What is the solution for 16y^{''}-40y^'+25y=0,y(0)=3,y^'(0)=-9/4 ?

The solution for 16y^{''}-40y^'+25y=0,y(0)=3,y^'(0)=-9/4 is y=3e^{(5t)/4}-6e^{(5t)/4}t

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16y^{''}-40y^'+25y=0,y(0)=3,y^'(0)=-9/4

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - \frac{9}{4}

y = 3 e^{\frac{5 t}{4}} - 6 e^{\frac{5 t}{4}} t

Solution steps

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

Solve linear ODE:

y = 3 e^{\frac{5 t}{4}} - 6 e^{\frac{5 t}{4}} t

y = 3 e^{\frac{5 t}{4}} - 6 e^{\frac{5 t}{4}} t

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

y^{^{'''}} - y^{^{''}} - 2 y^{^{'}} = 0

3 y^{^{''}} - 2 y^{^{'}} + y = 0

\frac{d ^{2} y}{d x ^{2}} - 5 y = 0

y^{^{''''}} + 8 y^{^{'''}} + 16 y^{^{''}} = 0

What is the solution for 16y^{''}-40y^'+25y=0,y(0)=3,y^'(0)=-9/4 ?
The solution for 16y^{''}-40y^'+25y=0,y(0)=3,y^'(0)=-9/4 is y=3e^{(5t)/4}-6e^{(5t)/4}t