What is the solution for (dx)/(dt)=tx-2x+t-2 ?

The solution for (dx)/(dt)=tx-2x+t-2 is x=e^{(t^2)/2-2t+c_{1}}-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dx)/(dt)=tx-2x+t-2

\frac{d x}{d t} = t x - 2 x + t - 2

x = e^{\frac{t ^{2}}{2} - 2 t + c_{1}} - 1

Solution steps

\frac{d x}{d t} = t x - 2 x + t - 2

Solve separable ODE:

x = e^{\frac{t ^{2}}{2} - 2 t + c_{1}} - 1

x = e^{\frac{t ^{2}}{2} - 2 t + c_{1}} - 1

\int \frac{x}{4 + 9 x ^{2}} d x

x \to 0 lim (\frac{1 - cot ( x )}{\frac{1}{x}})

5 + x^{2}

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

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

What is the solution for (dx)/(dt)=tx-2x+t-2 ?
The solution for (dx)/(dt)=tx-2x+t-2 is x=e^{(t^2)/2-2t+c_{1}}-1