en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'-6y=e^t,y(0)=1,y^'(0)=1

Solution

y^{^{''}} - y^{^{'}} - 6 y = e^{t}, y (0) = 1, y^{^{'}} (0) = 1

Solution

y = (- \frac{7}{15} + \frac{7}{6}) e^{3 t} + \frac{7}{15} e^{- 2 t} - \frac{1}{6} e^{t}

Solution steps

y^{''} (t) - y^{'} (t) - 6 y = e^{t}

Solve linear ODE:

y = (- \frac{7}{15} + \frac{7}{6}) e^{3 t} + \frac{7}{15} e^{- 2 t} - \frac{1}{6} e^{t}

y = (- \frac{7}{15} + \frac{7}{6}) e^{3 t} + \frac{7}{15} e^{- 2 t} - \frac{1}{6} e^{t}

Graph

Popular Examples

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

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

y^{''}+3y^'+2y=e^x,y(0)=0,y^'(0)=3

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{x}, y (0) = 0, y^{^{'}} (0) = 3

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

y^{^{''}} - y^{^{'}} + \frac{y}{4} = 3 + e^{\frac{x}{2}}

y^{''}+10y^'+16y=40cos(4x)

y^{^{''}} + 10 y^{^{'}} + 16 y = 40 cos (4 x)

y^{''}-8y^'+6y=3x^2-8

y^{^{''}} - 8 y^{^{'}} + 6 y = 3 x^{2} - 8

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-y^'-6y=e^t,y(0)=1,y^'(0)=1 ?
The solution for y^{''}-y^'-6y=e^t,y(0)=1,y^'(0)=1 is y=(-7/15+7/6)e^{3t}+7/15 e^{-2t}-1/6 e^t