What is the solution for (5t+2y+6)dt-dy=0 ?

The solution for (5t+2y+6)dt-dy=0 is y=-(5t)/2-17/4+c_{1}e^{2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(5t+2y+6)dt-dy=0

(5 t + 2 y + 6) d t - d y = 0

y = - \frac{5 t}{2} - \frac{17}{4} + c_{1} e^{2 t}

Solution steps

(5 t + 2 y + 6) d t - d y = 0

Solve linear ODE:

y = - \frac{5 t}{2} - \frac{17}{4} + c_{1} e^{2 t}

y = - \frac{5 t}{2} - \frac{17}{4} + c_{1} e^{2 t}

\frac{d}{d x} (\frac{d y}{d x}) + 1 = 0

\int \frac{x ^{3} + 1}{x ^{3}} \frac{1}{x ^{4}} d x

f (y) = ln (y)

\int_{0}^{l n (π)} 2 x e^{x^{2}} sin (e^{x^{2}}) d x

\frac{d}{d x} (ln ((tan (x))^{3}))

What is the solution for (5t+2y+6)dt-dy=0 ?
The solution for (5t+2y+6)dt-dy=0 is y=-(5t)/2-17/4+c_{1}e^{2t}