dy/dx=(y+(x^2+y^2)^0.5)/x

Question

solve diffrentional equation

Answer 1

Solve dy/dx=(y+(x^2+y^2)^0.5)/x

dy/dx = (y/x) + √(x^2 + y^2)/x

dy/dx = (y/x) + √(1 + (y/x)^2)   --------- (1)

Let v = y/x, or y = vx, then

dy/dx = v + x.dv/dx    -------------------- (2)

Substituting for v = y/x and dy/dx, from (2), into (1)

v + x.dv/dx = v + √(1 + v^2)

x.dv/dx = √(1 + v^2)

Rearranging,

dv/√(1 + v^2) = dx/x

integrating,

int dv/√(1 + v^2) = int dx/x

arcsinh(v) = ln(x) + ln(k) = ln(kx)

v = sinh(ln(kx))

y/x = sinh(ln(kx))

Answer: y(x) = x.sinh(ln(kx))