What is the solution for y^{''}-2/r y^'+2/(r^2)y=0,y(1)=1,y(2)=0 ?

The solution for y^{''}-2/r y^'+2/(r^2)y=0,y(1)=1,y(2)=0 is

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y^{''}-2/r y^'+2/(r^2)y=0,y(1)=1,y(2)=0

y^{^{''}} - \frac{2}{r} y^{^{'}} + \frac{2}{r ^{2}} y = 0, y (1) = 1, y (2) = 0

y = e^{\frac{t}{r}} (\frac{sin ( \frac{2}{r} ) cos ( \frac{t}{r} )}{e ^{\frac{1}{r}} ( sin ( \frac{2}{r} ) cos ( \frac{1}{r} ) - cos ( \frac{2}{r} ) sin ( \frac{1}{r} ) )} - \frac{cos ( \frac{2}{r} ) sin ( \frac{t}{r} )}{e ^{\frac{1}{r}} ( - cos ( \frac{2}{r} ) sin ( \frac{1}{r} ) + sin ( \frac{2}{r} ) cos ( \frac{1}{r} ) )})

Solution steps

y^{''} (t) - \frac{2}{r} y^{'} (t) + \frac{2}{r ^{2}} y = 0

Solve linear ODE:

y = e^{\frac{t}{r}} (\frac{sin ( \frac{2}{r} ) cos ( \frac{t}{r} )}{e ^{\frac{1}{r}} ( sin ( \frac{2}{r} ) cos ( \frac{1}{r} ) - cos ( \frac{2}{r} ) sin ( \frac{1}{r} ) )} - \frac{cos ( \frac{2}{r} ) sin ( \frac{t}{r} )}{e ^{\frac{1}{r}} ( - cos ( \frac{2}{r} ) sin ( \frac{1}{r} ) + sin ( \frac{2}{r} ) cos ( \frac{1}{r} ) )})

y = e^{\frac{t}{r}} (\frac{sin ( \frac{2}{r} ) cos ( \frac{t}{r} )}{e ^{\frac{1}{r}} ( sin ( \frac{2}{r} ) cos ( \frac{1}{r} ) - cos ( \frac{2}{r} ) sin ( \frac{1}{r} ) )} - \frac{cos ( \frac{2}{r} ) sin ( \frac{t}{r} )}{e ^{\frac{1}{r}} ( - cos ( \frac{2}{r} ) sin ( \frac{1}{r} ) + sin ( \frac{2}{r} ) cos ( \frac{1}{r} ) )})

\int 12 x^{2} - 48 x + 36 d x

x \to 3 lim (\frac{3 x ^{2} - 4 x - 15}{x ^{2} - 5 x + 6})

x^{2} + 2 x + 1, 3 x + 3

\frac{d}{d x} (\frac{c}{x ^{2}})

\frac{\partial}{\partial x} (3 y + 7)

What is the solution for y^{''}-2/r y^'+2/(r^2)y=0,y(1)=1,y(2)=0 ?
The solution for y^{''}-2/r y^'+2/(r^2)y=0,y(1)=1,y(2)=0 is