Solve by using the Laplace Transform Method?y′+y=6et2, y(0)=3

Given first order differential equation
${\displaystyle y^{\prime }+y=6e^{t^2},\dots \left(i\right)}$
${\displaystyle y\left(0\right)=3}$
claim- to check whether the differential equation can be solved using laplace transform or not
solution
applying laplace both sided
${\displaystyle L\left|y^{\prime }\right|+L\left|y\right|=L\left|6e^{t^2}\right|}$
as we know
${\displaystyle L\left|y^{\prime }\right|=sy\left(s\right)-y\left(0\right)}$
${\displaystyle L\left|y\right|=y\left(s\right)}$
equation (i) becomes,
${\displaystyle sy\left(s\right)-y\left(0\right)+y\left(s\right)=6L\left[e^{t^2}\right]}$
${\displaystyle L\left[e^{t^2}\right]=\frac{1}{2}\sqrt{\pi }{e}^{\frac{-s^2}{4}}erf\left(\frac{\pi }{2}\right)-\frac{1}{2}i\sqrt{\pi }{e}^{\frac{-s^2}{4}}}$
thus, we get that the first order differential equation cannot be solved using laplace transform, because the laplace transform of right hand side is not in the form of standard function.