Rule for solving homogeneous equation
Let the given equation be homogeneous. Then, by definition, the given equation can be put in the form
dy/dx = ƒ(y/x) ---------- (1)
Solution Method: -
To solve (1), let y/x = v
or, y = vx
Differentiating equation (1) with respect to x, we get
dy/dx = v + x (dv/dx)----------- (2)
So the equation becomes
v + x (dv/dx) = ƒ (v)
or, x (dv/dx) = ƒ (v) – v
Separating the variables x and v, we have
dx/x = dv/{ƒ (v) – v}--------------(3)
Integrating equation (3)
After integration, replace v by y/x and finally we will find the required solution.
Solve: (x3 + 3xy2)dx + (y3 + 3x2y)dy = 0
Given that, (x3 + 3xy2)dx + (y3 + 3x2y)dy = 0
or, dy/dx = - {(x3 + 3xy2)/ (y3 + 3x2y)}
or, dy/dx = - [{1 + 3(y/x)2}/{(y/x)3 + 3(y/x)}] -------(1)
Take, y/x = v
or, y = vx
Differentiating with respect to x, we get
dy/dx = v + x (dv/dx)----------- (2)
From equation (1) and (2)
v + x (dv/dx) = - (1 + 3v2)/(v3 + 3v)
or, x (dv/dx) = - (1 + 3v2)/(v3 + 3v) – v
or, x (dv/dx) = - (v4 + 6v2 + 1)/(v3 + 3v)
or 4(dx/x) = - {(4v3 + 12v)dv}/(v4 + 6v2 + 1)
Integrating,
4 lnx = - ln(v4 + 6v2 + 1) + lnc
or, lnx4 = ln{c/(v4 + 6v2 + 1)}
or, x4 (v4 + 6v2 + 1) = c
or, y4 + 6x2y2 + x4 = c [as y/x = v]
Answer:
Example:
Solve: dy/dx = y/x + tan(y/x)
Given that, dy/dx = y/x + tan(y/x) --------(1)
Let y/x = v
or, y = vx
Differentiating with respect to x, we get
dy/dx = v + x (dv/dx)----------- (2)
From equation (1) and (2)
v + x (dv/dx) = v + tanv
or, dx/x = cotv dv
Integrating,
ln x = ln sinv + ln c
or, x = c sin(y/x) [as y/x = v]
Answer:
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