Chapter 9 Differential Equations | class 12th | Important Question for Maths

Class 12 Maths Chapter 9 Important Extra Questions Differential Equations

Differential Equations Important Extra Questions Very Short Answer Type

Question 1.
Find the order and the degree of the differential equation: x2d2ydx2=[1+(dydx)2]4
(Delhi 2019)
Solution:
Here, order = 2 and degree = 1.

Question 2
Determine the order and the degree of the differential equation:(dydx)3+2yd2ydx2=0 (C.B.S.E. 2019 C)
Solution:
Order = 2 and Degree = 1.

Question 3.
Form the differential equation representing the family of curves: y = b (x + a), where « and b are arbitrary constants. (C.B.S.E. 2019 C)
Solution:
Wehave: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.

Question 4.
Write the general solution of differential equation:
dydx = e^x+y (C.B.S.E. Sample Paper 2019-20)
Solution:
We have: dydx = e^x+y
⇒ e^-y dy = e^x dx [Variables Separable
Integrating, ∫e−ydy+c=∫exdx
⇒ – e^-y + c = e^x
⇒ e^x + e^-y = c.

Question 5.
Find the integrating factor of the differential equation:
ydydx – 2x = y³e^-y
Solution:
The given equation can be written as.

Question 6.
Form the differential equation representing the family of curves y = a sin (3x – b), where a and b are arbitrary constants. (C.B.S.E. 2019C)
Solution:
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.

Differential Equations Important Extra Questions Short Answer Type

Question 1.
Determine the order and the degree of the differential equation:
(dydx)3+2yd2ydx2=0 (Outside Delhi 2019C)
Solution:
Order = 2 and Degree = 1.

Question 2.
Form the differential equation representing the family of curves: y = e^2x (a + bx), where ‘a’ and ‘h’ are arbitray constants. (Delhi 2019)
Solution:
We have : y = e^2x (a + bx) …(1)
Diff. w.r.t. x, dydx = e^2x (b) + 2e2x (a + bx)
⇒ dydx = be^2x + 2y ………….. (2)
Again diff. w.r.t. x,
d2ydx2 = 2be^2x + 2^2x
d2ydx2 = 2(dydx – 2y) + dydx
[Using (2)]
Hence, d2ydx2 -4 dydx + 4y = 0, which is the reqd. differential equation.

Question 3.
Solve the following differentia equation:
dydx + y = cos x – sin x (Outside Delhi 2019)
Solution:
The given differential equation is :
dydx + y = cos x – sin x dx Linear Equation
∴ I.F. = e^∫1dx = e^x
The solution is :
y.e^x = ∫ (cos x — sin x) e^x dx + C
⇒ y.e^x = e^x cos x + C
or y = cos x + C e^-x

Question 4.
Solve the following differential equation :
dxdy + x = (tan y + sec²y). (Outside Delhi 2019 C)
Solution:
The given differential equation is :
dxdy + x = (tany + sec²y).
Linear Equation
∵ I.F. = Jldy = ey
∴ The solution is :
x. ey = ∫ ey (tan y + sec² y)dy + c
⇒ x. ey = ey tan y + c
= x = tan y + c e-y, which is the reqd. solution.

Differential Equations Important Extra Questions Long Answer Type 1

Question 1.
Solve the differential equation
(x² – y²)dx + 2xydy = 0 (C.B.S.E. 2018)
Solution:

log x = -log (1 + v²) + log C
x(1 + v²) = C
x(1 + y2x2) = C
x² + y² = C.

Question 2.
Find the particular solution of the differential equation (1 + x²)dydx + 2xy = 11+x2, given that y = 0 when x = 1(C.B.S.E. 2018 C)
Solution:

Solution is y( 1 + x²) = ∫11+x2dx
= tan^-1 x + C
When y = 0,x = 1,
then 0 = π4 + C
C = π4
∴ y(1 + x²) = tan ^-1 x – π4
i.e, y = tan−1×1+x2−π4(1+x2)

Question 3.
Find the differential equation representing the family of curves y = ae^{bx + 5}, where ‘a’and ‘A’are arbitrary constants. {C.B.S.E. 2018)
Solution:
We have: y = ae^{bx + 5} + 5 …(1)
Diff. w.r.t. x, dydx = ae^{bx + 5}. (b)
dydx = dy ……(2) [Using (1)]]
Again diff. w.r.t x.,
d2ydx2=bdydx ………(3)

Dividing (3) by (2),

which is the required differential equation.

Question 4.
Find the particular solution of the differential equation x dx – ye^y 1+x2−−−−−√ dy = 0, given that y = 1 when x = 0. (C.B.S.E. 2019 C)
Solution:
The given differential equation is:

When x = 0,y = 1, ∴ 1 = c + c(0) ⇒ c = 1.
Putting in (2), 1+x2−−−−−√ = 1 + e^y(y -1),
which is the reqd. particular solution.

Question 5.
Obtain the differential equation of the family of circles, which touch the x-axis at the origin.   (N.C.E.R.T.; C.B.S.E. Sample Paper 2018)
Solution:
Let (0, α) be the centre of any member of the family of circles.

Then the equation of the family of circles is : x² + (y-α)² = α²
⇒ x² + y² – 2αy = 0 …(1)
Diff. w.r.t. x, 2x + 2y dydx 2α dydx = 0

which is the required differential equation.

Question 6.
Obtain the differential equation representing the family of parabolas having vertex at the origin and axis along the positive direction of x-axis. (N.C.E.R.T.)
Solution:
Let S (a, 0) be the focus of any member of the family of parabolas.
Then the equation of the family of curves is y² = 4 ax …………. (1)
Diff. w.r.t. x, 2ydydx = 4a ……………. (2)
Using (2) in (1), we get:
y² = (2ydydx)x

y² – 2xydydx = 0
which is the required differential equation.

Question 7.
Find the general solution of the differential equation:
(tan²x + 2 tanx + 5) dydx = 2 (1 + tanx) sec² x
Solution:
We have : (tan² x + 2 tan x + 5)dydx
= 2 (1 + tanx) sec²x
Integrating

Put t² + 2t + 5 = z
so that (2t+ 2) dt = dz.
∴ From (2),
1 = ∫dzz = log| z | = log |t² + 2t + 5|
= log |tan² x| + 2 tan x + 5|
From (1), y = log |tan² x| + 2 tan x + 5| + c,
which is the required general solution.

Question 8.
Solve the differential equation:
(x + 1) dydx = 2e^-y – 1 ; y(0) = 0. (C.B.S.E. 2019)
Solution:
The given equation can be written as :
dy2e−y−1=dxx+1
| Variables Separable

Integrating, ∫eydy2−ey=∫dxx+1
⇒ – log |2 – e^y| + log |C| = log |x + 1|
⇒ (2 – e^y) (x + 1) = C.
When x = 0, y = 0, then C = 1.
Hence, the solution is (2 – e^y) (x + 1) = 1.

Question 9.
Solve: (a) (i) dydx = 1 + x + y + xy
(ii) xyy’ = 1 + x + y + xy. (N.C.E.R.T.)
(b) Find the particular solution of the differential dy
equation dydx = 1 + x + y + xy, given that y = 0 when x = 1. (A.I.C.B.S.E. 2014)
Solution:
(a), (i) The given equation is :
dydx = 1 + x + y + xy
dydx = (1 +x) (1 +y)
d1+y = (1 + x)dx.
|Variables Separable

Integrating,
∫dy1+y = ∫(1 + x)dx + c
⇒ log |1 + y | = x + 12 x² + c,
which is the required solution.

(ii) The given equation is xyy’ = 1 + x + y + xy
⇒ xy∫dy1+y = (1 + x)(1 + y)

= y – log|1 + y| = log|x|+ x + c
= x + log |x (1 + y)|+ c,
which is the required solution.

(b) From part (a) (i),
Iog|1+y| = x + 12x² + c …(1)
When x = 1, y = 0, then:
log |1+0|= 1 + 12(1)² + c
log 1 = 1 + 12 + c
0 = 32 + c
c = −32
Putting in (1), log |1 + y| = x + 12x² – 32
which is the required particular solution.

Question 10.
Solve the differential equation:
dydx = 1 + x² + y² + x²y²
given thaty = 1 when x = 0. (Outside Delhi 2019)
Solution:
The given equation is dydx = 1 + x² + y² + x²y²
⇒ dydx = (1+x²)(1 +y²)
⇒ dy1+y2 (x² + 1) dx
|Variables Separable

Integrating, ∫dy1+y2 = ∫(x² + 1) dx + C
tan^-1y = x33 + x + C ………. (1)
Where x = 0,y = 1,
∴ tan^-1 (1) = C
C = π4
Putting in (1),
tan^-1 y = x33+x+π4
which is the reqd. particular solution.