d/(dt)((x)x)=sqrt(t^2+1)

Lösung

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

x = \frac{c _{1} + t 1 + t ^{2} + ln ( t + 1 + t ^{2} )}{2}, x = - \frac{c _{1} + t 1 + t ^{2} + ln ( t + 1 + t ^{2} )}{2}

Schritte zur Lösung

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

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

xx = \int t^{2} + 1 d t

\int t^{2} + 1 d t = \frac{1}{2} t 1 + t^{2} + \frac{1}{2} ln (t + 1 + t^{2}) + c_{1}

xx = \frac{1}{2} t 1 + t^{2} + \frac{1}{2} ln (t + 1 + t^{2}) + c_{1}

Isoliere

x : x = \frac{2 c _{1} + t 1 + t ^{2} + ln ( t + 1 + t ^{2} )}{2}, x = - \frac{2 c _{1} + t 1 + t ^{2} + ln ( t + 1 + t ^{2} )}{2}

x = \frac{2 c _{1} + t 1 + t ^{2} + ln ( t + 1 + t ^{2} )}{2}, x = - \frac{2 c _{1} + t 1 + t ^{2} + ln ( t + 1 + t ^{2} )}{2}

Vereinfache

x = \frac{c _{1} + t 1 + t ^{2} + ln ( t + 1 + t ^{2} )}{2}, x = - \frac{c _{1} + t 1 + t ^{2} + ln ( t + 1 + t ^{2} )}{2}

y^{^{'}} = 8 x sec (y)

(5 x + 4) + (5 y - 5) y^{^{'}} = 0

y^{^{'}} (x) = (1 - y^{2}) tan (x), y (0) = 3

\frac{d y}{d x} = \frac{- 0.9}{cos ( y )}, y (0) = \frac{π}{3}

e^{2 x} d y + 2 y d x = 0