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 = v^{2}

or, dv/dx = 4 + v^{2}

or, dx = dv/(4 + v^{2 })

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 cos^{2}(v/2) – 1

or, 1 + sinv + cosv = 2 cos^{2}(v/2) [ 1 + tan v/2 ]

So from equation (3)

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

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

Integrating,

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

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

Answer:

## No comments:

## Post a Comment