Question

find the general solution to the first order differential equation: (4-x)dy+2ydx=0

Answer 1

\((4-x)dy+2ydx=0\\(4-x)dy=-2ydx\\-\frac{1}{2y}dy=\frac{1}{4-x}dx\\-\int\frac{1}{2y}dy=\int\frac{1}{4-x}dx\implies-\frac{1}{2}\ln(y)=-\ln(4-x)+c\\\log(y)=2\log(x-4)+c_0\\y=e^{2\log(x-4)+c_0}=e^{2\log(x-4)}e^{c_0}=c_1e^{\log((x-4)^2)}=c_1(x-4)^2\)

Answer 2

thank u

Answer 3

np

Answer 4

wait....??....why -ln(4-x) but not ln(4-x)

Answer 5

\(\color{red}{EDIT*}\)
\((4-x)dy+2ydx=0\\(4-x)dy=-2ydx\\-\frac{1}{2y}dy=\frac{1}{4-x}dx\\-\int\frac{1}{2y}dy=\int\frac{1}{4-x}dx\implies-\frac{1}{2}\ln(y)=-\ln(4-x)+c\\\log(y)=2\log(4-x)+c_0\\y=e^{2\log(4-x)+c_0}=e^{2\log(4-x)}e^{c_0}=c_1e^{\log((4-x)^2)}=c_1(4-x)^2\)

Answer 6

ok...I got it...thank u again

Answer 7

np

Answer 8

\[\int\limits_{e}^{4e} 1/x dx\]
Can u help me with that?.

Answer 9

evaluate

Answer 10

\(\log(4e)-\log(e)=\log(4)+\log(e)-\log(e)=\log(4)\)