What is the solution for 16y^{''}-40y^'+25y=0,y(0)=3,y^'(0)=-94 ?

The solution for 16y^{''}-40y^'+25y=0,y(0)=3,y^'(0)=-94 is y=3e^{(5t)/4}-(391e^{(5t)/4}t)/(4)

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16y^{''}-40y^'+25y=0,y(0)=3,y^'(0)=-94

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - 94

y = 3 e^{\frac{5 t}{4}} - \frac{391 e ^{\frac{5 t}{4}} t}{4}

Solution steps

16 y^{''} (t) - 40 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = 3 e^{\frac{5 t}{4}} - \frac{391 e ^{\frac{5 t}{4}} t}{4}

y = 3 e^{\frac{5 t}{4}} - \frac{391 e ^{\frac{5 t}{4}} t}{4}

x \to 0 lim (x \cdot sin (\frac{π}{x}))

m = \frac{9}{5 + h} - \frac{9}{5}

n = 1 \sum \infty e^{- n} + 1

\frac{d}{d x} (1 + \frac{1}{4 x})

\frac{1}{10} e^{4 t} + \frac{9}{10} e^{- 6 t}

What is the solution for 16y^{''}-40y^'+25y=0,y(0)=3,y^'(0)=-94 ?
The solution for 16y^{''}-40y^'+25y=0,y(0)=3,y^'(0)=-94 is y=3e^{(5t)/4}-(391e^{(5t)/4}t)/(4)