(t^2+x^2)dt+(tx)dx=0

Lösung

(t^{2} + x^{2}) d t + (t x) d x = 0

x = \frac{- t ^{4} + c _{1}}{2 t}, x = - \frac{- t ^{4} + c _{1}}{2 t}

Schritte zur Lösung

(t^{2} + x^{2}) d t + (t x) d x = 0

Solve Bernoulli ODE:

x = \frac{- t ^{4} + c _{1}}{2 t}, x = - \frac{- t ^{4} + c _{1}}{2 t}

x = \frac{- t ^{4} + c _{1}}{2 t}, x = - \frac{- t ^{4} + c _{1}}{2 t}

\frac{d}{d x} (ln (x x^{2} - 7))

in v erse l a pl a ce \frac{1}{s ^{2} \cdot ( s ^{2} + 4 )}

\frac{\partial}{\partial x} (sin (5 x - 9 y))

\frac{d}{d x} (x^{5} + 3)

y = x^{2} + x + 1