What is the solution for (t^2+4)y^'+2ty=t^2(t^2+4) ?

The solution for (t^2+4)y^'+2ty=t^2(t^2+4) is y=(t^5)/(5(t^2+4))+(4t^3)/(3(t^2+4))+(c_{1})/(t^2+4)

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(t^2+4)y^'+2ty=t^2(t^2+4)

(t^{2} + 4) y^{^{'}} + 2 t y = t^{2} (t^{2} + 4)

y = \frac{t ^{5}}{5 ( t ^{2} + 4 )} + \frac{4 t ^{3}}{3 ( t ^{2} + 4 )} + \frac{c _{1}}{t ^{2} + 4}

Solution steps

(t^{2} + 4) y^{'} (t) + 2 t y = t^{2} (t^{2} + 4)

Solve linear ODE:

y = \frac{t ^{5}}{5 ( t ^{2} + 4 )} + \frac{4 t ^{3}}{3 ( t ^{2} + 4 )} + \frac{c _{1}}{t ^{2} + 4}

y = \frac{t ^{5}}{5 ( t ^{2} + 4 )} + \frac{4 t ^{3}}{3 ( t ^{2} + 4 )} + \frac{c _{1}}{t ^{2} + 4}

y = x^{2} + 3, (4, 19)

\int \frac{2 x ^{2} + x - 1}{x ( x ^{2} + 1 )} d x

\frac{d}{d y} (\frac{2 x ^{2} - x y + 2 y ^{2}}{x - y})

x^{2} + 2 y^{2} - 2 x - 8 y + 6 = 0 (0.3)

\frac{1}{2 s ^{2} + 1}

What is the solution for (t^2+4)y^'+2ty=t^2(t^2+4) ?
The solution for (t^2+4)y^'+2ty=t^2(t^2+4) is y=(t^5)/(5(t^2+4))+(4t^3)/(3(t^2+4))+(c_{1})/(t^2+4)