What is the solution for y^{''}-4y^'=e^x ?

The solution for y^{''}-4y^'=e^x is y=c_{1}e^{4x}+c_{2}-1/3 e^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y^'=e^x

y^{^{''}} - 4 y^{^{'}} = e^{x}

y = c_{1} e^{4 x} + c_{2} - \frac{1}{3} e^{x}

Solution steps

y^{''} (x) - 4 y^{'} (x) = e^{x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} - \frac{1}{3} e^{x}

y = c_{1} e^{4 x} + c_{2} - \frac{1}{3} e^{x}

y^{^{''}} + 3 y^{^{'}} + 2 y = t H (t - 3), y (0) = 0, y^{^{'}} (0) = 1

4 \cdot y^{^{''}} - 11 \cdot y^{^{'}} - 3 y = 5 \cdot e^{3 x}

\frac{d ^{2} x}{d t ^{2}} + x = 6 cos (2 t), x (0) = 3

y^{^{''}} + y = 3 sin (2 x)

y^{^{''}} - 12 y^{^{'}} + 36 y = 30 x + 4

What is the solution for y^{''}-4y^'=e^x ?
The solution for y^{''}-4y^'=e^x is y=c_{1}e^{4x}+c_{2}-1/3 e^x