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Saturday, February 21, 2009

Rule for solving homogeneous equation

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.

Example:

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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