What is the solution for y^{''}-10y^'+25y=4,-e^ty(0)=1,y(1)=0 ?

The solution for y^{''}-10y^'+25y=4,-e^ty(0)=1,y(1)=0 is y=c_{1}e^{5t}+c_{2}te^{5t}+4/25

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-10y^'+25y=4,-e^ty(0)=1,y(1)=0

y^{^{''}} - 10 y^{^{'}} + 25 y = 4, - e^{t} y (0) = 1, y (1) = 0

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{4}{25}

Solution steps

y^{''} (t) - 10 y^{'} (t) + 25 y = 4

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{4}{25}

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{4}{25}

y^{^{''}} + 16 y = 3 e^{3 t}

y^{^{''}} - y^{^{'}} + 2 y = cos (x), y (0) = 1, y^{^{'}} (0) = 0

y^{^{'''}} + 7 y^{^{''}} - 6 y^{^{'}} - 72 y = - 360

y^{^{''}} + 4 y^{^{'}} = 6 e^{3 t} - 3 e^{- t}

y^{^{''}} + 4 y = 2 sin (t)

What is the solution for y^{''}-10y^'+25y=4,-e^ty(0)=1,y(1)=0 ?
The solution for y^{''}-10y^'+25y=4,-e^ty(0)=1,y(1)=0 is y=c_{1}e^{5t}+c_{2}te^{5t}+4/25