Here, putting x = kx and y = ky
Therefore, the given differential equation is homogeneous.
To solve it we make the substitution.
y = vx
Differentiating eq. with respect to x, we get
Integrating both sides, we get
Put 1 + v2 = t
2vdv = dt
∴ log(1 + v2) = -logx + logC (∴ From (i) eq.)
The required solution of the differential equation.
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