en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'-y^2-1=0,y(0)=1

Solution

y^{^{'}} - y^{2} - 1 = 0, y (0) = 1

Solution

y = tan (t + \frac{π}{4})

Solution steps

y^{'} (t) - y^{2} - 1 = 0

Solve separable ODE:

y = tan (t + \frac{π}{4})

y = tan (t + \frac{π}{4})

Popular Examples

integral of 7sec^6(t)

\int 7 sec^{6} (t) d t

implicit (dy)/(dx),x=ye^y implicit derivative

\frac{d y}{d x}, x = y e^{y}

sum from n=1 to infinity of 3(-2/3)^n

n = 1 \sum \infty 3 (- \frac{2}{3})^{n}

(dy)/(dx)=(2+y^2)/(2xy),y(1)=1

\frac{d y}{d x} = \frac{2 + y ^{2}}{2 x y}, y (1) = 1

integral of xsqrt(x^2+3)

\int x x^{2} + 3 d x

Frequently Asked Questions (FAQ)

What is the solution for y^'-y^2-1=0,y(0)=1 ?
The solution for y^'-y^2-1=0,y(0)=1 is y=tan(t+pi/4)