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
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:
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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