Differential Equations -Important Questions
& Solutions For Exam Practice
Q.1. Write the general solution of differential equation:
dy/dx = ex+y
Solution:
We have: dy/dx = ex+y
⇒ e-y dy = ex dx [Variables Separable
Integrating, ∫e−ydy+c=∫exdx
⇒ – e-y + c = ex
⇒ ex + e-y = c.
Q.2. Form the differential equation representing the family of curves y = a sin
(3x – b), where a and b are arbitrary constants
Solution:
We have: y – a sin (3x – b) …(1)
Diff. W.r.t y dy/dx = a cos (3x – b) .3
= 3a cos (3x – b)
d2y/dx2 = -3a sin (3x – b) 3
= -9a sin (3x – b)
= -9y [Using (1)]
d2y/dx2 + 9y = 0,m
which in the reqd. differential equation.
Q.3. Find the integrating factor of the differential equation:
ydy/dx – 2x = y3e-y
Solution:
The given equation can be written as
, Q.4. Find the particular solution of the differential equation (1 + x2)dy/dx +
2xy = 1/1+x2, given that y = 0 when x = 1
Solution:
Solution is y( 1 + x2) = ∫1/1+x2dx
= tan-1 x + C
When y = 0,x = 1,
then 0 = π/4 + C
C = π/4
∴ y(1 + x2) = tan -1 x – π/4
i.e, y = tan−1x/1+x2−π/4(1+x2)
Q.5. Find the differential equation representing the family of curves y= aebx + 5,
where ‘a’and ‘A’are arbitrary constants.
Solution:
We have: y = aebx + 5 + 5 …(1)
Diff. w.r.t. x, dy/dx = aebx + 5. (b)
dy/dx = dy ……(2) [Using (1)]]
Again diff. w.r.t x.,
d2y/dx2=bdy/dx ………(3)
Dividing equations (3) by (2)
which is the required differential equation.
& Solutions For Exam Practice
Q.1. Write the general solution of differential equation:
dy/dx = ex+y
Solution:
We have: dy/dx = ex+y
⇒ e-y dy = ex dx [Variables Separable
Integrating, ∫e−ydy+c=∫exdx
⇒ – e-y + c = ex
⇒ ex + e-y = c.
Q.2. Form the differential equation representing the family of curves y = a sin
(3x – b), where a and b are arbitrary constants
Solution:
We have: y – a sin (3x – b) …(1)
Diff. W.r.t y dy/dx = a cos (3x – b) .3
= 3a cos (3x – b)
d2y/dx2 = -3a sin (3x – b) 3
= -9a sin (3x – b)
= -9y [Using (1)]
d2y/dx2 + 9y = 0,m
which in the reqd. differential equation.
Q.3. Find the integrating factor of the differential equation:
ydy/dx – 2x = y3e-y
Solution:
The given equation can be written as
, Q.4. Find the particular solution of the differential equation (1 + x2)dy/dx +
2xy = 1/1+x2, given that y = 0 when x = 1
Solution:
Solution is y( 1 + x2) = ∫1/1+x2dx
= tan-1 x + C
When y = 0,x = 1,
then 0 = π/4 + C
C = π/4
∴ y(1 + x2) = tan -1 x – π/4
i.e, y = tan−1x/1+x2−π/4(1+x2)
Q.5. Find the differential equation representing the family of curves y= aebx + 5,
where ‘a’and ‘A’are arbitrary constants.
Solution:
We have: y = aebx + 5 + 5 …(1)
Diff. w.r.t. x, dy/dx = aebx + 5. (b)
dy/dx = dy ……(2) [Using (1)]]
Again diff. w.r.t x.,
d2y/dx2=bdy/dx ………(3)
Dividing equations (3) by (2)
which is the required differential equation.