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}+(7e^{(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{7 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{7 e ^{\frac{4 t}{5}} t}{5}

y = 2 e^{\frac{4 t}{5}} + \frac{7 e ^{\frac{4 t}{5}} t}{5}

\int \frac{24 x ^{2}}{36 - x ^{2}} d x

x \to 1 lim (6 x + 4)

\frac{( e ^{\frac{s}{2}} )}{s ^{2}}

x = y^{2} - 7, x = 1 - y^{2}, - 2 \leq y \leq 2

\frac{d y}{d t} t + y = 2 t

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}+(7e^{(4t)/5}t)/(5)