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: (x^{3} + 3xy^{2})dx + (y^{3} + 3x^{2}y)dy = 0

Given that, (x^{3} + 3xy^{2})dx + (y^{3} + 3x^{2}y)dy = 0

or, dy/dx = - {(x^{3} + 3xy^{2})/ (y^{3} + 3x^{2}y)}

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 + 3v^{2})/(v^{3} + 3v)

or, x (dv/dx) = - (1 + 3v^{2})/(v^{3} + 3v) – v

or, x (dv/dx) = - (v^{4} + 6v^{2} + 1)/(v^{3} + 3v)

or 4(dx/x) = - {(4v^{3} + 12v)dv}/(v^{4} + 6v^{2} + 1)

Integrating,

4 lnx = - ln(v^{4 }+ 6v^{2} + 1) + lnc

or, lnx^{4 }= ln{c/(v^{4} + 6v^{2} + 1)}

or, x^{4 }(v^{4} + 6v^{2} + 1) = c

or, y^{4} + 6x^{2}y^{2} + x^{4} = 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: