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All questions of Chapter 9 - Differential Equations for Commerce Exam

Low Haemoglobin causes________.
  • a)
    Malaria
  • b)
    Anaemia
  • c)
    Cancer
  • d)
    Cholera
Correct answer is option 'B'. Can you explain this answer?

Priyanka Sharma answered
Hemoglobin, the substance that gives color to red blood cells, is the substance that allows for the transport of oxygen throughout the body. Low hemoglobin levels lead to anemia, which causes symptoms like fatigue and trouble breathing.
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The solution of the initial value problem  edy/dx = x + 1, y(0) = 3 is :
  • a)
    y = (x – 1)log(x + 1) + 3
  • b)
    y = (x + 1)log(x + 1) -x+3
  • c)
    y = (x + 1)log(x + 1) + 4
  • d)
    y = (x + 1)log(x + 1) – 1
Correct answer is option 'B'. Can you explain this answer?

Gunjan Lakhani answered
 edy/dx = x + 1
⇒ dy/dx = log(x+1)
⇒ dy = log(x+1)dx
Integrating both sides, we get
⇒ ∫dy =  ∫log(x+1)dx
y = log(x + 1)  ∫1dx - ∫[d/dx{log(x + 1)} ∫1dx]dx
y = xlog(x + 1) -  ∫1/(x+1)xdx
y = xlog(x + 1) -  ∫(1 - 1/(x + 1)dx
y = xlog(x + 1) -  ∫dx +  ∫1/(x+1)dx
y = xlog(x + 1) - x + log|x + 1| + c
y = (x + 1)log|x + 1| - x + c
Here c is 3
y = (x + 1)log|x + 1| - x + 3 

The differential equation,is a:
  • a)
    second order, third degree equation.
  • b)
    second order, second degree equation. 
  • c)
    third order, third degree equation.
  • d)
    second order, first degree equation
Correct answer is option 'D'. Can you explain this answer?

Divey Sethi answered
To calculate the degree or the order of a differential equation, the powers of derivatives should be an integer.
On squaring both sides, we get a differential equation with the integral power of derivatives.
⇒ Order (the highest derivative) = 2
⇒ Degree (the power of highest degree) = 1

The differential equation  is a:
  • a)
    Third order, third degree equation
  • b)
    Second order, second degree equation
  • c)
    Second order, first degree equation
  • d)
    Second order, third degree equation
Correct answer is option 'B'. Can you explain this answer?

Suresh Iyer answered
3*(d2y/dx2) = [1+(dy/dx)2]3/2
On squaring both side,
9*(d2y/dx2)2 = [1+(dy/dx)2]3
The order of the equation is 2. The power of the term determining the order determines the degree.
So, the degree is also 2.

The differential equation for the equation  is :
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Sushil Kumar answered
y = Acos(αx) + Bsin(αx)
dy/dx = -Aαsin(αx) + Bαcos(αx)
d2y/dx2 = -Aα2cos(αx) - Bα2sin(αx)
= -α2(Acos(αx) + Bsin(αx))
= -α2 * y
d2y/dx2 + α2*y = 0

The degree of the differential equation 
  • a)
    2
  • b)
    1
  • c)
    3
  • d)
    0
Correct answer is option 'C'. Can you explain this answer?

Sushil Kumar answered
Given equation is : 
(dy/dx)2 + 1/(dy/dx) = 1
((dy/dx)3 + 1)/(dy/dx) = 1
(dy/dx)3 +1 = dy/dx
So, final equation is
(dy/dx)3 - dy/dx + 1 = 0
So, degree = 3

The solution of the differential equation is :
  • a)
    2y = x2(log x + 1) + c
  • b)
  • c)
  • d)
    y = x2(log x + 1) + c
Correct answer is option 'B'. Can you explain this answer?

Poonam Reddy answered
dy/dx = x logx
=> ∫dy = ∫x logx dx
y = logx . x2/2 - ∫1/x . x2/2 dx + c
y = x2/2 log x -½ ∫x dx + c
y = x2/2 log x - ½ . x2/2 + c
y = x2/2 log x - x2/4 + c

The solution of the differential equation x dy = (2y + 2x4 + x2) dx is:​
  • a)
    y = x4 + x2 log |x2 + cx2
  • b)
    y = x3 + 2x + cx2
  • c)
    y = x3 + x log |x| + cx
  • d)
    y = x4 + x2 log |x| + cx2
Correct answer is option 'D'. Can you explain this answer?

Sai Kulkarni answered
xdy = (2y + 2x4 + x2)dx
→ dy/dx − (2x)y = 2x3 + x
This differential is of the form y′+P(x)y=Q(x) which is the general first order linear differential equation, where P(x) and Q(x) are continuous function defined on an interval.
The general solution for this is y∙I.F = ∫I.F × Q(x)dx
Where I.F = e∫P(x)dx is the integrating factor of the differential equation.
I.F = e∫P(x)dx
= e∫−2/xdx
= e(−2∙lnx)
= eln(x−2)
= x−2
Thus y(1/x2) = ∫1/x2(2x3 + x)dx
=∫(2x + 1/x)dx
= x2 + lnx + C
⟹ y = x4 + x2lnx + c

The integrating factor of differential equation is :
  • a)
    -tan x
  • b)
    sec2x
  • c)
    -sec2x
  • d)
    tan x
Correct answer is option 'B'. Can you explain this answer?

Vikas Kapoor answered
 dy/dx+2ytanx=sinx
This is in the form of dy/dx + py = θ
where p=2tanx,θ=sinx
∴ finding If e∫pdx = e∫2tanxdx 
=e(2log secx)
=e(log sec2x)
=sec2x

The solution of the differential equation is :
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Sushil Kumar answered
The given differential equation may be written as
dy/dx + (2x/(x2+1)y = √x2+4/(x2+1)  ... (i)
This is of the form dy/dx + Py = Q,
 where P=2x/(x2+1) and Q=(√x2+4)/(x2+1)
Thus, the given differential equation is linear.
IF=e(∫Pdx)
= e(∫2x(x2+1)dx)
= e(log(x2+1) = (x2+1)
So, the required solution is given by
y × IF = ∫{Q×IF}dx + C,
i.e., y(x2+1)=∫(√x2+4)/(x2+1)×(x2+1)dx
⇒y(x2+1)=∫(√x2+4)dx
=1/2x (√x2+4) +1/2 × (2)2 × log|x+(√x2+4)| + C
=1/2x (√x2+4) + 2log|x+(√x2+4) + C.
Hence, y(x2+1) = 1/2x(√x2+4) + 2log|x+√x2+4| + C is the required solution.

The solution of the differential equation is :
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Aryan Khanna answered
y/x cos y/x dx − (x/y sin y/x + cosy/x) dy = 0
⇒dy/dx = (y/x cos y/x)/(x/y sin y/x + cos y/x)
This is a homogeneous differential equation.
Putting y = vx and dy/dx = v+xdv/dx, we have v+xdv/dx = (v2cosv)/(sinv+vcosv)
⇒ xdv/dx = (v2cosv)/(sinv+vcosv) - v
⇒x dv/dx = (v2 cosv - v2 cosv - vsinv )/(sinv+vcosv) 
⇒ x dv/dx = - [v sinv/(sinv + vcosv)]
⇒ ∫[(sinv + vcosv)/v sinv]dv = ∫dx/x
Integrating both the sides, we get
∫(cot v + 1/v)dv = - ln(x) + c
ln(sin v) + ln(v) = -ln(x) + c
ln(sin(y/x) + ln(y/x) + ln(x) = c
ln(y/x sin y/x * x) + c
y sin(y/x) = c

Identify the type of differential equation
  • a)
    Variable separable
  • b)
    Homogeneous
  • c)
    Linear
  • d)
    Both Variable separable and Linear
Correct answer is option 'B'. Can you explain this answer?

Tejas Verma answered
(x−y)dy/dx = x + 2y
⇒ dy/dx = (x + 2y)/(x−y)
F (x,y) = (x + 2y)/(x−y)
F(Ax, Ay) = (Ax + 2Ay)/(Ax−Ay) 
= A(x + 2y)/A(x−y) 
= (x + 2y)/(x − y) = F (x,y)
Hence, the equation is homogenous.

The order of a differential equation representing a family of curves is same as:​
  • a)
    The number of arbitrary constants present in the equation.
  • b)
    The number of variables in the equation
  • c)
    The degree of the equation
  • d)
    The number of curves in the family
Correct answer is option 'A'. Can you explain this answer?

Palak Datta answered
The Order of Differential Equations
In the context of differential equations, the order refers to the highest derivative present in the equation. When dealing with a family of curves, the order of the differential equation is closely linked to the number of arbitrary constants it contains.
Understanding Arbitrary Constants
- Arbitrary constants arise when integrating a function, representing a family of solutions or curves.
- For instance, if we have a first-order differential equation, it typically yields one arbitrary constant, leading to a family of curves parameterized by that constant.
Connection to Differential Equations
- The order of the differential equation mirrors the number of arbitrary constants.
- Thus, a second-order differential equation will introduce two arbitrary constants, allowing for a broader family of curves.
Other Options Explained
- Number of Variables: The number of variables in the equation does not directly correlate with the order. A single variable can lead to various orders based on derivatives.
- Degree of the Equation: The degree refers to the power of the highest derivative and is distinct from the order, which is purely about the highest derivative itself.
- Number of Curves in the Family: The number of curves is influenced by the number of arbitrary constants but is not the same as the order.
Conclusion
The correct answer, option 'A', is justified by the direct relationship between the order of a differential equation and the number of arbitrary constants present. This relationship is fundamental in understanding the behavior and characteristics of families of curves represented by differential equations.

The solution of the differential equation is :
  • a)
    x3 + 10y3 = x2
  • b)
    x3 + y3 = 10x2
  • c)
    x3 + y2 = 10x3
  • d)
    x3 -10y3 = x2
Correct answer is option 'C'. Can you explain this answer?

Om Desai answered
dy/dx = (x2 + 3y2)/2xy………….(1)
Let y = vx
dy/dx = v + xdv/dx
Substitute the value of y and dy/dx in (1)
v + x dv/dx = (1+3v2)/2v
x dv/dx = (1+3v2)/2v - v
x dv/dx = (1 + 3v2 - 2v2)/2
x dv/dx = (1+ v2)/2v
2v/(1+v2) dv = dx/x…………(2)
Integrating both the sides
∫2v/(1+v2) dv = ∫dx/x
Put t = 1 + v2
dt = 2vdv
∫dt/t  = ∫dx/x
=> log|t| = log|x| + log|c|
=> log|t/x| = log|c|
t/x = +- c
(1+v2)/x = +-c
(1 + (y2)/(x2))/x = +-c
x2 + y2 = Cx3……….(3)
y(1) = 3
1 + 9 = c(1)3
c = 10
From eq(3), we get x
3
+ y2 = 10x3

The solution of the equation x3dx + (y + 1)2 dy = 0
  • a)
    12[x4 - (y +1)3] = c
  • b)
    12[x4 + (y +1)3] = c
  • c)
    3x4 + 4(y +1)3 = c
  • d)
    3x4 + 3(y +1)3 = c
Correct answer is option 'C'. Can you explain this answer?

Om Desai answered
x3dx + (y + 1)dy = 0
=> d(x4/4) + d((y + 1)3/3) = 0
=> d(x4/4 + ((y + 1)3/3)) = 0
=> (x4/4 + ((y + 1)3/3)) = constt
=> 12 *(x4/4 + ((y + 1)3/3)) = 12 * constt
=> 3x4 + 4(y + 1)3 = constt

Write the order and degree of the given differential equation:
​
  • a)
    2,3
  • b)
    1,1
  • c)
    2,1
  • d)
    3, 1
Correct answer is option 'D'. Can you explain this answer?

Gaurav Kumar answered
Order = 3 and Degree = 1
The order of a differential equation is determined by the highest-order derivative; the degree is determined by the highest power on a variable. The higher the order of the differential equation, the more arbitrary constants need to be added to the general solution.

The first order, first degree differential equation y’ = f(x,y)  is said to be homogeneous, if​
  • a)
    f(x , y) is a homogeneous function of degree zero.
  • b)
    f(x , y) is a homogeneous function of second degree.
  • c)
    f(x , y) is a homogeneous function of first degree.
  • d)
    f(x , y) is a homogeneous function of third degree.
Correct answer is option 'A'. Can you explain this answer?

Arshiya Mehta answered
The correct answer is option A, because a first-order, first-degree differential equation is said to be homogeneous if the function f(x,y) is a homogeneous function of degree zero.
A function f(x,y) is said to be homogeneous of degree n if it satisfies the equation:
f(λx, λy) = λ^n * f(x,y)
where λ is a scalar. This means that if we multiply both the independent and dependent variables by a constant λ, the value of the function is scaled by a factor of λ^n.
In the case of a first-order, first-degree differential equation, the function f(x,y) is a function of two variables: x and y. For this type of equation, f(x,y) is said to be homogeneous if it is a homogeneous function of degree zero, which means that it satisfies the equation:
f(λx, λy) = f(x,y)
In other words, if we multiply both the independent and dependent variables by a constant λ, the value of the function does not change. This means that f(x,y) does not depend on the magnitude of the variables x and y, but only on their relative values.
For example, if f(x,y) = xy, then f(λx, λy) = λxy = xy, so f(x,y) is a homogeneous function of degree zero. On the other hand, if f(x,y) = x^2 + y^2, then f(λx, λy) = λ^2x^2 + λ^2y^2 = x^2 + y^2, so f(x,y) is not a homogeneous function.
Therefore, the correct answer is option A, because a first-order, first-degree differential equation is said to be homogeneous if the function f(x,y) is a homogeneous function of degree zero.

Formation of the differential equation corresponding to the ellipse major axis 2a and minor axis 2b is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Geetika Shah answered
Equation of ellipse :
 x2/a2 + y2/b2 = 1
Differentiation by x,
2x/a2 + (dy/dx)*(2y/b2) = 0
dy/dx = -(b2/a2)(x/y)
-(b2/a^2) = (dy/dx)*(y/x) ----- eqn 1
Again differentiating by x,
d2y/dx2 = -(b2/a2)*((y-x(dy/dx))/y2)
Substituting value of -b2/a2 from eqn 1
d2y/dx2 = (dy/dx)*(y/x)*((y-x(dy/dx))/y2)
d2y/dx2 = (dy/dx)*((y-x*(dy/dx))/xy)
(xy)*(d2y/dx2) = y*(dy/dx) - x*(dy/dx)2
(xy)*(d2y/dx2) + x*(dy/dx)2- y*(dy/dx) = 0

A differential equation of the form y' = F(x,y) is homogeneous if
  • a)
    F(x,y) is a homogeneous function of degree one
  • b)
    F(x,y) is a homogeneous function of degree three
  • c)
    F(x,y) is a homogeneous function of degree two
  • d)
    F(x,y) is a homogeneous function of degree zero
Correct answer is option 'D'. Can you explain this answer?

Arka Chavan answered
Understanding Homogeneous Differential Equations
Homogeneous differential equations have specific characteristics that define them. The equation y = F(x, y) is considered homogeneous if the function F(x, y) exhibits a particular type of behavior concerning scaling.
Definition of Homogeneous Functions
A function F(x, y) is termed homogeneous of degree n if, for any scalar t:
- F(tx, ty) = t^n * F(x, y)
This means that if we scale both variables x and y by the same factor t, the function F is scaled by t raised to the power of n.
Homogeneity in Differential Equations
For the differential equation y = F(x, y) to be classified as homogeneous, the function F(x, y) must satisfy the following condition:
- F(x, y) is a homogeneous function of degree zero.
This implies:
- F(tx, ty) = F(x, y) for any scalar t.
Why Degree Zero is Key
- Degree Zero Function: When F(x, y) is homogeneous of degree zero, it signifies that the function does not change in value when both x and y are multiplied by the same non-zero scalar. This reflects the property that the relationship between x and y remains invariant under scaling.
- Implication: This invariance is crucial for the analysis of solutions to the differential equation, helping to identify unique solutions based on initial conditions.
Conclusion
In summary, for the differential equation y = F(x, y) to be homogeneous, F(x, y) must specifically be a homogeneous function of degree zero. This characteristic is essential in the context of solving differential equations and understanding their behavior under transformations.

The order and degree of the differential equation: (y”)2 + (y”)3 + (y’)4 + y5 = 0 is:​
  • a)
    2, 4
  • b)
    3, 5
  • c)
    2, 5
  • d)
    2, 3
Correct answer is option 'D'. Can you explain this answer?

Naina Bansal answered
Toolbox:
The highest order derivative present in the differential equation determines its order.
The highest power raise to the derivative determines its degree.

(y′′′)^2+(y′′)^3+(y′)^4+y^5 = 0
The order of the given equation is 3.
The degree corresponding to the highest order is 2

The order of the differential equation: 
  • a)
    4
  • b)
    2
  • c)
    3
  • d)
    1
Correct answer is option 'C'. Can you explain this answer?

Anuj Seth answered
Order is seen from highest efficiency of differentiability I.e. in hindi , kisi bhi equation me jisme sabse jyada differential power hai wo jeeta

In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (loge2 = 0.6931).
  • a)
    7.93%
  • b)
    8.93%
  • c)
    9.93%
  • d)
    6.93%
Correct answer is option 'D'. Can you explain this answer?

Anirban Desai answered
To solve this problem, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:
A is the final amount
P is the principal amount (initial amount)
r is the annual interest rate (as a decimal)
n is the number of times the interest is compounded per year
t is the number of years

In this case, the principal doubles itself in 10 years. So, the final amount A is 2 times the principal P.

2P = P(1 + r/n)^(nt)

Since we are given that the principal increases continuously, we can take the limit as n approaches infinity. This simplifies the formula to:

2 = e^(rt)

Where e is the base of natural logarithms and is approximately equal to 2.71828.

Now, we can take the natural logarithm of both sides to solve for r:

ln(2) = rt

Dividing both sides by t:

r = ln(2)/t

Given that t = 10 years, we can substitute this value into the equation:

r = ln(2)/10

Using the given value of loge2 = 0.6931, we can calculate r:

r = 0.6931/10 = 0.06931 = 6.93%

Therefore, the value of r is 6.93%.

So, the correct answer is option D.

Differential equation of the family of ellipses having foci on y-axis and centre at origin is
  • a)
    xy′′ + x(y′)2  yy′ = 0
  • b)
    xyy′′ + x(y′)2  yy′ = 0
  • c)
    yy′′ + x(y′)2  yy′ = 0
  • d)
    xyy′′ −x(y′)2 +yy′ = 0
Correct answer is option 'B'. Can you explain this answer?

Palak Banerjee answered
The differential equation of the family of ellipses having foci on the y-axis and center at the origin can be found as follows:

Let's consider an ellipse with foci (0, c) and (0, -c), where c is a positive constant. The distance between the foci is 2c.

The general equation of an ellipse centered at the origin is given by:

x^2/a^2 + y^2/b^2 = 1,

where a and b are positive constants representing the semi-major and semi-minor axes, respectively.

Since the foci lie on the y-axis, the equation of the ellipse becomes:

x^2/a^2 + (y - c)^2/b^2 = 1.

We know that the distance between the foci is 2c, so we have:

2c = 2b^2/a.

Simplifying this equation, we get:

b^2 = ac.

Now, differentiating both sides of the equation with respect to x, we have:

2b(b') = a'c + ac',

where b' and a' represent the derivatives of b and a with respect to x, respectively.

Since the derivatives of a and b are unknown, we cannot solve for the differential equation in terms of x and y directly. However, we can eliminate a' and b' by using the relationship b^2 = ac, which gives:

2b(b') = 2b(c'/b) + ac'.

Simplifying this equation, we find:

b' = c'/b + ac'/(2b^2).

Rearranging terms, we get:

b'(2b^2) = c' + ac'.

Substituting b^2 = ac, we obtain:

b'(2b^2) = c' + b^2c'.

Finally, dividing both sides by 2b^2, we get the differential equation:

b' = (c' + b^2c')/(2b^2).

Therefore, the differential equation of the family of ellipses having foci on the y-axis and center at the origin is:

b' = (c' + b^2c')/(2b^2).

Find the differential equation of all the straight lines touching the circle x2 + y2 = r2.
  • a)
    r2 (1+(dy/dx))2
  • b)
    3r2 (1+(dx/dy))2
  • c)
    2r2 (1+(dy/dx))2
  • d)
    r2 (1+(dx/dy))2
Correct answer is option 'A'. Can you explain this answer?

Defence Exams answered
  1. Let y = mx + c be the equation of all the straight lines touching the circle.
    Given : The equation of the circle is x2 + y2 = r2----------> (1)
    The tangent to the circle is c2 = r2(1+m2)
    c = r√(1+m2)
    we know that y = mx + c---------->(2)
    y = mx + r√(1+m2) ---------->(3)
    y - mx = r√(1+m2)
    Differentiating wrt x we get dy/dx -m =0
    dy/dx = m
    Substituting this in equation (3)
    y - (dy/dx . x) = r√(1+(dy/dx)2)
    Squaring on both sides, we get
    y2 - (dy/dx . x)2 = [ r√(1+(dy/dx)2)]2
    [y - x(dy/dx)]2 = r2 (1+(dy/dx))2 is the required differential equation.
    Answer: The differential equation of all the straight lines touching the circle x2 + y2 = r2 is [y - x(dy/dx)]2 = r2 (1+(dy/dx))2​​​​​​
 
 

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