Solve differential equation dy/dx=x^3*y/(x-1)^4

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You need to separate the variables dividing both sides by `(dx/y)`  such that:

`(dy)/y = (x^3dx)/(x-1)^4`

You may find the general solution to this equation integrating both sides such that:

`int (dy)/y = int (x^3dx)/(x-1)^4`

`ln|y| =int (x^3dx)/(x-1)^4`

You should write `x^3`  in terms of `(x-1)`  such that:

`x^3 = [(x-1)+1]^3`

Expanding the cube to the right side yields:

`x^3 = (x-1)^3 + 1^3 + 3(x-1)(x-1+1)`

`x^3 = (x-1)^3 + 1 + 3x(x-1)`

You need to substitute `(x-1)^3 + 1 + 3x(x-1)`  for `x^3`  such that:

`int (x^3dx)/(x-1)^4 = int (((x-1)^3 + 1 + 3x(x-1))dx)/(x-1)^4`

You need to split the general integral in simpler integrals such that:

`int (x^3dx)/(x-1)^4 = int ((x-1)^3dx)/(x-1)^4 + int (dx)/(x-1)^4 + int (3x(x-1)dx)/(x-1)^4`

`int (x^3dx)/(x-1)^4 = int (dx)/(x-1) + int (dx)/(x-1)^4 + int (3xdx)/(x-1)^3`

`int (x^3dx)/(x-1)^4 = ln|x-1| + int (dx)/(x-1)^4 + 3 int ((x-1)dx)/(x-1)^3 + 3int (dx)/(x-1)^4`

`int (x^3dx)/(x-1)^4 = ln|x-1| + 4int (dx)/(x-1)^4 + 3 int (dx)/(x-1)^2`

You should come up with the substitution `x - 1 =p =gt dx = dp`  such that:

`int (x^3dx)/(x-1)^4 = ln|x-1| + 4int (dp)/(p)^4 + 3 int (dp)/(p)^2` `int (x^3dx)/(x-1)^4 = ln|x-1|- 4/(3p^3) - 3/p + c`

You need to substitute `x-` 1 for p in `ln|x-1| - 4/(3p^3) - 3/p + c ` such that:

`int (x^3dx)/(x-1)^4 = ln|x-1| - 4/(3(x-1)^3) - 3/(x-1) + c`

Hence, since `ln|y| = int (x^3dx)/(x-1)^4` , thus`ln|y| = ln|x-1| - 4/(3(x-1)^3) - 3/(x-1) + c`

`` `y = e^(ln|x-1| - 4/(3(x-1)^3) - 3/(x-1)) + c`

Hence, evaluating the general solution to differential equation yields `y = e^(ln|x-1| - 4/(3(x-1)^3) - 3/(x-1)) + c` .

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