What is the solution for (x(1-x)y^{''}-5)/(2y^'-y)=0 ?

The solution for (x(1-x)y^{''}-5)/(2y^'-y)=0 is y=5(xln(x)-xln(x-1)+ln(x-1)-1)+c_{1}x+c_{2}

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(x(1-x)y^{''}-5)/(2y^'-y)=0

\frac{x ( 1 - x ) y ^{^{''}} - 5}{2 y ^{^{'}} - y} = 0

y = 5 (x ln (x) - x ln (x - 1) + ln (x - 1) - 1) + c_{1} x + c_{2}

Solution steps

\frac{x ( 1 - x ) y ^{''} ( x ) - 5}{2 y ^{'} ( x ) - y} = 0

Isolate

y^{''} (x) : y^{''} (x) = \frac{5}{x ( 1 - x )}

Solve

y^{''} (x) = \frac{5}{x ( 1 - x )} : y^{'} (x) = 5 (ln (x) - ln (x - 1)) + c_{1}

Solve

y^{'} (x) = 5 (ln (x) - ln (x - 1)) + c_{1} : y = 5 (x ln (x) - x ln (x - 1) + ln (x - 1) - 1) + c_{1} x + c_{2}

y = 5 (x ln (x) - x ln (x - 1) + ln (x - 1) - 1) + c_{1} x + c_{2}

y^{^{''}} - 4 y + 4 y = 2 e^{2} x

f^{^{'}} (x) = 3 x^{2} + 8 x + 2

f^{^{'}} (x) = \frac{2 cos ^{2} ( x )}{a ^{9}}

y^{^{''}} = 2 x + 1

y^{^{''}} = 2 x^{2}

What is the solution for (x(1-x)y^{''}-5)/(2y^'-y)=0 ?
The solution for (x(1-x)y^{''}-5)/(2y^'-y)=0 is y=5(xln(x)-xln(x-1)+ln(x-1)-1)+c_{1}x+c_{2}