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Answer 1 · 2016-12-19T17:13:11

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

#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!