5y-1-y-2-3-2-please-solve-the-differtial-equation-

Question Number 71516 by oyemi kemewari last updated on 16/Oct/19

{5 y}^{″} = {(1 + y^{' 2})}^{\frac{3}{2}}

please solve the differtial equation

Answered by mind is power last updated on 17/Oct/19

let z = y^{'}

{5 z}^{'} = {(1 + z^{2})}^{\frac{3}{2}}

\Rightarrow \frac{dz}{{(1 + z^{2})}^{\frac{3}{2}}} = dx

z = tg (u)

dz = (1 + {tg}^{2} (u)) du

\Rightarrow \frac{du}{\sqrt{1 + {tg}^{2} (u)}} = \frac{1}{5} dx

\Rightarrow \cos (u) = \frac{x}{5} + v

u = arcos (\frac{x}{5} + v)

z = tg (u) = \frac{\sqrt{1 - {(\frac{x}{5} + v)}^{2}}}{\frac{x}{5} + v}

y = \int zdx = \int \frac{\sqrt{1 - (\frac{x}{5} + v)}}{\frac{x}{5} + v} dx

= 5 \int \frac{\sqrt{1 - u}}{u} du, \sqrt{1 - u} = r \Rightarrow du = - 2 rdr

5 \int \frac{- {2 r}^{2}}{1 - r^{2}} dr = - 10 \int - 1 + \frac{1}{1 - r^{2}} dr = 10 r - 10 \int (\frac{1}{2 (1 - r)} + \frac{1}{2 (1 + r)}) dr

= 10 r - 5 \ln ∣ 1 - r^{2} ∣

y = 10 \sqrt{1 - (\frac{x}{5} + v)} - 5 \ln ∣ \frac{x}{5} + v ∣ + c