What is the solution for 36y^{''}-60y^'+26y=0,y(4)=2,y^'(4)=5 ?

The solution for 36y^{''}-60y^'+26y=0,y(4)=2,y^'(4)=5 is

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36y^{''}-60y^'+26y=0,y(4)=2,y^'(4)=5

36 y^{^{''}} - 60 y^{^{'}} + 26 y = 0, y (4) = 2, y^{^{'}} (4) = 5

y = e^{\frac{5 t}{6}} (\frac{( 2 ( cos ^{2} ( \frac{2}{3} ) + sin ^{2} ( \frac{2}{3} ) ) - sin ( \frac{2}{3} ) ( 20 cos ( \frac{2}{3} ) + 2 sin ( \frac{2}{3} ) ) ) cos ( \frac{t}{6} )}{e ^{\frac{10}{3}} cos ( \frac{2}{3} ) ( cos ^{2} ( \frac{2}{3} ) + sin ^{2} ( \frac{2}{3} ) )} + \frac{( 20 cos ( \frac{2}{3} ) + 2 sin ( \frac{2}{3} ) ) sin ( \frac{t}{6} )}{e ^{\frac{10}{3}} ( cos ^{2} ( \frac{2}{3} ) + sin ^{2} ( \frac{2}{3} ) )})

Solution steps

36 y^{''} (t) - 60 y^{'} (t) + 26 y = 0

Solve linear ODE:

y = e^{\frac{5 t}{6}} (\frac{( 2 ( cos ^{2} ( \frac{2}{3} ) + sin ^{2} ( \frac{2}{3} ) ) - sin ( \frac{2}{3} ) ( 20 cos ( \frac{2}{3} ) + 2 sin ( \frac{2}{3} ) ) ) cos ( \frac{t}{6} )}{e ^{\frac{10}{3}} cos ( \frac{2}{3} ) ( cos ^{2} ( \frac{2}{3} ) + sin ^{2} ( \frac{2}{3} ) )} + \frac{( 20 cos ( \frac{2}{3} ) + 2 sin ( \frac{2}{3} ) ) sin ( \frac{t}{6} )}{e ^{\frac{10}{3}} ( cos ^{2} ( \frac{2}{3} ) + sin ^{2} ( \frac{2}{3} ) )})

y = e^{\frac{5 t}{6}} (\frac{( 2 ( cos ^{2} ( \frac{2}{3} ) + sin ^{2} ( \frac{2}{3} ) ) - sin ( \frac{2}{3} ) ( 20 cos ( \frac{2}{3} ) + 2 sin ( \frac{2}{3} ) ) ) cos ( \frac{t}{6} )}{e ^{\frac{10}{3}} cos ( \frac{2}{3} ) ( cos ^{2} ( \frac{2}{3} ) + sin ^{2} ( \frac{2}{3} ) )} + \frac{( 20 cos ( \frac{2}{3} ) + 2 sin ( \frac{2}{3} ) ) sin ( \frac{t}{6} )}{e ^{\frac{10}{3}} ( cos ^{2} ( \frac{2}{3} ) + sin ^{2} ( \frac{2}{3} ) )})

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

\int 250 (e^{- 0.2 x} - 1) d x

\frac{d}{d x} (\frac{x}{( x ^{2} + y ^{2} )})

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

(x^{2} + 49) \frac{d y}{d x} = 7

What is the solution for 36y^{''}-60y^'+26y=0,y(4)=2,y^'(4)=5 ?
The solution for 36y^{''}-60y^'+26y=0,y(4)=2,y^'(4)=5 is