`y' - y = e^xroot3(x)` Solve the Bernoulli differential equation.

Given ,

`y' - y = e^xroot3(x)` ----------(1)

this equation is already in linear form of first order ,so

this is of the form

`y' +Py=Q` ------------(2)

then the general solution `y=((int (I.F) (Q) dx)+c)/(I.F)`

where I.F is the integration function =` e^(int p dx)`

so on comapring the equations (1) and (2) we get,

`P =-1 ,  Q= e^xroot3(x)`

so the `I.F = e^(int p dx) = e^(int -1 dx)= e^(-x)`

now...

Given ,

`y' - y = e^xroot3(x)` ----------(1)

this equation is already in linear form of first order ,so

this is of the form

`y' +Py=Q` ------------(2)

then the general solution `y=((int (I.F) (Q) dx)+c)/(I.F)`

where I.F is the integration function =` e^(int p dx)`

so on comapring the equations (1) and (2) we get,

`P =-1 ,  Q= e^xroot3(x)`

so the `I.F = e^(int p dx) = e^(int -1 dx)= e^(-x)`

now the general solution is

`y=((int (I.F) *Q dx )+c)/(I.F)`

=>`y= ((int (e^(-x)) (e^xroot3(x)) dx)+c)/(e^(-x))`

 `= (int root3(x) dx + c)/e^(-x) `

`= ((x^(4/3))/(4/3) +c)/(e^(-x))`

`=((x^(4/3))/(4/3) +c)*e^(x)`

is the general solution.