
Differential-Equations
DIFFERENTIAL EQUATIONS
Multiple Choice questions-
(a) 3
(b) 2
(c) 1
(d) not defined.
2. The order of the differential equation:
(a) 2
(b) 1
(c) 0
(d) not defined.
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.
4. The number of arbitrary constants in the particular solution of a differential equation of third order is:
(a) 3
(b) 2
(c) 1
(d) 0.
5. Which of the following differential equations has y = c1 ex + c2 e-x as the general solution?
(a) d2ydx2 + y = 0
(b) d2ydx2 – y = 0
(c) d2ydx2 + 1 = 0
(d) d2ydx2 – 1 = 0
6. Which of the following differential equations has y = x as one of its particular solutions?
(a) d2ydx2 – x² dydx + xy = x
(b) d2ydx2 + x dydx + xy = x
(c) d2ydx2 – x² dydx + xy = 0
(d) d2ydx2 + x dydx + xy = 0
7. The general solution of the differential equation dydx = ex+y is
(a) ex + e-y = c
(b) ex + ey = c
(c) e-x + ey = c
(d) e-x + e-y = c.
8. Which of the following differential equations cannot be solved, using variable separable method?
(a) dydx + ex+y + e-x+y
(b) (y² – 2xy) dx = (x² – 2xy) dy
(c) xydydx = 1 + x + y + xy
(d) dydx + y = 2.
9. A homogeneous differential equation of the form dydx = h(xy) can be solved by making the substitution.
(a) y = vx
(b) v = yx
(c) x = vy
(d) x = v
10. Which of the following is a homogeneous differential equation?
(a) (4x + 6y + 5)dy – (3y + 2x + 4)dx = 0
(b) xy dx – (x³ + y²)dy = Q
(c) (x³ + 2y²) dx + 2xy dy = 0
(d) y² dx + (x² – xy – y²)dy = 0.
Very Short Questions:
dydx = ex+y (C.B.S.E. Sample Paper 2019-20)
ydydx – 2x = y3e-y
Short Questions:
dydx + y = cos x – sin x (Outside Delhi 2019)
dxdy + x = (tan y + sec2y). (Outside Delhi 2019 C)
Long Questions:
x2 + y2 = a2. (N.C.E.R.T.)
{(x,y): x2 + y2< 8, x2< 2y}. (C.B.S.E. Sample Paper 2018-19)
Case Study Questions:
1. If the equation is of the form where P, Q are functions of x, then the solution of the differential equation is given by ye
is called the integrating factor (I.F.).
Based on the above information, answer the following questions.
2. If the equation is of the form where f(x, y), g(x, y) are homogeneous functions of the same degree in x and y, then put y = vx And
so that the dependent variable y is changed to another variable v and then apply variable separable method.
Based on the above information, answer the following questions.
Answer Key-
Multiple Choice questions-
Very Short Answer:
We have: y= b(x + a) …(1)
Diff. w.r.t. x, b.
Again diff. w.r.t. x, d2ydx2 = 0,
which is the reqd. differential equation.
We have: dydx = ex+y
⇒ e–ydy = ex dx [Variables Separable
⇒– e-y + c = ex
⇒ ex + e-y = c.
The given equation can be written as.
We have: y – a sin (3x – b) …(1)
Diff. W.r.t y dydx = a cos (3x – b) .3
= 3a cos (3x – b)
d2ydx2 = -3a sin (3x – b) 3
= -9a sin (3x – b)
= -9y [Using (1)]
d2ydx2 + 9y = 0,m
which in the reqd. differential equation.
Short Answer:
We have: y = e2x (a + bx) …(1)
Diff. w.r.t. x, dydx = e2x (b) + 2e2x (a + bx)
⇒dydx = be2x + 2y ………….. (2)
Again diff. w.r.t. x,
d2ydx2 = 2be2x + 22x
d2ydx2 = 2 (dydx–2y)+dydx
[Using (2)]
Hence, d2ydx2 -4 dydx + 4y = 0, which is the reqd. differential equation.
The given differential equation is:
dydx + y = cos x – sin x dx Linear Equation
∴ I.F. = e∫1dx = ex
The solution is :
y.ex = ∫ (cos x — sin x) ex dx + C
⇒y.ex = ex cos x + C
or y = cos x + C e-x
The given differential equation is:
dxdy + x = (tany + sec2y).
Linear Equation
∵ I.F. = Jldy = ey
∴ The solution is:
x. ey = ∫ ey (tan y + sec2y)dy + c
⇒ x. ey = ey tan y + c
= x = tan y + c e-y, which is the reqd. solution.
Long Answer:
log x = -log (1 + v2) + log C
x(1 + v2) = C
x(1 + y2x2)=C
x2 + y2 = C.
= tan-1 x + C
When y = 0, x = 1,
then 0 = π4 + C
C = π4
∴ y (1 + x2) = tan -1 x –π4
We have: y = aebx + 5 + 5 …(1)
Diff. w.r.t. x, dydx = aebx + 5. (b)
dydx = dy ……(2) [Using (1)]]
Again diff. w.r.t x.,
d2ydx2 = b dydx ………(3)
Dividing (3) by (2),
which is the required differential equation.
The given differential equation is:
When x = 0, y = 1, ∴ 1 = c + c(0) ⇒ c = 1.
Putting in (2), √1 + x2= 1 + ey(y -1),
which is the reqd. particular solution.
Case Study Answers:
1. Answer :
2. Answer :