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

The solution for x^2y^{''}-3xy^'+5y=0,y(1)=-1,y^'(1)=4 is y=x^2(-cos(ln(x))+6sin(ln(x)))

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x^2y^{''}-3xy^'+5y=0,y(1)=-1,y^'(1)=4

x^{2} y^{^{''}} - 3 x y^{^{'}} + 5 y = 0, y (1) = - 1, y^{^{'}} (1) = 4

y = x^{2} (- cos (ln (x)) + 6 sin (ln (x)))

Solution steps

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

Solve linear ODE:

y = x^{2} (- cos (ln (x)) + 6 sin (ln (x)))

y = x^{2} (- cos (ln (x)) + 6 sin (ln (x)))

What is the solution for x^2y^{''}-3xy^'+5y=0,y(1)=-1,y^'(1)=4 ?
The solution for x^2y^{''}-3xy^'+5y=0,y(1)=-1,y^'(1)=4 is y=x^2(-cos(ln(x))+6sin(ln(x)))