<s>[INST] How to solve the given differential equation? 
Given $T$ is a function in $t$ $(x>0)$. And $T(@t=0)=40$. And which satisfies,
$$ \frac{dt}{4}= \frac{3\cdot dT}{2} + \frac{T\cdot dt}{400+t}$$
Then find $T$

I tried putting $400+t=u$ and then,
$$ \frac{du}{4}= \frac{3\cdot dT}{2} + \frac{T\cdot du}{u}$$
And multiplying by $4u$ both sides gives,
$$u\cdot du= 6\cdot u\cdot dT+ 4\cdot T \cdot du$$
And by integrating by parts bay taking common $4$ gives,
$$ \frac{u^2}{2}=4\cdot u \cdot T+ \int 2\cdot u \cdot dT+c$$
But now I am stuck here. Can someone help me solving this?
 [/INST] Under your substitution, the ODE becomes homogeneous
$$\frac32 \, dT = \left(\frac14 - \frac T{400+t}\right) \, dt \implies \frac{dT}{du} = \frac16 \left(1-\frac{4T}u\right)$$
Further substituting $U=\frac Tu$ and $\frac{dU}{du}=\frac{u\frac{dT}{du} - T}{u^2}$ gives the separable ODE,
$$u\,\frac{dU}{du} + U = \frac{1-4U}6 \implies \frac{dU}{1-10U} = \frac{du}{6u}$$
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