What is the solution for y^{''}-4y^'=3t ?

The solution for y^{''}-4y^'=3t is y=c_{1}e^{4t}+c_{2}-(3t^2)/8-(3t)/(16)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y^'=3t

y^{^{''}} - 4 y^{^{'}} = 3 t

y = c_{1} e^{4 t} + c_{2} - \frac{3 t ^{2}}{8} - \frac{3 t}{16}

Solution steps

y^{''} (t) - 4 y^{'} (t) = 3 t

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} - \frac{3 t ^{2}}{8} - \frac{3 t}{16}

y = c_{1} e^{4 t} + c_{2} - \frac{3 t ^{2}}{8} - \frac{3 t}{16}

\int 1 - cos (x) d x

\int t^{4} + 1 d t

\int csc^{2} (9 θ + 2) d θ

\int sin (4 t) cos (5 t) d t

\int \frac{x + 18}{x ^{2} + 6 x + 13} d x

What is the solution for y^{''}-4y^'=3t ?
The solution for y^{''}-4y^'=3t is y=c_{1}e^{4t}+c_{2}-(3t^2)/8-(3t)/(16)