What is the solution for x^2y^{''}+3xy^'+2y=0,y(1)=0.2,y^'(1)=0.3 ?

The solution for x^2y^{''}+3xy^'+2y=0,y(1)=0.2,y^'(1)=0.3 is y=(0.2cos(ln(x))+0.5sin(ln(x)))/x

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x^2y^{''}+3xy^'+2y=0,y(1)=0.2,y^'(1)=0.3

x^{2} y^{^{''}} + 3 x y^{^{'}} + 2 y = 0, y (1) = 0.2, y^{^{'}} (1) = 0.3

y = \frac{0.2 cos ( ln ( x ) ) + 0.5 sin ( ln ( x ) )}{x}

Solution steps

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

Solve linear ODE:

y = \frac{0.2 cos ( ln ( x ) ) + 0.5 sin ( ln ( x ) )}{x}

y = \frac{0.2 cos ( ln ( x ) ) + 0.5 sin ( ln ( x ) )}{x}

\frac{d}{d x} (cos (sin (tan (π) x)))

3 \cdot sin (3 t + 2)

\int cos (X) d X

\frac{d}{d x} (x (7 x + 3)^{\frac{1}{3}})

1 1^{x}

What is the solution for x^2y^{''}+3xy^'+2y=0,y(1)=0.2,y^'(1)=0.3 ?
The solution for x^2y^{''}+3xy^'+2y=0,y(1)=0.2,y^'(1)=0.3 is y=(0.2cos(ln(x))+0.5sin(ln(x)))/x