y^'=t^{1/4}-y

Lösung

y^{^{'}} = t^{\frac{1}{4}} - y

y = \frac{Γ ( \frac{1}{\frac{4}{5}} , - ( ( t ^{\frac{1}{4}} ) ^{5} ) ^{\frac{4}{5}} )}{( - 1 ) ^{\frac{1}{4}} e ^{t}} + \frac{c _{1}}{e ^{t}}

Schritte zur Lösung

y^{'} (t) = t^{\frac{1}{4}} - y

Solve linear ODE:

y = \frac{Γ ( \frac{1}{\frac{4}{5}} , - ( ( t ^{\frac{1}{4}} ) ^{5} ) ^{\frac{4}{5}} )}{( - 1 ) ^{\frac{1}{4}} e ^{t}} + \frac{c _{1}}{e ^{t}}

y = \frac{Γ ( \frac{1}{\frac{4}{5}} , - ( ( t ^{\frac{1}{4}} ) ^{5} ) ^{\frac{4}{5}} )}{( - 1 ) ^{\frac{1}{4}} e ^{t}} + \frac{c _{1}}{e ^{t}}

\int_{- 7}^{7} (49 - x^{2}) d x

\frac{d}{d x} (ln (\frac{x + 3}{e ^{3 x}}))

\frac{\partial}{\partial x} (\frac{x}{9 + y})

d er i v a t i v e y = 7 x^{3} - 6 x^{2} - \frac{8}{x}

4 y^{^{''}} - y^{^{'}} + y = 0