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\begin{align*} \newcommand{\dd}{\; \mathrm{d}} (4+t)^2\frac{\dd y}{\dd t} + 2ty &= 4t \tag{1}\\ \implies (4+t^2)\frac{\dd y}{\dd t} + y\frac{\dd(4+t^2)}{\dd t} &= 4t \tag{2}\\ \implies \frac{\dd}{\dd t}[(4+t^2)]y &= 4t \tag{3}\\ \implies (4+t^2) y &= \int 4t \dd t\\ \implies (4+t^2) y &= 2t^2 + C \\ \implies y &= \frac{2t^2 + C}{t^2+4} \end{align*}

I don't know how number 1 and number 2 are same...

Also, how number 2 and 3 are same... ?

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    Try going backwards and see if that enlightens. Use product rule to go from 3 to 2 and take a derivative to go from 2 to 1. Also, please use mathjax if you're going to post on this site. – 2017-02-06
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    You've here our [tutorial](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) to write nice mathematics, also with the mouse right click Show Math As> Tex Commands you can explore formulas. – 2017-02-06
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    For some basic information about writing math at this site see e.g. [here](/help/notation), [here](//meta.math.stackexchange.com/q/5020), [here](//meta.stackexchange.com/a/70559) and [here](//meta.math.stackexchange.com/q/1773). I hope the way I rewrote it is approximately what you wanted - feel free to edit it further. – 2017-02-06

2 Answers 2

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Why are one and two the same: $$ \frac{d}{dt}(4+t^2)=2t $$ Why are two and three the same? The product rule: $$ \frac{d}{dt}[(4+t^2)y]=(\frac{d}{dt}[4+t^2])y+(\frac{d}{dt}[y])(4+t^2)=2ty+\frac{dy}{dt}(4+t^2) $$

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Work backwards:

$$\frac{d}{dt}[(4+t^2)y]=2ty+(4+t^2)\frac{dy}{dt}.$$

(You can fill the intermediate step.)