Question

Solve dy/dx+y/x=y² by bernollies equation ?

Answer 1

The general solution is `y=1/(-xlnx+Cx)`

Explanation:

The Bernouilli ODE is of the form

`y'+p(x)y=q(x)y^n`

The general solution is obtained by substituting

`v=y^(1-n)`

and solving

`1/(1-n)v'+p(x)v=q(x)`

Here,

The equation is

`dy/dx+y/x=y^2`

`p(x)=1/x`

`q(x)=1`

`n=2`

Divide both sides by `y^2`

`1/y^2dy/dx+1/(yx)=1`.....................`(1)`

Let `v=1/y`, `=>`, `yv=1`

Differentiating both side wrt `x`

`vdy/dx+y(dv)/dx=0`

`dy/dx=-y/v(dv)/dx`

Substituting in `(1)`

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

`(dv)/dx-v/x=-1`...................`(2)`

The integrating factor is

`IF=e^(int(-1/x)dx) =e^(-lnx)=1/x`

Multiply `(2)` by `IF`

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

`d/dx(v/x)=-1/x`

Integrating both sides

`v/x=-lnx+C`

`v=-xlnx+Cx`

Substituting back `v=1/y`

`=>`, `1/y=-xlnx+Cx`

`y=1/(-xlnx+Cx)`

Answer 2

`y = 1/(x(C - ln absx ))`

Explanation:

Bernoulli General Form:

• `y'+P(x)y=Q(x)y^n qquad "with "{(P = 1/x),(Q = 1),(n = 2):}`

Standard substitution:

• `z(x) = y^(1-n) = 1/y qquad :. z' = -1/y^2 y'`

So the DE `y'+y/x=y^2` becomes:

`- y^2 \ z' + y/x = y^2`

`z' - 1/(xy) = - 1 implies z' - z/x = - 1`

This is now linear, and can be solved by Integration factor:

• `exp (int -1/x dx) = 1/x`

`1/x(z' - z/x) = - 1/x`

`(1/xz)^' = - 1/x`

`1/xz = - ln absx + C`

`z = x(C - ln absx) = 1/y`

`y = 1/(x(C - ln absx ))`