t^4*y^'=7y+4

Lösung

t^{4} \cdot y^{^{'}} = 7 y + 4

y = \frac{e ^{\frac{- 7 + 21 c _{1} t ^{3}}{3 t ^{3}}}}{7} - \frac{4}{7}

Schritte zur Lösung

t^{4} \cdot y^{'} (t) = 7 y + 4

Solve separable ODE:

y = \frac{e ^{\frac{- 7 + 21 c _{1} t ^{3}}{3 t ^{3}}}}{7} - \frac{4}{7}

y = \frac{e ^{\frac{- 7 + 21 c _{1} t ^{3}}{3 t ^{3}}}}{7} - \frac{4}{7}

\frac{d y}{d x} y = x^{2} + 4

\frac{d y}{d x} = \frac{5 e ^{x^{2}} x + 1}{y}

y - 2 y^{^{'}} + 6 y = 0

\frac{d y}{d x} = \frac{sin ( x )}{y ^{2}}

x^{2} \cdot y^{^{'}} - y = 0