Question

Given the differential Equation (dy/dx)+(2/x)y=x^2y^3 ;solve this equation using the Bernoulli method; Final answer should be (1/y^2)=?

Answer 1

`(1)/(y^2)=2x^3+Cx^4`.

Explanation:

Given differential equation

`\frac{\mathrm{d}y}{\mathrm{d}x}+(2)/(x)y=x^2y^3`

Differential equation can be write as

`y^(-3)\frac{\mathrm{d}y}{\mathrm{d}x}+(2)/(x)y^(-2)=x^2`

By Bernoulli method

Susbstitute
`y^(-2)=t`.....{equationI}

Differentiate equation I w.r.t x then we get

`\frac{\mathrm{d}t}{\mathrm{d}x}=-2y^(-3)\frac{\mathrm{d}y}{\mathrm{d}x}`

`-(1)/(2)\frac{\mathrm{d}t}{\mathrm{d}x}=y^(-3)\frac{\mathrm{d}y}{\mathrm{d}x}`

Susbstitute the values in the given differential equation then we get

`-(1)/(2)\frac{\mathrm{d}t}{\mathrm{d}x}+(2)/(x)t=x^2`

`\frac{\mathrm{d}t}{\mathrm{d}x}-(4)/(x)t=-2x^2`

It is first order linear differential equation and compare with the first order linear differential equation
`\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x)`

Then we get P(x)=
`-(4)/(x)` and Q(x)=
`-2x^2`

Integration factor=
`e^\intP(x)dx`

Integration factor=
`e^{-\int(4)/(x)dx`

Integration factor=
`e^(-4lnx)=e^{lnx^(-4)}=x^(-4)`.

Using
`e^(logb)=b`

`t* (1)/(x^4)=\int{-2x^2}*(1)/(x^4)dx+C`

`t=-2x^4{\intx^(-2)dx+C}`

`t=2x^4*(1)/(x)+Cx^4`

`t=2x^3+Cx^4`

Substitute
`t=(1)/(y^2)` then we get

`(1)/(y^2)=2x^3+Cx^4`.

Answer:
`(1)/(y^2)=2x^3+Cx^4`.