What is the solution for (x-1)y^{''}-xy^'+y=0 ?

The solution for (x-1)y^{''}-xy^'+y=0 is y=e^{x+c_{1}-1}+c_{2}x

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(x-1)y^{''}-xy^'+y=0

(x - 1) y^{^{''}} - x y^{^{'}} + y = 0

y = e^{x + c_{1} - 1} + c_{2} x

Solution steps

(x - 1) y^{''} (x) - x y^{'} (x) + y = 0

Solve using substitution:

y = e^{x + c_{1} - 1} + c_{2} x

y = e^{x + c_{1} - 1} + c_{2} x

k = 1 \sum \infty \frac{5}{2 ^{k}}

\frac{d y}{d x} = e^{x} + y

\int \frac{4}{x ^{5}} - 5 x^{4} d x

\frac{\partial}{\partial y} (ln (x^{6} y))

n = 0 \sum \infty (\frac{2}{3})^{n}

What is the solution for (x-1)y^{''}-xy^'+y=0 ?
The solution for (x-1)y^{''}-xy^'+y=0 is y=e^{x+c_{1}-1}+c_{2}x