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)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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}

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = e^{2 x}

y^{^{''}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (0) = α, y^{^{'}} (0) = 5

y^{^{''}} - 10 y^{^{'}} + 61 y = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = 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}+(7e^{(4t)/5}t)/(5)