CBSE Class 12 Math 2008 Solved Paper

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Question : 18 of 29
 
Marks: +1, -0
Solve the following differential equation:
(x2y2) dx + 2xy dy = 0
given that y = 1 when x = 1
OR
Solve the following differential equation:
dydx = 2yx2y+x , if y = 1 when x = 1
Solution:
(x2y2)dx + 2xydy = 0
dydx = y2x22xy ... (1)
It is a homogeneous differential equation.
Let y = vx ...(2)
dydx = v + x dvdx ... (3)
Substituting (2) and (3) in (1), we get:
v + x dvdx = v2x2x22xvx
v + x dvdx = x2v212vx2 - v212v
2v2+2vxdvdx = v2 - 1
2vx dvdx = - v2 - 1
(2vv2+1) dv = - dxx
Integrating both sides, we get:
2vv2+1 dv = - ∫ (1x) dx
log |v2+1| = - log |x| + log C
log |v2+1| = log |Cx|
v2 + 1 = Cx
xv2 + 1 = C
x |(yx)2+1| = C
y2+x2 = Cx ... 4
It is given that when x = 1, y = 1
(1)2+(1)2 = C(1)
C = 2
Thus, the required solution is y2+x2 = 2x.
OR
We need to solve the following differential equation
dydx = x(2yx)x(2y+x)
dydx = 2yx2y+x ... (1)
It is a homogeneous differential equation.
Let y = vx ...(2)
dydx = v + x dvdx ... 3
Substituting (2) and (3) in (1), we get:
v + x dvdx = x(2v1)x(2v+1)
x dvdx = v12v+1 - v
x dvdx = 2v2+v12v+1
(2v+12v2+v1) dv = (1x) dx
(2v+12v2v+1) dv = (1x) dx
Integrating both sides,
12(4v1+32v2v+1) dv = ∫ (1x) dx
12(4v12v2v+1) dv + ∫ 32(12v2v+1) dv = ∫ (1x) dx
12(4v12v2v+1) dv + ∫ 34|1v2v2+12| dv = f (1x) dx
1/2 log |2v2v+1| + 34(1v2v2+116+716) dv = - log |x| + C
1/2 log |2v2v+1| + 34dv(v14)2+(74)2 = - log |x| + C
1/2 log |2v2v+1| + 34×47tan1(v1474) = - log |x| + C
1/2 log |2v2v+1| + 37tan1(4v17) = C - log |x|
Put v = yx
12 log |2(yx)2(yx)+1| + 37tan1(4yx17) = C - log |x|
12 log |2y2xy+x2x2| + 37tan1(4yx7x) = C - log |x| ... 4
Now y = 1 when x = 1
12 log |(2)(1)21(1)+1212| + 37tan1|(4)(1)17(1)| = C - log |1|
12 log 2 + 37tan1(37) = C ... (5)
Therefore, form (4) and (5) we get:
12 log |2y2xy+x2x2| + 37tan1(4yx7x) = 12 log 2 + 37tan1(37) - log |x|
12 log |2y2xy+x2x2| - 12 log 2 + log |x| = 37[tan137tan1(4yx7x)]
12 log |2y2xy+x22x2.x2| = 37tan1(3x4y+x7x1+3(4yx)7x)
12 log |2y2xy+x22| = 37tan1(4(xy)7x7x+12y3x7x)
12 log |2y2xy+x22| = 37tan1(7(xy)x+3y)
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