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Friday, February 20, 2009

Transformation of some equation in the form in which variables are separable.

Transformation of some equation in the form in which variables are separable. Equation of the form

Dy/dx = ƒ (ax + by + c)

or, dy/dx = ƒ (ax + by)

can be reduced to an equation in which variables can be separated. For this purpose we use the substitution ax + by + c = v or, ax + by = v

Example:

Solve: dy/dx = (4x + y +1)2

Given that, dy/dx = (4x + y +1)2 ----------- (1)

Let, 4x + y +1 = v -------------- (2)

Differentiating equation (1) with respect to x, we get

4 + dy/dx = dv/dx

or, dy/dx = dv/dx – 4 --------------- (3)

We get from equation (2) & (3),

dv/dx – 4 = v2

or, dv/dx = 4 + v2

or, dx = dv/(4 + v2 )

Integrating,

x + c = (1/2) tan -1 (v/2)

or, 2x + 2c = tan -1 (v/2)

or, v = 2 tan (2x + 2c)

or, 4x + y +1 = 2 tan (2x + 2c)

Answer:

Example:

Solve: dy/dx = sin (x + y) + cos (x + y)

Given that, dy/dx = sin (x + y) + cos (x + y) ------------- (1)

Let, x + y = v ------------(2)

Differentiating equation (1) with respect to x, we get

1 + dy/dx = dv/dx

or, dy/dx = dv/dx – 1 --------------- (3)

We get from equation (2) & (3),

dv/dx – 1 = sinv + cosv

or, dv/dx = 1 + sinv + cosv -------------(4)

but, 1 + sinv + cosv = 1 + 2 sin(v/2)cos(v/2) + 2 cos2(v/2) – 1

or, 1 + sinv + cosv = 2 cos2(v/2) [ 1 + tan v/2 ]

So from equation (3)

dx = dv/[2 cos2(v/2) { 1 + tan v/2 }]

or, dx = {(1/2)sec2 (v/2)}/ {1 + tan (v/2)}

Integrating,

x + c = ln {1 + tan (v/2)}

or, x + c = ln [1 + tan{(x + y)/2}]

Answer:

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