What is the solution for y^{''}-4y=(2x+3)e^{2x} ?

The solution for y^{''}-4y=(2x+3)e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-2x}+(e^{2x}x^2)/4+(5e^{2x}x)/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} x ^{2}}{4} + \frac{5 e ^{2 x} x}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} x ^{2}}{4} + \frac{5 e ^{2 x} x}{8}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} x ^{2}}{4} + \frac{5 e ^{2 x} x}{8}

x^{^{''}} + 16 x = cos (4 t), x (0) = 0, x^{^{'}} (0) = 1

\frac{1}{4} y^{^{''}} + 4 y = tan (4 t) - \frac{1}{4} e^{t}

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

(D^{2} - 1) y = x e^{3 x}

y^{^{''}} - 2 y^{^{'}} + y = 15 e^{x}

What is the solution for y^{''}-4y=(2x+3)e^{2x} ?
The solution for y^{''}-4y=(2x+3)e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-2x}+(e^{2x}x^2)/4+(5e^{2x}x)/8