What is the solution for x^2y^{''}-3xy^'+3y=0,y(1)=5,y^'(1)=3 ?

The solution for x^2y^{''}-3xy^'+3y=0,y(1)=5,y^'(1)=3 is y=-x^3+6x

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x^2y^{''}-3xy^'+3y=0,y(1)=5,y^'(1)=3

x^{2} y^{^{''}} - 3 x y^{^{'}} + 3 y = 0, y (1) = 5, y^{^{'}} (1) = 3

y = - x^{3} + 6 x

Solution steps

x^{2} y^{''} (x) - 3 x y^{'} (x) + 3 y = 0

Solve linear ODE:

y = - x^{3} + 6 x

y = - x^{3} + 6 x

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

y = ln (x^{5}) + e^{x} - ln (e^{2})

y = (3 x + 1)^{2}

\frac{d}{d x} (- y + x + 2)

\int \frac{sec ^{2} ( x )}{2 x} d x

What is the solution for x^2y^{''}-3xy^'+3y=0,y(1)=5,y^'(1)=3 ?
The solution for x^2y^{''}-3xy^'+3y=0,y(1)=5,y^'(1)=3 is y=-x^3+6x