en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

y = c_{1} e^{x} + c_{2} + e^{x} x

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} + e^{x} x

y = c_{1} e^{x} + c_{2} + e^{x} x

Popular Examples

y^{''}+4y^'+4y=8e^{-2x}

y^{^{''}} + 4 y^{^{'}} + 4 y = 8 e^{- 2 x}

y^{''}-8y^'+16y=e^{4x}ln(9+x^2)

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{4 x} ln (9 + x^{2})

(d^2y)/(dx^2)-8(dy)/(dx)+16y=e^{4x}

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 16 y = e^{4 x}

y^{''}+y=5xsin(x)

y^{^{''}} + y = 5 x sin (x)

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

Frequently Asked Questions (FAQ)

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