Question

please help me answering this.. P= C1et/1+C1et ;dP/dt= P(1-P)

Answer 1

start by separating the variables...P on 1 side, t on other side
\[\rightarrow \frac{dP}{P(1-P)} = dt\]
then you can integrate both sides

Answer 2

split fraction into 2 fractions
\[\frac{1}{P(1-P)} = \frac{1}{P} + \frac{1}{1-P}\]
\[\int\limits \frac{1}{P} + \frac{1}{1-P} dP = \int\limits dt\]
\[\ln P - \ln (1-P) = t+C\]

Answer 3

can you get answer from here?

Answer 4

$$\frac{dP}{dt}=P(1-P)\\\frac1{P(1-P)}dP=dt$$Observe we can use partial fractions:$$\frac1{P(1-P)}=\frac{A}P+\frac{B}{1-P}=\frac{A(1-P)+B(P)}{P(1-P)}$$so we know \(A(1-P)+B(P)=1\). Allowing \(P=0\) we observe \(A=1\); similarly, allowing \(P=1\) we observe \(B=1\). So our decomposition tells us:$$\frac1{P(1-P)}=\frac1P+\frac1{1-P}$$

Answer 5

this is a solution to the differential equations.. the answer to the first equation should be equal to the second equation.. thanks

Answer 6

Y= e3xcos2x ; y”-6y2+13y=0

Y=6/5 - 6/5 e-20t ; dy/dt +20y=24

Y=C1x-1 + C2x+ C3xlnx + 4x2 ; x3d3y/dx3 + 2x2 d2y/ dx +y =12x2

-2x2y+y2=1 ; 2xydx + (x2-y)dy=0