What is the solution for x^2y^{''}-xy^'+y=ln(x) ?

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

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x^2y^{''}-xy^'+y=ln(x)

x^{2} y^{^{''}} - x y^{^{'}} + y = ln (x)

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

Solution steps

x^{2} y^{''} (x) - x y^{'} (x) + y = ln (x)

Solve linear ODE:

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

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

f^{^{'}} (x) = 6 x, f (0) = 8

t y^{^{'}} + 2 y = 3 t^{2} - 3 t

\int t d t

\frac{d ^{2} y}{d t ^{2}} - 15 \frac{d y}{d t} + 50 y = 0

\frac{\partial}{\partial x} (9 e^{x y})

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