What is the solution for y^'=(ty^2-1)y,y(1)= 1/(sqrt(2)) ?

The solution for y^'=(ty^2-1)y,y(1)= 1/(sqrt(2)) is y=sqrt(2/(2t+1+e^{2(t-1)))}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(ty^2-1)y,y(1)= 1/(sqrt(2))

y^{^{'}} = (t y^{2} - 1) y, y (1) = \frac{1}{2}

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

Solution steps

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

Solve Bernoulli ODE:

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

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

\int 1 - e^{3 x} d x

f (x) = \frac{3}{x}, (3, 1)

x^{2} y^{^{'}} + 4 x y = \frac{sin ( 2 x )}{x ^{2}}

\frac{d}{d x} (C e^{- x})

\int \frac{w}{( 1 + w ) ^{\frac{3}{4}}} d w

What is the solution for y^'=(ty^2-1)y,y(1)= 1/(sqrt(2)) ?
The solution for y^'=(ty^2-1)y,y(1)= 1/(sqrt(2)) is y=sqrt(2/(2t+1+e^{2(t-1)))}