School's out! Exams are over! Enjoy this problem!

${ \quad \quad \quad \quad \quad y }^{ 2 }dt\quad +\quad 2{ t }^{ 2 }ydt\quad +\quad 2{ t }^{ 3 }dy\quad -tydy\quad =\quad 0\\ \Longrightarrow ({ y }^{ 2 }dt\quad -\quad tydy)\quad +\quad (2{ t }^{ 2 }ydt\quad +{ 2t }^{ 3 }dy)\quad =\quad 0\\ \Longrightarrow y{ t }^{ 2 }(\frac { ydt-tdy }{ { t }^{ 2 } } )\quad +\quad 2{ t }^{ 2 }(ydt\quad +\quad tdy)\quad =0\\ \Longrightarrow -y{ t }^{ 2 }(\frac { tdy-ydt }{ { t }^{ 2 } } )\quad +\quad 2{ t }^{ 2 }(ydt\quad +\quad tdy)\quad =0\\ \Longrightarrow -y{ t }^{ 2 }d(\frac { y }{ t } )\quad +(2{ t }^{ 2 })d(yt)=0\\ \Longrightarrow yd(\frac { y }{ t } )\quad =2\quad d(yt) .....(1)$

Now let $\frac { y }{ t } =u\\ yt=v\\ \Longrightarrow { y }^{ 2 }=uv\\ \Longrightarrow y=\sqrt { u } \sqrt { v }$

Substituting the above in equation (1)

$\sqrt { u } \sqrt { v } du=2dv\\ \\ \Longrightarrow \int { \sqrt { u } } du=\int { \frac { 2 }{ \sqrt { v } } } dv\quad \\ \Longrightarrow \frac { 2 }{ 3 } { u }^{ 3/2 }\quad +\quad A=4\sqrt { v } \quad \\ \Longrightarrow 4({ yt) }^{ 1/2 }-\frac { 2 }{ 3 } { (\frac { y }{ t } })^{ 3/2 }+A\quad =0$

Hence the value of a+d+e+f+g+h = $\boxed { 16 }$

$This\quad differential\quad equation\quad (DE)\quad resembles\quad an\quad exact\quad \\ differential\quad equation\quad but\quad it\quad isn't\quad one.\quad That's\quad because\quad it\\ is\quad missing\quad its\quad integration\quad factor\quad (IF).\quad In\quad this\quad problem,\\ you\quad multiply\quad the\quad entire\quad equation\quad by\quad { t }^{ h }{ y }^{ k }\quad and\quad then\quad you\quad \\ test\quad it\quad for\quad matching\quad the\quad exact\quad DE\quad criteria.\\ \\ So,\quad if\quad the\quad equation\quad is\quad like\quad Mdy+Ndt=0\quad right\quad now\quad and\quad \\ you\quad multiply\quad it\quad on\quad both\quad sides\quad by\quad the\quad IF,\quad you\quad should\quad get\\ (Mdy+Ndt){ t }^{ h }{ y }^{ k }=0.\\ This\quad is\quad an\quad exact\quad DE\quad iff\quad \\ \frac { \partial (M{ t }^{ h }{ y }^{ k }) }{ \partial t } =\frac { \partial (N{ t }^{ h }{ y }^{ k }) }{ \partial y } .\\ If\quad you\quad solve\quad this\quad equation,\quad you\quad will\quad get\quad the\quad values\quad for\\ h\quad and\quad k.\\Substitute\quad the \quad values \quad and \quad solve\quad it\quad like\quad an\quad exact\quad differential.\\ If\quad you\quad do\quad it\quad right,\quad you\quad should\quad get:\\ a=4;\quad b=nothing\quad because\quad I\quad forgot\quad to\quad use\quad it.\\ c=1\quad (but\quad it\quad is\quad not\quad included\quad in\quad the\quad sum\quad of\quad a+d+...)\\ d=2;\quad e=2;\quad f=3;\quad g=3;\quad h=2.\quad So\quad the\quad solution\quad should\\ look\quad like:\\ 4\sqrt { ty } -\frac { 2 }{ 3 } (\frac { y }{ t } )^{ { 3 }/{ 2 } }=A.$

Note: Here, M and N are functions of y and t.

I have seen this problem before so i knew the answer , but i m not able to solve it

i got stuck at this step $\frac{2}{y}d(yt)=d(\frac{y}{t})$

Any hint plz suggest vishnu c