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

The solution for x^2y^{''}+3xy^'-3y=0,y(1)=0,y^'(1)=5 is y=(5x)/4-5/(4x^3)

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

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

y = \frac{5 x}{4} - \frac{5}{4 x ^{3}}

Solution steps

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

Solve linear ODE:

y = \frac{5 x}{4} - \frac{5}{4 x ^{3}}

y = \frac{5 x}{4} - \frac{5}{4 x ^{3}}

\int_{0}^{\frac{π}{4}} x sin (12 x) d x

\int \frac{4}{3 x ^{2}} d x

y^{^{''}} - 4 y^{^{'}} - 5 y = - 2, y (0) = 0, y^{^{'}} (0) = 0

\frac{d}{d x} (e^{3 x} cos (x))

\frac{d}{d x} (\frac{( x ^{2} + 1 )}{x ^{2} - 1})

What is the solution for x^2y^{''}+3xy^'-3y=0,y(1)=0,y^'(1)=5 ?
The solution for x^2y^{''}+3xy^'-3y=0,y(1)=0,y^'(1)=5 is y=(5x)/4-5/(4x^3)