What is the solution for y^{''}-3y^'=e^{3t} ?

The solution for y^{''}-3y^'=e^{3t} is y=c_{1}e^{3t}+c_{2}+(e^{3t}t)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-3y^'=e^{3t}

y^{^{''}} - 3 y^{^{'}} = e^{3 t}

y = c_{1} e^{3 t} + c_{2} + \frac{e ^{3 t} t}{3}

Solution steps

y^{''} (t) - 3 y^{'} (t) = e^{3 t}

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} + \frac{e ^{3 t} t}{3}

y = c_{1} e^{3 t} + c_{2} + \frac{e ^{3 t} t}{3}

k = 1 \sum \infty \frac{5 + ( - 2 ) ^{k}}{4 ^{k + 1}}

\int \frac{sin ^{7} ( 12 x )}{cos ^{4} ( 12 x )} d x

f (x) = ln (\frac{7 ( x + 3 )}{x})

\int x^{2} e^{- \frac{x}{3}} d x

\int_{1}^{5} \frac{1}{4 x + 5} d x

What is the solution for y^{''}-3y^'=e^{3t} ?
The solution for y^{''}-3y^'=e^{3t} is y=c_{1}e^{3t}+c_{2}+(e^{3t}t)/3