CBSE Class 12 Math 2008 Solved Paper
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Question : 18 of 29
Marks:
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Solve the following differential equation:
() dx + 2xy dy = 0
given that y = 1 when x = 1
OR
Solve the following differential equation:
= , if y = 1 when x = 1
() dx + 2xy dy = 0
given that y = 1 when x = 1
OR
Solve the following differential equation:
= , if y = 1 when x = 1
Solution:
()dx + 2xydy = 0
⇒ = ... (1)
It is a homogeneous differential equation.
Let y = vx ...(2)
∴ = v + x ... (3)
Substituting (2) and (3) in (1), we get:
v + x =
v + x = -
= - 1
2vx = - - 1
dv = -
Integrating both sides, we get:
∫ dv = - ∫ dx
log = - log |x| + log C
log = log
+ 1 =
+ 1 = C
x = C
= Cx ... 4
It is given that when x = 1, y = 1
= C(1)
C = 2
Thus, the required solution is = 2x.
OR
We need to solve the following differential equation
=
= ... (1)
It is a homogeneous differential equation.
Let y = vx ...(2)
∴ = v + x ... 3
Substituting (2) and (3) in (1), we get:
v + x =
x = - v
x =
dv = dx
dv = dx
Integrating both sides,
dv = ∫ dx
∫ dv + ∫ dv = ∫ dx
∫ dv + ∫ dv = f dx
1/2 log + dv = - log |x| + C
1/2 log + ∫ = - log |x| + C
1/2 log + = - log |x| + C
1/2 log + = C - log |x|
Put v =
log + = C - log |x|
log + = C - log |x| ... 4
Now y = 1 when x = 1
log + = C - log |1|
log 2 + = C ... (5)
Therefore, form (4) and (5) we get:
log + = log 2 + - log |x|
log - log 2 + log |x| =
log =
log =
log =
⇒ = ... (1)
It is a homogeneous differential equation.
Let y = vx ...(2)
∴ = v + x ... (3)
Substituting (2) and (3) in (1), we get:
v + x =
v + x = -
= - 1
2vx = - - 1
dv = -
Integrating both sides, we get:
∫ dv = - ∫ dx
log = - log |x| + log C
log = log
+ 1 =
+ 1 = C
x = C
= Cx ... 4
It is given that when x = 1, y = 1
= C(1)
C = 2
Thus, the required solution is = 2x.
OR
We need to solve the following differential equation
=
= ... (1)
It is a homogeneous differential equation.
Let y = vx ...(2)
∴ = v + x ... 3
Substituting (2) and (3) in (1), we get:
v + x =
x = - v
x =
dv = dx
dv = dx
Integrating both sides,
dv = ∫ dx
∫ dv + ∫ dv = ∫ dx
∫ dv + ∫ dv = f dx
1/2 log + dv = - log |x| + C
1/2 log + ∫ = - log |x| + C
1/2 log + = - log |x| + C
1/2 log + = C - log |x|
Put v =
log + = C - log |x|
log + = C - log |x| ... 4
Now y = 1 when x = 1
log + = C - log |1|
log 2 + = C ... (5)
Therefore, form (4) and (5) we get:
log + = log 2 + - log |x|
log - log 2 + log |x| =
log =
log =
log =
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