Question

First order linear differential equation

Answer 1

can some help my find the solution to\[0=\gamma \frac{ d \Psi (t) }{ dt }\left[ (1- \gamma)\left( r+\frac{ 1 }{ 2 } \frac{ \lambda^2 }{ \gamma }\right) -\rho\right]\Psi(t)+\gamma\]

Answer 2

where \[\Psi(T)=0\]

Answer 3

arrh sorry

Answer 4

Mathway.com

Answer 5

\[0=\gamma \frac{ d \Psi (t) }{ dt }+\left[ (1- \gamma)\left( r+\frac{ 1 }{ 2 } \frac{ \lambda^2 }{ \gamma }\right) -\rho\right]\Psi(t)+\gamma\]

Answer 6

I forgot a "+"

Answer 7

-.- take a snap :o

Answer 8

set \((1- \gamma)\left( r+\frac{ 1 }{ 2 } \frac{ \lambda^2 }{ \gamma }\right) -\rho = m\)

Answer 9

\(\large 0=\gamma \frac{ d \Psi (t) }{ dt }+m\Psi(t)+\gamma\)

Answer 10

change it to standard form by dividing gamma and subtracting 1

Answer 11

\[0=\gamma \frac{ d \Psi (t) }{ dt }+m\Psi(t)+\gamma\]

Answer 12

\[0=\frac{ d \Psi (t) }{ dt }+m\Psi(t)/\gamma-1\] like this?

Answer 13

\(\large 0=\gamma \frac{ d \Psi (t) }{ dt }+m\Psi(t)+\gamma\)

\(\large 0=\frac{ d \Psi (t) }{ dt }+\frac{m}{\gamma}\Psi(t)+1\)

\(\large -1=\frac{ d \Psi (t) }{ dt }+\frac{m}{\gamma}\Psi(t)\)

Answer 14

\(\large \frac{ d \Psi (t) }{ dt }+\frac{m}{\gamma}\Psi(t) = -1\)

Answer 15

we can separate variables right ?

Answer 16

no need of IF and all

Answer 17

\(\large 0=\gamma \frac{ d \Psi (t) }{ dt }+m\Psi(t)+\gamma\)

\(\large 0=\frac{ d \Psi (t) }{ dt }+\frac{m}{\gamma}\Psi(t)+1\)

\(\large \frac{ d \Psi (t) }{ dt } = -1-\frac{m}{\gamma}\Psi(t)\)

\(\large \frac{ d \Psi (t) }{1+\frac{m}{\gamma}\Psi(t) } = -dt\)

Answer 18

integrate both sides