Explain this to me please??

Question

x(x+1)du/dx=u^2,u(1)=1?

Answer 1

Given data

x(x+1)du/dx = u²

u(1) = 1

Step1 : First separate the variables

Multiply with dx / x(x+1)u² both sides

(du/u²) = dx(1/x(x+1))

(1/u²)du = (1/x(x+1))dx

Substitute 1/x(x+1) = (1/x)-(1/(x+1))

1/u²)du = (1/x)-(1/(x+1))dx

Step2 : Apply integration on both sides

∫ (1/u²)du = ∫ [ (1/x)-(1/(x+1))dx ]

(-1/u) = lnx - ln(x+1) + c

(-1/u) = lnx - ln(x+1) + c  -------------(1)

Substitute : u(1) = 1 ⇒ u = 1 at x = 1

(-1/1) = ln1 - ln(1+1) + c

-1 = - ln2 + c

c = ln2 - 1

Substitute : c = ln2 - 1 in equation (1)

(-1/u) = lnx - ln(x+1) + ln2 - 1

(-1/u) = lnx + ln2 - ln(x+1) - 1

Apply formula : lnA + lnB = lnAB

(-1/u) = ln2x - ln(x+1) - 1

Apply formula : lnA - lnB = lnA/B

(-1/u) = ln2x/(x+1) - 1

u = -1 / [ ln2x/(x+1) - 1 ]

The solution is u = -1 / [ ln2x/(x+1) - 1 ]