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

The solution for 25y^{''}-40y^'+16y=0,y(0)=4,y^'(0)=-3 is y=4e^{(4t)/5}-(31e^{(4t)/5}t)/(5)

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

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

y = 4 e^{\frac{4 t}{5}} - \frac{31 e ^{\frac{4 t}{5}} t}{5}

Solution steps

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

Solve linear ODE:

y = 4 e^{\frac{4 t}{5}} - \frac{31 e ^{\frac{4 t}{5}} t}{5}

y = 4 e^{\frac{4 t}{5}} - \frac{31 e ^{\frac{4 t}{5}} t}{5}

3 y^{^{'''}} + 16 y^{^{''}} + 26 y^{^{'}} + 7 y = 0

(y^{^{''}} + a y^{^{'}}) = b y

y^{^{''}} - 16 y = 0, y (0) = 3, y^{^{'}} (0) = 5

\frac{d ^{4} y}{d x ^{4}} - \frac{d ^{2} y}{d x ^{2}} + 4 y = 0

\frac{d ^{2} y}{d x ^{2}} = 0.15 y, y (0) = 240, y (10) = 150

What is the solution for 25y^{''}-40y^'+16y=0,y(0)=4,y^'(0)=-3 ?
The solution for 25y^{''}-40y^'+16y=0,y(0)=4,y^'(0)=-3 is y=4e^{(4t)/5}-(31e^{(4t)/5}t)/(5)