Answer to Question #136406 in Differential Equations for Arun

Question

y'+2ty=4t^3

Answer 1

$y'+2ty=4t^3$

Multiply both parts of the equation by $e^{t^2}$:

$e^{t^2}y'+2te^{t^2}y=4t^3e^{t^2}$

$(e^{t^2}y)'=4t^3e^{t^2}$

$e^{t^2}y=4\int t^3e^{t^2}dt=2\int t^2e^{t^2}dt^2=$

$\ \ \ \ \ |t^2=x|$

$=2\int xe^{x}dx=$

$\ \ \ \ \ |u=x, dv=e^xdx, du=dx, v=e^x|$

$=2xe^x-2\int e^xdx+C=2xe^x-2e^x+C=$

$\ \ \ \ |x=t^2|$

$=2e^{t^2}(t^2-1)+C$

Therefore, we have the following equation:

$e^{t^2}y=2e^{t^2}(t^2-1)+C$

Finally, the general solution of the differential equation is the following:

$y=2(t^2-1)+Ce^{-t^2}$