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}+c_{2}e^{(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}} + c_{2} 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}} + c_{2} e^{\frac{5 t}{4}} t

y = 3 e^{\frac{5 t}{4}} + c_{2} e^{\frac{5 t}{4}} t

y^{^{''}} - 2 y^{^{'}} + 1 y = 0, y (0) = 7, y^{^{'}} (0) = 4

\frac{d ^{2} y}{d x ^{2}} = a \frac{d y}{d x}

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

u^{^{''}} + a H (t) = 0

y^{^{''}} + 2 y^{^{'}} + 37 y = 0, y (0) = 2, y^{^{'}} (0) = - 26

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}+c_{2}e^{(5t)/4}t