What is the y^'(t)= t/((1+y^2)),y(0)=0 ?

The y^'(t)= t/((1+y^2)),y(0)=0 is t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'(t)= t/((1+y^2)),y(0)=0

y^{^{'}} (t) = \frac{t}{( 1 + y ^{2} )}, y (0) = 0

t

Solution steps

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

Constant function results in a constant

y (0) = 0

= \frac{t}{1 + 0 ^{2}}

Simplify

\frac{t}{1 + 0 ^{2}} : t

= t

y d x + (x - 9) d y = 0, y (10) = 49

\frac{d y}{d t} = 12 - \frac{y}{50}

v y ln (v) d y - y (v d y + y d v) = 0

y^{^{'}} = 0.01398 y, y (0) = 150

\frac{d y}{d x} = (- 2 x + 1) y, y (0) = 6.8