X = (x – y/(dy/dx))
Where it meets the y axis, X = 0 and Y = (y – x/(dy/dx))
Also, the area of the triangle formed by (1) with the coordinate axes is 2, so that,
(x – y/(dy/dx))* (y – x/(dy/dx)) = 4
Or, (y – x/(dy/dx))2 – 4dy/dx = 0
Or, x2(dy/dx)2 – 2(xy – 2)dy/dx + y2 = 0
Solving for dy/dx we get,
dy/dx = [(xy – 2) ± √(1 – xy)]/ x2
Let, 1 – xy = t2
⇒ x(dy/dx) + y = -2t(dt/dx)
⇒ x2(dy/dx) = t2 – 1 – 2tx(dt/dx), so that (3) gives
t(x(dt/dx) – (t ± 1)) = 0
Hence, either t = 0
⇒ xy = 1 which is satisfied by (1, 1)
Or, x dt/dx = t ± 1
⇒ dx/x = dt/t ± 1
⇒ t ± 1 = cx
For x = 1, y = 1 and t = 0
⇒ c = ± 1, so the solution is
t = ± (x – 1) => t2 = (x – 1)2
Or, 1 – xy = x2 – 2x + 1
Or, x + y = 2
Thus, the two curves that satisfies are xy = 1 and x + y = 2

Mathematics: CUET Mock Test - 4 - Question 16

A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What is the differential equation?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 16

The equation of tangent to the curve y = f(x), at point (x, y), is
Y – y = dy/dx * (X – x) …..(1)
Where it meets the x axis, Y = 0 and X = (x – y/(dy/dx))
Where it meets the y axis, X = 0 and Y = (y – x/(dy/dx))
Also, the area of the triangle formed by (1) with the coordinate axes is 2, so that,
(x – y/(dy/dx))* (y – x/(dy/dx)) = 4
Or, (y – x/(dy/dx))2 – 4dy/dx = 0
Or, x2(dy/dx)2 – 2(xy – 2)dy/dx + y2 = 0
Solving for dy/dx we get,
dy/dx = [(xy – 2) ± √(1 – xy)]/ x2

Mathematics: CUET Mock Test - 4 - Question 17

What will be the value of dy/dx – a/x * y = (x + 1)/x?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 17

dy/dx – a/x * y = (x + 1)/x …….(1)
Multiplying both sides of equation (1) by
e∫-a/xdx
= e-a log x
= elog x-a
= x-a
We get, x-ady/dx – x-a (a/x)y = x-a (x + 1)/x
Or, d/dx(y . x-a) = x-a + x-a – 1 …….(2)
Integrating both sides of (2) we get,
y. x-a = x-a + 1/(-a + 1) + x-a – 1 + 1/(-a -1 + 1) + c
= x-a.x/(1 – a) + x-a/-a + c
Or, y = x/(1 – a) – 1/a + cxa

Mathematics: CUET Mock Test - 4 - Question 18

What is the solution of dy/dx = (6x + 9y – 7)/(2x + 3y – 6)?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 18

dy/dx = (6x + 9y – 7)/(2x + 3y – 6)
So, dy/dx = (3(2x + 3y) – 7)/(2x + 3x – 6) ……….(1)
Now, we put, 2x + 3y = z
Therefore, 2 + 3dy/dx = dz/dx [differentiating with respect to x]
Or, dy/dx = 1/3(dz/dx – 2)
Therefore, from (1) we get,
1/3(dz/dx – 2) = (3z – 7)/(z – 6)
Or, dz/dx = 2 + (3(3z – 7))/(z – 6)
= 11(z – 3)/(z – 6)
Or, (z – 6)/(z – 3) dz = 11 dx
Or, ∫(z – 6)/(z – 3) dz = ∫11 dx
Or, ∫(1 – 3/(z – 3)) dz = 11x + c
Or, z – log |z – 3| = 11x + c
Or, 2x + 3y – 11x – 3log|2x + 3y -3| = c
Or, 3y – 9x – 3log|2x + 3y – 3| = c
Or, 3x – y + log|2x + 3y – 3| = -c/3

Mathematics: CUET Mock Test - 4 - Question 19

A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t2 – 5t)cm/sec2. What will be the distance from the origin at the end of 4 seconds?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 19

Let x cm be the distance of the moving particle from the origin and v cm/sec be its velocity at the end of t seconds. Then the acceleration of the particle at time t seconds = dv/dt and its velocity at that time = v = dx/dt.
By question, dv/dt = 3t2 – 5t
Or, dv = 3t2 dt – 5t dt
Or, ∫dv = 3∫t2 dt – 5∫t dt
Or, v = t3 – (5/2)t2 + c ……….(1)
Given, v = 5, when t = 0; hence putting these values in equation (1) we get, c = 5
Thus v = t3 – (5/2)t2 + 5
Or, dx/dt = t3 – (5/2)t2 + 5 ………..(2)
Or, dx = t3 dt – (5/2)t2 dt + 5 dt
Integrating this we get,
x = (1/4)t4 – (5/2)t3/3 + 5t + k ……….(3)
By the problem, x = 0, when t = 0; hence, from (3) we get, k = 0.
Thus, x = (1/4)t4 – (5/6)t3 + 5t ……….(4)
Thus, the velocity of the particle at the end of 4 seconds,
= [x]t = 4 = (1/4)44 – (5/6)43 + 5(4) [putting t = 4 in (4)]
= 30(2/3) cm

Mathematics: CUET Mock Test - 4 - Question 20

What is the equation of the curve passing through (1, 0) of (y(dy/dx) + 2x)2 = (y2 + 2x2)[1 + (dy/dx)2]?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 20

Here, y2(dy/dx)2 + 4x2 + 4xy(dy/dx) = (y2 + 2x2)[1 + (dy/dx)2]
⇒ dy/dx = y/x ± √(1/2(y/x)2) + 1
Let, y = vx
⇒ v + x dv/dx = v ± √(1/2(v)2) + 1
Integrating both sides,
±∫dv/(√(1/2(v)2) + 1) = ∫dx/x
cx±1/√2 = y/x + √(y2 + 2x2)/x2 (put v/√2 = tan t)
putting x = 1, y = 0, we get c = √2
So, the curve is given by,
√2x±1/√2 = y/x + √(y2 + 2x2)/x2

Mathematics: CUET Mock Test - 4 - Question 21

Find the general solution of the differential equation (x, y≠3).

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 21

Given that,
Separating the variables, we get

log⁡(y - 3) = log⁡(x - 3) + log⁡C1
log⁡(y - 3) - log⁡(x - 3) = log⁡C1
= log⁡C1
= 0
y - 3 = 0 is the general solution for the given differential equation.

Mathematics: CUET Mock Test - 4 - Question 22

Find the general solution of the differential equation .

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 22

Given that,

Separating the variables, we get
2 cos⁡y dy = 3 sin⁡x dx
Integrating both sides, we get
∫ 2 cos⁡y dy = ∫ 3 sin⁡x dx
2 sin⁡y = 3(-cos⁡x) + C
3 cos⁡x + 2 sin⁡y = C

Mathematics: CUET Mock Test - 4 - Question 23

Find the general solution of the differential equation

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 23

Given that,
Separating the variables, we get
dy = (3e+ 2)dx
Integrating both sides, we get
-----(1)
y = 3e+ 2x + C which is the general solution of the given differential equation.

Mathematics: CUET Mock Test - 4 - Question 24

Find the particular solution of the differential equation  given that y = 1/3 when x = 1.

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 24

Given that,

Separating the variables, we get
dy = (4x - 2)2 dx
Integrating both sides, we get

Given that, y = 1/3 when x = 1
Therefore, equation (1) becomes,

Hence, the particular solution for the given differential solution is y =

Mathematics: CUET Mock Test - 4 - Question 25

Find the particular solution of the differential equation .

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 25

Separating the variables, we get

Integrating both sides, we get

First, for integrating logy/y
Let log⁡y = t
Differentiating w.r.t y, we get
1/y dy=dt
∴

Hence, equation (1), becomes

Given y = 2, we get x = 2
Substituting the values in equation (2), we get

C = 0
Therefore, the equation becomes

∴ (log⁡y)2  - (log⁡x)2 = 0.

Mathematics: CUET Mock Test - 4 - Question 26

Find the particular solution for the differential equation given that, y = 1 when x = 1.

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 26

Given that,
Separating the variables, we get
7y dy = 3x2 dx
Integrating both sides, we get

Given that y = 1, when x = 1
Substituting the values in equation (1), we get

Hence, the particular solution of the given differential equation is:

⇒ 7y= 2x+ 5

Mathematics: CUET Mock Test - 4 - Question 27

Which of the following functions is the solution of the differential equation dydx + 2y = 0?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 27

Consider the function y = e-2x
Differentiating both sides w.r.t x, we get

dy/dx = -2y
⇒

Mathematics: CUET Mock Test - 4 - Question 28

Which of the following functions is a solution for the differential equation xy’ - y = 0?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 28

Consider the function y = 2x
Differentiating w.r.t x, we get
y’= dy/dx = 2
Substituting in the equation xy’-y, we get
xy’- y = x(2) - 2x = 2x - 2x = 0
Therefore, the function y = 2x is a solution for the differential equation xy’ - y = 0.

Mathematics: CUET Mock Test - 4 - Question 29

Which of the following functions is a solution for the differential equation y” + 6y = 0?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 29

Consider the function y = 5 cos⁡3x
Differentiating w.r.t x, we get
y’ = dy/dx = -15 sin⁡3x
Differentiating again w.r.t x, we get
y” = d2y/dx2 = -30 cos⁡3x
⇒ y” + 6y=0.
Hence, the function y = 5 cos⁡3x is a solution for the differential equation y” + 6y = 0.

Mathematics: CUET Mock Test - 4 - Question 30

Which of the following differential equations given below has the solution y = log⁡x?

Detailed Solution for Mathematics: CUET Mock Test - 4 - Question 30

Consider the function y = log⁡x
Differentiating w.r.t x, we get

Differentiating (1) w.r.t x, we get

∴

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