y^'=(1+t)*sqrt(y),y(0)=1

Lösung

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

y = t + \frac{t ^{4}}{16} + \frac{t ^{3}}{4} + \frac{3 t ^{2}}{4} + 1

Schritte zur Lösung

y^{'} (t) = (1 + t) \cdot y

Solve separable ODE:

y = t + \frac{t ^{4}}{16} + \frac{t ^{3}}{4} + \frac{3 t ^{2}}{4} + 1

y = t + \frac{t ^{4}}{16} + \frac{t ^{3}}{4} + \frac{3 t ^{2}}{4} + 1

\int \frac{1}{sec ( x )} d x

\frac{d}{d x} ((x^{2} + 1) (x^{3} + x^{2} + 1))

\frac{\partial}{\partial x} (e^{6 x})

x \to 0 - lim (\frac{0}{x})

\frac{\partial}{\partial x} (y^{4} e^{- x})