What is the solution for y^{''}-4y^'+5y=1+t,y(0)=0,y^'(0)=4 ?

The solution for y^{''}-4y^'+5y=1+t,y(0)=0,y^'(0)=4 is y=e^{2t}(-9/25 cos(t)+113/25 sin(t))+t/5+9/25

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y^'+5y=1+t,y(0)=0,y^'(0)=4

y^{^{''}} - 4 y^{^{'}} + 5 y = 1 + t, y (0) = 0, y^{^{'}} (0) = 4

y = e^{2 t} (- \frac{9}{25} cos (t) + \frac{113}{25} sin (t)) + \frac{t}{5} + \frac{9}{25}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 5 y = 1 + t

Solve linear ODE:

y = e^{2 t} (- \frac{9}{25} cos (t) + \frac{113}{25} sin (t)) + \frac{t}{5} + \frac{9}{25}

y = e^{2 t} (- \frac{9}{25} cos (t) + \frac{113}{25} sin (t)) + \frac{t}{5} + \frac{9}{25}

y^{^{''}} - 2 y + 3 = 0

(d + 1)^{2} \cdot y = 2 e^{- 2} x

y^{^{''}} + 4 y^{^{'}} + 5 y = 35 e^{- 4} x

y^{^{''}} + 9 y = sin^{2} (t)

(d^{5} - d) y = 2 x + 2^{x}

What is the solution for y^{''}-4y^'+5y=1+t,y(0)=0,y^'(0)=4 ?
The solution for y^{''}-4y^'+5y=1+t,y(0)=0,y^'(0)=4 is y=e^{2t}(-9/25 cos(t)+113/25 sin(t))+t/5+9/25