Question

help me in this kind of linear differential equation, plsssssssssssssssss

Answer 1

problem is \[(10-t)dQ +(2Q+4t -80) dt = 0\]

Answer 2

@baru

Answer 3

\[dQ/dt=(2Q+4t-8)/t-10\]

Answer 4

then what to do

Answer 5

\[dQ/dt=\frac{(2Q+4t-8)}{t-10}\\ \frac{dQ}{dt} = \frac{2Q}{t-10}+ \frac{ 4t-8}{t-10}\]

Answer 6

\[\frac{ DQ }{ dt } -\frac{ 2Q }{ t-10}=\frac{ 4t-8 }{ t-10 }\]

Answer 7

so \[p= \frac{ -2 }{ t-10 },q=\frac{ 4t }{ t-10 }\]

Answer 8

IF is \[e ^{\int\limits \frac{ -2 }{ 10-t }dt}\]

Answer 9

yes

Answer 10

no