What is the solution for t^2y^{''}+11ty^'+25y=0,y(1)=4,y^'(1)=9 ?

The solution for t^2y^{''}+11ty^'+25y=0,y(1)=4,y^'(1)=9 is y= 4/(t^5)+(29ln(t))/(t^5)

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t^2y^{''}+11ty^'+25y=0,y(1)=4,y^'(1)=9

t^{2} y^{^{''}} + 11 t y^{^{'}} + 25 y = 0, y (1) = 4, y^{^{'}} (1) = 9

y = \frac{4}{t ^{5}} + \frac{29 ln ( t )}{t ^{5}}

Solution steps

t^{2} y^{''} (t) + 11 t y^{'} (t) + 25 y = 0

Solve linear ODE:

y = \frac{4}{t ^{5}} + \frac{29 ln ( t )}{t ^{5}}

y = \frac{4}{t ^{5}} + \frac{29 ln ( t )}{t ^{5}}

\int 18 - \frac{54}{15 - 2 x} d x

\frac{d}{d x} (\frac{( x ^{2} - 4 ) ^{2}}{( x ^{2} + 2 ) ^{3}})

\int \frac{1}{y - 1} d y

x \to 1 lim (1^{2} - 3)

x y^{^{'}} - 2 y = - x^{2}

What is the solution for t^2y^{''}+11ty^'+25y=0,y(1)=4,y^'(1)=9 ?
The solution for t^2y^{''}+11ty^'+25y=0,y(1)=4,y^'(1)=9 is y= 4/(t^5)+(29ln(t))/(t^5)