# Solve the second order Differential equation xy” + y’ = 0; where y1 = ln x is a solution.?

given that the solution is Y=ln x * v(x)

Find V(x)

• xy'' + y' = 0

y1 = lnx

Y = v(x) * y1

Y = v ln(x) ( i wanna make it simpler but keeping in mind that v is a function of x)

dY/dx = v (1/x) + ln(x) *dV/dx

Y' = v/x + v'ln(x)

Y'' = 2v'/x - v/x² + v''ln(x)

replace in the original diff equa; xy'' + y' =- 0

xy'' + y' = 0

x{2v'/x - v/x² + v''ln(x) } + v/x + v'ln(x) = 0

2v' - v/x + xv''ln(x) + v/x + v'ln(x) = 0

2v' + xv''ln(x) + v'ln(x) = 0

let u = v'

 keeping in mind that u = u(x)

u' = v''

2v' + xv''ln(x) + v'ln(x) = 0

2u + xu' ln(x) + uln(x) = 0

xu'lnx = -ulnx - 2u

 separable

u' (xlnx) = -u(2 + lnx)

u' / u = -(2 + lnx) / (xlnx)

(1/u) du/dx = -(2 + lnx) / (xlnx)

(1/u) du = -(2 + lnx) / (xlnx) dx

integrate both sides to get

∫ (1/u) du = -∫ (2 + lnx) / (xlnx) dx

u = C / (xln²x)

but u = v'

so

v(x) = A + B/lnx

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