What is expand ((dy)/(dt)+2/t y=t^2-t+1)t^2 ?

The solution to expand ((dy)/(dt)+2/t y=t^2-t+1)t^2 is t^2(dy)/(dt)+2ty=-t^3+2t^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand ((dy)/(dt)+2/t y=t^2-t+1)t^2

expand

(\frac{d y}{d t} + \frac{2}{t} y = t^{2} - t + 1) t^{2}

t^{2} \frac{d y}{d t} + 2 t y = - t^{3} + 2 t^{2}

Solution steps

(\frac{d y}{d t} + \frac{2}{t} y = t^{2} - t + 1) t^{2}

= t^{2} (\frac{d y}{d t} + = \frac{2}{t} t^{2} y - t + 1)

Distribute parentheses

t^{2} \frac{d y}{d t} + t^{2} \frac{2}{t} y = t^{2} + t^{2} (- t) + t^{2} \cdot 1

Apply minus-plus rules

+ (- a) = - a

t^{2} \frac{d y}{d t} + \frac{2}{t} t^{2} y = t^{2} - t^{2} t + 1 \cdot t^{2}

Simplify

t^{2} \frac{d y}{d t} + \frac{2}{t} t^{2} y = t^{2} - t^{2} t + 1 \cdot t^{2} : t^{2} \frac{d y}{d t} + 2 t y = - t^{3} + 2 t^{2}

t^{2} \frac{d y}{d t} + 2 t y = - t^{3} + 2 t^{2}

(2 - \frac{1}{3} x)^{4}

A (- 7, 3) B (\frac{1}{2}, - \frac{3}{4})

- (\frac{1}{8}) (y - 5 - 5025 e^{- 8 x})

\frac{- 11 x - 35}{( x + 2 ) ( 5 x - 4 )}

(3 + 4 sin (2000 π t))^{2}

What is expand ((dy)/(dt)+2/t y=t^2-t+1)t^2 ?
The solution to expand ((dy)/(dt)+2/t y=t^2-t+1)t^2 is t^2(dy)/(dt)+2ty=-t^3+2t^2