Question

Solve the differential equation (x² + y²) dx - 2xy dy = 0 by completing the following activity.

Solution: (x² + y²) dx - 2xy dy = 0

∴ `dy/dx=(x^2+y^2)/(2xy)`                      ...(1)

Puty = vx

∴ `dy/dx=square`

∴ equation (1) becomes

`x(dv)/dx = square`

∴ `square  dv = dx/x`

On integrating, we get

`int(2v)/(1-v^2) dv =intdx/x`

∴ `-log|1-v^2|=log|x|+c_1`

∴ `log|x| + log|1-v^2|=logc       ...["where" - c_1 = log c]`

∴ x(1 - v²) = c

By putting the value of v, the general solution of the D.E. is `square`= cx

Answer 1

(x² + y²) dx - 2xy dy = 0

∴ `dy/dx=(x^2+y^2)/(2xy)`                      ...(1)

Puty = vx

∴ `dy/dx=bb(v+x(dv)/dx)`

∴ equation (1) becomes

`v+x(dv)/dx = (x^2 + v^2x^2)/(2x.vx) = (1+v^2)/(2v)`

∴ `x(dv)/dx= (1+v^2)/(2v) - v = (1+v^2-2v^2)/(2v)`

`x(dv)/dx = bb((1+v^2)/(2v))`

∴ `bb((2v)/(1-v^2))  dv = dx/x`

On integrating, we get

`int(2v)/(1-v^2) dv =intdx/x`

`-int(-2v)/(1-v^2) dv =intdx/x`

∴ `-log|1-v^2|=log|x|+c_1       ...[∵d/dx(1-v^2) = -2v  "and"  int(f^'(x))/f(x) dx = log|f(x)|+c]`

∴ `log|x| + log|1-v^2|=logc       ...["where" - c_1 = log c]`

∴  `log|x(1-v^2)|=logc`

∴ x(1 - v²) = c

By putting the value of v, the general solution of the D.E. is

`x(1-y^2/x^2)=c`

∴`x((x^2 - y^2)/x^2) = c`

`bb(x^2 - y^2)` = cx