Important Questions For Class 12 Maths Chapter 9 Differential Equations helps in preparing for CBSE Board exams. All the questions are prepared by our Maths experts. Apart from studying and practising problems on the Differential equation from the NCERT book, students shall also practice these important questions.
Class 12 Maths Chapter 9 contains many formulas and methods for solving given differential equations. Students can go through with the important questions for all Maths Class 12 chapters here to revise all the chapters and score good marks in the CBSE final examination.
Important Questions & Answers For Class 12 Maths Chapter 9 Differential Equations
Q. No.1: Determine order and degree (if defined) of differential equation (y′′′)2 + (y″)3 + (y′)4 + y5 = 0
Solution:
Given differential equation is (y′′′)2 + (y″)3 + (y′)4 + y5 = 0
The highest order derivative present in the differential equation is y′′′.
Therefore, its order is 3.
The given differential equation is a polynomial equation in y′′′, y′′, and y′.
The highest power raised to y′′′ is 2.
Hence, its degree is 2.
Q. No. 2: Verify that the function y = a cos x + b sin x, where, a, b ∈ R is a solution of the differential equation d2y/dx2 + y=0.
Solution:
The given function is y = a cos x + b sin x … (1)
Differentiating both sides of equation (1) with respect to x,
dy/dx = – a sinx + b cos x
d2y/dx2 = – a cos x – b sinx
LHS = d2y/dx2 + y
= – a cos x – b sinx + a cos x + b sin x
= 0
= RHS
Hence, the given function is a solution to the given differential equation.
Q. No. 3: The number of arbitrary constants in the general solution of a differential equation of fourth order is:
(A) 0 (B) 2 (C) 3 (D) 4
Solution:
We know that the number of constants in the general solution of a differential equation of order n is equal to its order.
Therefore, the number of constants in the general equation of the fourth-order differential equation is four.
Hence, the correct answer is D.
Note: The number of constants in the general solution of a differential equation of order n is equal to zero.
Q. No. 4: Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constants.
Solution:
Given,
y = a sin (x + b) … (1)
Differentiating both sides of equation (1) with respect to x,
dy/dx = a cos (x + b) … (2)
Differentiating again on both sides with respect to x,
d2y/dx2 = – a sin (x + b) … (3)
Eliminating a and b from equations (1), (2) and (3),
d2y/ dx2 + y = 0 … (4)
The above equation is free from the arbitrary constants a and b.
This the required differential equation.
Q. No. 5: Find the differential equation of the family of lines through the origin.
Solution:
Let y = mx be the family of lines through the origin.
Therefore, dy/dx = m
Eliminating m, (substituting m = y/x)
y = (dy/dx) . x
or
x. dy/dx – y = 0
Q. No. 6: Form the differential equation of the family of circles having a centre on y-axis and radius 3 units.
Solution:
The general equation of the family of circles having a centre on the y-axis is x2 + (y – b)2 = r2
Given the radius of the circle is 3 units.
The differential; equation of the family of circles having a centre on the y-axis and radius 3 units is as below:
x2 + (y – b)2 = 32
x2 + (y – b)2 = 9 ……(i)
Differentiating (i) with respect to x,
2x + 2(y – b).y′ = 0
⇒ (y – b). y′ = -x
⇒ (y – b) = -x/y′ …….(ii)
Substituting (ii) in (i),
x2 + (-x/y′)2 = 9
⇒ x2[1 + 1/(y′)2] = 9
⇒ x2 [(y′)2 + 1) = 9 (y′)2
⇒ (x2 – 9) (y′)2 + x2 = 0
Hence, this is the required differential equation.
Q. No. 7: Find the general solution of the differential equation dy/dx =1+y2/1+x2.
Solution:
Given differential equation is dy/dx =1+y2/1+x2
Since 1 + y2 ≠0, therefore by separating the variables, the given differential equation can be written as:
dy/1+y2 = dx/1+x2 …….(i)
Integrating equation (i) on both sides,
tan-1y = tan-1x + C
This is the general solution of the given differential equation.
Q. No. 8: For each of the given differential equation, find a particular solution satisfying the given condition:
dy/dx = y tan x ; y = 1 when x = 0
Solution:
dy/dx = y tan x
dy/y = tan x dx
Integrating on both sides,
log y = log (sec x) +C
log y = log (C sec x)
⇒ y = C sec x ……..(i)
Now consider y = 1 when x= 0.
1 = C sec 0
1 = C (1)
C = 1
Substituting C = 1 in (i)
y = sec x
Hence, this is the required particular solution of the given differential equation.
Q. No. 9: Find the equation of a curve passing through (1, Ï€/4) if the slope of the tangent to the curve at any point P (x, y) is y/x – cos2(y/x).
Solution:
According to the given condition,
dy/dx = y/x – cos2(y/x) ………….(i)
This is a homogeneous differential equation.
Substituting y = vx in (i),
v + (x) dv/dx = v – cos2v
⇒ (x)dv/dx = – cos2v
⇒ sec2v dv = – dx/x
By integrating on both the sides,
⇒ ∫sec2v dv = – ∫dx/x
⇒ tan v = – log x + c
⇒ tan (y/x) + log x = c ……….(ii)
Substituting x = 1 and y = π/4,
⇒ tan (π/4) + log 1 = c
⇒ 1 + 0 = c
⇒ c = 1
Substituting c = 1 in (ii),
tan (y/x) + log x = 1
Q. No. 10: Integrating factor of the differential equation (1 – x2)dy/dx – xy = 1 is
(A) -x
(B) x/ (1 + x2)
(C) √(1- x2)
(D) ½ log (1 – x2)
Solution:
Given differential equation is (1 – x2)dy/dx – xy = 1
(1 – x2)dy/dx = 1 + xy
dy/dx = (1/1 – x2) + (x/1 – x2)y
dy/dx – (x/1 – x2)y = 1/1-x2
This is of the form dy/dx + Py = Q
We can get the integrating factor as below:
Let 1 – x2 = t
Differentiating with respect to x
-2x dx = dt
-x dx = dt/2
Now,
I.F = √t = √(1- x2)
Hence, option C is the correct answer.
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Practice Questions For Class 12 Maths Chapter 9 Differential Equations
- Solve dy/dx + y = 1.
- Solve (x +1) dy/dx = 2xy
- In a bank principal increases at the rate of 6% per year. In how many years Rs 2500 double itself.
- Solve: dy/dx = cos(x+y)
- Solve the initial value problem: cos(a+b)dy =dx, y(0) = 0.
- Find the equation of a curve passing through (2, 1) if the slope of the tangent to the curve at any point (x, y) is (x2 + y2)/2xy.
- In culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?
- At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).
- Solve the differential equation:
- Given that dy/dx = e-2y and y = 0 when x = 5. Find the value of x when y = 3.
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