What is the solution for t^4*y^'=7y+4 ?

The solution for t^4*y^'=7y+4 is y=(e^{(-7+21c_{1}t^3)/(3t^3)})/7-4/7

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

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

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

Solution steps

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} y = x^{3} e^{x}

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

What is the solution for t^4*y^'=7y+4 ?
The solution for t^4*y^'=7y+4 is y=(e^{(-7+21c_{1}t^3)/(3t^3)})/7-4/7