What is the solution for ydy+(1+y^2)sin(t)dt=0 ?

The solution for ydy+(1+y^2)sin(t)dt=0 is y=sqrt(e^{2cos(t)+c_{1)}-1},y=-sqrt(e^{2cos(t)+c_{1)}-1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

ydy+(1+y^2)sin(t)dt=0

y d y + (1 + y^{2}) sin (t) d t = 0

y = e^{2 c o s (t) + c_{1}} - 1, y = - e^{2 c o s (t) + c_{1}} - 1

Solution steps

y d y + (1 + y^{2}) sin (t) d t = 0

Solve separable ODE:

y = e^{2 c o s (t) + c_{1}} - 1, y = - e^{2 c o s (t) + c_{1}} - 1

y = e^{2 c o s (t) + c_{1}} - 1, y = - e^{2 c o s (t) + c_{1}} - 1

\frac{d}{d x} (e x)

\frac{d}{d x} ((3 x - 2)^{2})

y = 1 + x

y = x, y = 2 x, x = 1

\int 14 tan^{3} (x) sec (x) d x

What is the solution for ydy+(1+y^2)sin(t)dt=0 ?
The solution for ydy+(1+y^2)sin(t)dt=0 is y=sqrt(e^{2cos(t)+c_{1)}-1},y=-sqrt(e^{2cos(t)+c_{1)}-1}