What is the solution for 16y^{''}-40y^'+25y=0,y(0)=3,y^'(0)=-9/4 ?

The solution for 16y^{''}-40y^'+25y=0,y(0)=3,y^'(0)=-9/4 is y=3e^{(5t)/4}-6e^{(5t)/4}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

16y^{''}-40y^'+25y=0,y(0)=3,y^'(0)=-9/4

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - \frac{9}{4}

y = 3 e^{\frac{5 t}{4}} - 6 e^{\frac{5 t}{4}} t

Solution steps

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

Solve linear ODE:

y = 3 e^{\frac{5 t}{4}} - 6 e^{\frac{5 t}{4}} t

y = 3 e^{\frac{5 t}{4}} - 6 e^{\frac{5 t}{4}} t

What is the solution for 16y^{''}-40y^'+25y=0,y(0)=3,y^'(0)=-9/4 ?
The solution for 16y^{''}-40y^'+25y=0,y(0)=3,y^'(0)=-9/4 is y=3e^{(5t)/4}-6e^{(5t)/4}t