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

The solution for 25y^{''}-40y^'+16y=0,y(0)=3,y^'(0)=-2 is y=3e^{(4t)/5}-(22e^{(4t)/5}t)/(5)

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

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

y = 3 e^{\frac{4 t}{5}} - \frac{22 e ^{\frac{4 t}{5}} t}{5}

Solution steps

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

Solve linear ODE:

y = 3 e^{\frac{4 t}{5}} - \frac{22 e ^{\frac{4 t}{5}} t}{5}

y = 3 e^{\frac{4 t}{5}} - \frac{22 e ^{\frac{4 t}{5}} t}{5}

\int cos (5 x) d (5 x) d x

f (x) = \frac{5}{x ^{4}}

\frac{d}{d x} (4 - 2 sin (2 x - \frac{π}{4}))

\frac{\partial}{\partial x} (f (x) g (x))

\frac{\partial}{\partial y} (x^{\frac{1}{4}} y^{\frac{3}{4}})

What is the solution for 25y^{''}-40y^'+16y=0,y(0)=3,y^'(0)=-2 ?
The solution for 25y^{''}-40y^'+16y=0,y(0)=3,y^'(0)=-2 is y=3e^{(4t)/5}-(22e^{(4t)/5}t)/(5)