What is the solution for y^{''}-y^'=0,y(0)=10 ?

The solution for y^{''}-y^'=0,y(0)=10 is y=(10-c_{2})e^t+c_{2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'=0,y(0)=10

y^{^{''}} - y^{^{'}} = 0, y (0) = 10

y = (10 - c_{2}) e^{t} + c_{2}

Solution steps

y^{''} (t) - y^{'} (t) = 0

Solve linear ODE:

y = (10 - c_{2}) e^{t} + c_{2}

y = (10 - c_{2}) e^{t} + c_{2}

\int \frac{4}{3 - x} d x

\frac{\partial}{\partial x} (x^{2} - x y - y^{2})

\frac{d y}{d x} = x y + 3 x + 5 y + 15

\frac{d}{d x} (\frac{5 ^{x}}{1 + 5 ^{2 x}})

\int_{\frac{1}{7}}^{5} 16 x ln (7 x) d x

What is the solution for y^{''}-y^'=0,y(0)=10 ?
The solution for y^{''}-y^'=0,y(0)=10 is y=(10-c_{2})e^t+c_{2}