What is the solution for (2+t)y^'=3y ?

The solution for (2+t)y^'=3y is y=8e^{c_{1}}+12e^{c_{1}}t+6e^{c_{1}}t^2+e^{c_{1}}t^3

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(2+t)y^'=3y

(2 + t) y^{^{'}} = 3 y

y = 8 e^{c_{1}} + 12 e^{c_{1}} t + 6 e^{c_{1}} t^{2} + e^{c_{1}} t^{3}

Solution steps

(2 + t) y^{'} (t) = 3 y

Solve separable ODE:

y = 8 e^{c_{1}} + 12 e^{c_{1}} t + 6 e^{c_{1}} t^{2} + e^{c_{1}} t^{3}

y = 8 e^{c_{1}} + 12 e^{c_{1}} t + 6 e^{c_{1}} t^{2} + e^{c_{1}} t^{3}

y^{^{'}} = \frac{3 x 16 + y ^{2}}{y}

y^{^{'}} = \frac{6 sin ( x )}{sin ( y )}

(\frac{d x}{d t}) = - x

2 v^{2} y^{2} d y + 2 v y^{3} d v + v^{2} y^{2} d y + y^{2} d y = 0

\frac{d v}{d t} = 4 - 7 v

What is the solution for (2+t)y^'=3y ?
The solution for (2+t)y^'=3y is y=8e^{c_{1}}+12e^{c_{1}}t+6e^{c_{1}}t^2+e^{c_{1}}t^3