What is the solution for y^{''}-4y=t^2+1 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y=t^2+1

y^{^{''}} - 4 y = t^{2} + 1

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

Solution steps

y^{''} (t) - 4 y = t^{2} + 1

Solve linear ODE:

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

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

5 y^{^{''}} - 10 y^{^{'}} + 10 y = e x sec (x)

y^{^{''}} + 3 y^{^{'}} + 2 y = 20 e^{- t}

y^{^{''}} + 9 y = 2 x^{2}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 8 y = cos (x)

2 \frac{d ^{2} y}{d x ^{2}} + 1 \frac{d y}{d x} + 1 y = 2

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