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Solve the given initial-value problem. The DE is homogeneous, xy2 dy/dx = y3 - x3; y(1) = 3.
Solution:
Given that, xy2 dy/dx = y3 - x3
xy2 dy/dx = y3 - x3
dy/dx = (y3 - x3)/ xy2
dy/dx = [(y/x)3-1]/ (y2/x2)............(1)
Put y/x = v ⇒ y = xv
Differentiate w.r.t x
dy/dx = x. dv/dx + v
∴ Equation (1) becomes,
x. dv/dx + v = (v3-1)/v2
x. dv/dx = [(v3-1)/v2] - v
x. dv/dx = (v3-1-v3)/ v2
v2. dv = -1/x .dx
v3/3 = -logx + C
[1/3 (y3/x3 )]+ logx= C …………………(2)
Now, given that y(1) = 3
∴ Substituting y =3 and x =1 in equation(2)
[(1/3) × (33/13)]+ log 1 = C
C = 9
∴The required ‘Differential Equation' is,
(y3/3x3)+ logx = 9
y3+ 3x3logx = 27x3
Solve the given initial-value problem. The DE is homogeneous, xy2 dy/dx = y3 - x3; y(1) = 3.
Summary:
The ‘Differential Equation’ for the given initial-value problem xy2 dy/dx = y3 - x3; y(1) = 3 is y3+ 3x3logx = 27x3.
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