en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

y = e^{t} (- \frac{7}{25} cos (2 t) + \frac{51}{25} sin (2 t)) + \frac{t}{5} + \frac{7}{25}

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 1 + t

Solve linear ODE:

y = e^{t} (- \frac{7}{25} cos (2 t) + \frac{51}{25} sin (2 t)) + \frac{t}{5} + \frac{7}{25}

y = e^{t} (- \frac{7}{25} cos (2 t) + \frac{51}{25} sin (2 t)) + \frac{t}{5} + \frac{7}{25}

Popular Examples

x^{''}+4x^'+5x=13sin(2t)

x^{^{''}} + 4 x^{^{'}} + 5 x = 13 sin (2 t)

y^{''}-3y^'+4y=16x-12e^{2x}

y^{^{''}} - 3 y^{^{'}} + 4 y = 16 x - 12 e^{2 x}

y^{''}-7y^'=(x-1)^2

y^{^{''}} - 7 y^{^{'}} = (x - 1)^{2}

y^{''}-2y^'+y=cos(x)

y^{^{''}} - 2 y^{^{'}} + y = cos (x)

y^{''}+y=8cos(2x)-4sin(x)

y^{^{''}} + y = 8 cos (2 x) - 4 sin (x)

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-2y^'+5y=1+t,y(0)=0,y^'(0)=4 ?
The solution for y^{''}-2y^'+5y=1+t,y(0)=0,y^'(0)=4 is y=e^t(-7/25 cos(2t)+51/25 sin(2t))+t/5+7/25