Explanation
Given, $${ y }'-4{ y '}-12y=t{ e }^{ 4t }$$
$$\Rightarrow -{ 3y }'-12y=t{ e }^{ 4t }\\ \Rightarrow { y }'+4y=-\dfrac{t{ e }^{ 4t }}{3}\\ \Rightarrow \dfrac { dy }{ dt } +4y=-\dfrac{1}{3}t{ e }^{ 4t }$$
This is a linear form of differential equation with
$$\displaystyle{\text{I.F.}={ e }^{ \int 4dt }={ e }^{ 4t }}$$
Solution of the differential equation is
$$\displaystyle{y(\text{I.F.})=\int { Q( } \text{I.F.})dt\\ y{ e }^{ 4t }=\int { \dfrac { -t{ e }^{ 4t } }{ 3 } { e }^{ 4t } } dt\\ y{ e }^{ 4t }=\dfrac { -1 }{ 3 } \int { t{ e }^{ 8t } } dt\\ y{ e }^{ 4t }=\dfrac { -1 }{ 3 } \left\{ t\int { { e }^{ 8t } } -\int { \dfrac { d }{ dt } (t)\int { { e }^{ 8t } } } \right\} \\ y{ e }^{ 4t }=\dfrac { -1 }{ 3 } \left\{ \dfrac { t{ e }^{ 8t } }{ 8 } -\int { \dfrac { { e }^{ 8t } }{ 8 } } \right\} \\ y{ e }^{ 4t }=\dfrac { -1 }{ 3 } \left\{ \dfrac { t{ e }^{ 8t } }{ 8 } -\dfrac { { e }^{ 8t } }{ 64 } \right\} \\ y{ e }^{ 4t }=\dfrac { -{ e }^{ 8t } }{ 3 } \left\{ \dfrac { 8t-1 }{ 64 } \right\} \\ y=\dfrac { -{ e }^{ 4t } }{ 192 } (8t-1)}$$
So, option D is correct.
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