What is the solution for 25y^{''}-40y^'+16y=0,y(0)=2,y^'(0)=-3 ?

The solution for 25y^{''}-40y^'+16y=0,y(0)=2,y^'(0)=-3 is y=2e^{(4t)/5}-(23e^{(4t)/5}t)/(5)

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25y^{''}-40y^'+16y=0,y(0)=2,y^'(0)=-3

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = - 3

y = 2 e^{\frac{4 t}{5}} - \frac{23 e ^{\frac{4 t}{5}} t}{5}

Solution steps

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

Solve linear ODE:

y = 2 e^{\frac{4 t}{5}} - \frac{23 e ^{\frac{4 t}{5}} t}{5}

y = 2 e^{\frac{4 t}{5}} - \frac{23 e ^{\frac{4 t}{5}} t}{5}

\int 5 sin^{3} (x) cos^{7} (x) d x

\int_{1}^{2} ln x^{2} + 1 d x

\frac{d}{d x} (ln (4 x - 1))

\int_{4}^{5} \frac{1}{5 x - 1} d x

x y^{^{'}} + 3 y = \frac{4 e ^{2 x}}{x ^{2}}

What is the solution for 25y^{''}-40y^'+16y=0,y(0)=2,y^'(0)=-3 ?
The solution for 25y^{''}-40y^'+16y=0,y(0)=2,y^'(0)=-3 is y=2e^{(4t)/5}-(23e^{(4t)/5}t)/(5)