Question

Question #a16d8

Answer 1

The solution is `y= x - 8ln|x| + 1`

Explanation:

`dy = (x- 8)/x dx`

`dy = (x- 8)x^-1dx`

`int(dy)= int(x - 8)x^-1dx`

Integrate by parts on the right-hand side. Let `dv= 1/x` and `u = x - 8`. Then `v = ln|x|` and `du = 1dx`.

The integration by parts formula states that `int(u dv) = uv - int(vdu)`.

Therefore:

`int(x - 8)x^-1 = (x- 8)ln|x| - int(ln|x|)`

We will now need to reintegrate using integration by parts.

We let `u = ln|x|` and `dv = 1`. `du = 1/x dx` and `v = x`.

`int(ln|x|) = xln|x| - int(1/x xx x)`

`" "= xln|x| - x`

`" "=x(ln|x| - 1) + C`

So, the complete integration, after integrating both sides, will be:

`y = (x -8)ln|x| - (x(ln|x| -1)) + C`

`y= (x- 8)ln|x| - xln|x| + x + C`

`y = xln|x| - 8ln|x| - xln|x| + x + C`

`y= x - 8ln|x| + C`

We want the equation that passes through `(1, 2)`. Thus:

`2 = (1- 8)ln|1| - 1(ln|1|) + 1 + C`

`2 = -7(0) - 1(0) + 1 + C`

`2 - 1 = C`

`C = 1`

The solution to the differential equation is therefore `y= x - 8ln|x| + 1`

Hopefully this helps!