Question

Please solve problem below​

Answer 1

(a) This is a Bernoulli equation:

`(dy)/(dx) + (x)/(2(x^2+1)) y = x^3y^5`

`\implies y^(-5) (dy)/(dx) + (x)/(2(x^2+1)) y^(-4) = x^3`

Substitute
`z=y^(-4)` and
`(dz)/(dx) = -4y^(-5)(dy)/(dx)` to transform the ODE to

`-\frac14 (dz)/(dx) + (x)/(2(x^2+1)) z = x^3`

`\implies (dz)/(dx) - (2x)/(x^2+1) z = -4x^3`

which is now linear in
`z`. Using the integrating factor method, the I.F. is

`\mu = \displaystyle \exp\left(\int -(x)/(x^2+1) \, dx\right) = \exp\left(-\frac12 \ln(1+x^2)\right) = \frac1{√(1+x^2)}`

Distribute
`\mu` on both sides to get a derivative of a product on the left side.

`\frac1{√(1+x^2)} (dz)/(dx) - (2x)/((x^2+1)^(3/2)) z = -(4x^3)/(√(1+x^2))`

`\implies (d)/(dx) \left(\frac1{√(1+x^2)} z\right) = -(4x^3)/(√(1+x^2))`

Integrate both sides (the integral on the right can be done by parts) to get

`\displaystyle \frac1{√(1+x^2)} z = -4 \int (x^3)/(√(1+x^2)) \, dx = -\frac43 (x^2-2) √(1+x^2) + C`

Solve for
`z`.

`\displaystyle \frac1{√(1+x^2)} z = -\frac43 (x^2-2) √(1+x^2) + C`

`\implies z = -\frac43 (x^2-2) (1+x^2) + C √(1+x^2)`

Solve for
`y`.

`\frac1{y^4} = -\frac43 (x^2-2) (1+x^2) + C √(1+x^2)`

`\implies \boxed{y^4 = -\frac3{4(x^2-2) (1+x^2) + C √(1+x^2)}}`

You could go on to solve explicitly for
`y` if you like.

(b) This is also a Bernoulli equation:

`x^2y' + 2xy - y^3 = 0`

`\implies x^2 y^(-3) y' + 2xy^(-2) = 1`

Substitute
`z=y^(-2)` and
`z' = -2y^(-3)y'`.

`-\frac{x^2}2 z' + 2xz = 1`

`\implies z' - \frac4x z = -\frac2{x^2}`

Now repeat the method from (a) to solve for
`y`.

`\mu = \exp\left(-\displaystyle \int \frac4x \, dx\right) = \frac1{x^4}`

`\implies \frac1{x^4} z' - \frac4{x^5} z = -\frac2{x^6}`

`\implies \left(\frac1{x^4} z\right)' = -\frac2{x^6}`

`\displaystyle \implies \frac1{x^4} z = -2 \int (dx)/(x^6)`

`\implies \frac1{x^4} z = \frac2{5x^5} + C`

`\implies z = \frac2{5x} + Cx^4`

`\implies \frac1{y^2} = \frac2{5x} + Cx^4`

`\implies \boxed{y^2 = (5x)/(2 + Cx^5)}`