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

解

\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}

解答ステップ

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

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

であれば

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}

分離する

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}

簡素化

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}