What is the solution for (t^2+9)y^'+2ty=t^2(t^2+9) ?

The solution for (t^2+9)y^'+2ty=t^2(t^2+9) is y=(t^5)/(5(t^2+9))+(3t^3)/(t^2+9)+(c_{1})/(t^2+9)

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(t^2+9)y^'+2ty=t^2(t^2+9)

(t^{2} + 9) y^{^{'}} + 2 t y = t^{2} (t^{2} + 9)

y = \frac{t ^{5}}{5 ( t ^{2} + 9 )} + \frac{3 t ^{3}}{t ^{2} + 9} + \frac{c _{1}}{t ^{2} + 9}

Solution steps

(t^{2} + 9) y^{'} (t) + 2 t y = t^{2} (t^{2} + 9)

Solve linear ODE:

y = \frac{t ^{5}}{5 ( t ^{2} + 9 )} + \frac{3 t ^{3}}{t ^{2} + 9} + \frac{c _{1}}{t ^{2} + 9}

y = \frac{t ^{5}}{5 ( t ^{2} + 9 )} + \frac{3 t ^{3}}{t ^{2} + 9} + \frac{c _{1}}{t ^{2} + 9}

What is the solution for (t^2+9)y^'+2ty=t^2(t^2+9) ?
The solution for (t^2+9)y^'+2ty=t^2(t^2+9) is y=(t^5)/(5(t^2+9))+(3t^3)/(t^2+9)+(c_{1})/(t^2+9)