What is the solution for 25y^{''}-40y^'+16y=0,y(0)=4,y^'(0)=-5 ?

The solution for 25y^{''}-40y^'+16y=0,y(0)=4,y^'(0)=-5 is y=4e^{(4t)/5}-(41e^{(4t)/5}t)/(5)

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25y^{''}-40y^'+16y=0,y(0)=4,y^'(0)=-5

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (0) = 4, y^{^{'}} (0) = - 5

y = 4 e^{\frac{4 t}{5}} - \frac{41 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{41 e ^{\frac{4 t}{5}} t}{5}

y = 4 e^{\frac{4 t}{5}} - \frac{41 e ^{\frac{4 t}{5}} t}{5}

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What is the solution for 25y^{''}-40y^'+16y=0,y(0)=4,y^'(0)=-5 ?
The solution for 25y^{''}-40y^'+16y=0,y(0)=4,y^'(0)=-5 is y=4e^{(4t)/5}-(41e^{(4t)/5}t)/(5)