What is the solution for y^'+(10)/(40+2t)y=0 ?

The solution for y^'+(10)/(40+2t)y=0 is y=(c_{1})/(t^5+100t^4+4000t^3+80000t^2+800000t+3200000)

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y^'+(10)/(40+2t)y=0

y^{^{'}} + \frac{10}{40 + 2 t} y = 0

y = \frac{c _{1}}{t ^{5} + 100 t ^{4} + 4000 t ^{3} + 80000 t ^{2} + 800000 t + 3200000}

Solution steps

y^{'} (t) + \frac{10}{40 + 2 t} y = 0

Solve separable ODE:

y = \frac{c _{1}}{t ^{5} + 100 t ^{4} + 4000 t ^{3} + 80000 t ^{2} + 800000 t + 3200000}

y = \frac{c _{1}}{t ^{5} + 100 t ^{4} + 4000 t ^{3} + 80000 t ^{2} + 800000 t + 3200000}

\frac{d y}{d x} = \frac{x + 8}{x y}

(1 - t) \frac{d y}{d t} - y = 0

\frac{d x}{2 x} + \frac{d y}{2 y} = 0

\frac{d y}{d t} = - \frac{1}{50} (y - 25)

e^{y} y^{^{'}} = 4, y (0) = 2

What is the solution for y^'+(10)/(40+2t)y=0 ?
The solution for y^'+(10)/(40+2t)y=0 is y=(c_{1})/(t^5+100t^4+4000t^3+80000t^2+800000t+3200000)