Solve the separable initial value problem dydx xx2 49 y2y0

Solve the separable initial value problem dy/dx = x/x^2 + 49 y^2,y(0) = 1 dy/dx = x/x^2 + 49 y^2,y(0) = 1

Solution

dy/dx = {x/(x^2 +49) }y^2 with y(0) = 1

separating the terms: dy/y^2 = (x/x^2 +49)dx

Integrating both sides\"

Integral of dy/y^2 = -1/y

Integral of (x/x^2 +49)dx , Let x^2 +49 = t

dt/dx = 2x ----> dt/2 = dx.x

So ,Integral of (x/x^2 +49)dx = Integral of dt/(2t)

= ln|t| /2 = {ln| x^2 +49|}/2

So, -1/y =  {ln| x^2 +49|}/2 + lnC

use the intial value problem : y(0) =1 to find constant C

-1 = ln(42)/2 +C

C = -1 -ln(42)/2 = -2.868

So, -1/y = ln|x^2 +49| -2.868