Question

Find the general solution of the Bernoulli differential equation: y' + 4y = xy^4. Use lower case c for the constant in answer.

Answer 1

Answer with Step-by-step explanation:

The given equation is

`(dy)/(dx)+4y=xy^4`

Dividing throughout by
`y^4` we get

`(dy)/(y^4dx)+(4)/(y^3)=x\\\\y^(-4)(dy)/(dx)+4y^(-3)=x\\\\`

Substituting
`t=y^(-3)` in the above equation we get

`dt=-3y^(-4)\cdot (dy)/(dx)\\\\y^(-4)\cdot {dy}=(-dt)/(3)`

Thus we get

`(-dt)/(dx)+12t=3x`

Which is a linear differential equation of the form

`t'+p(x)t=q(x)dx}`[/tex]

whose solution is given by

`te^(\int p(x)dx)=\int (e^(\int p(x))q(x)dx`

Solving we get

`te^(12x)=\int 3e^(12x)* xdx\\\\te^(12x)=3x\cdot (e^(12x))/(12)-\int 3* (e^(12x))/(12)dx\\\\te^(12x)=(xe^(12x))/(4)-(e^(12x))/(48)\\\\t=(x)/(4)-(1)/(48)\\\\(1)/(y^3)=(x)/(4)-(1)/(48)+c`