Question

First order linear differential equation

Answer 1

Can someone help me through this step:

STEP 1: \[0=\gamma \frac{ d \Psi(t) }{ dt }+a*\Psi(t)+\gamma\]
With a boundary condition \[\Psi(T)=0\]
STEP 2:
This is a standard first order linear differential equation with a well-known solution:
\[\Psi(t)=\frac{ 1 }{ a }(1-e^{a(t-T)})\]

Answer 2

Some math give me
\[0=\gamma \frac{ d \Psi(t) }{ dt }+a*\Psi(t)+\gamma <=> -1=\frac{ d \Psi(t) }{ dt }+a*\Psi(t)/\gamma\]
i.e. \[Q(t)=-1 \] and \[P(t)=\frac{ \Psi(t) }{ \gamma }\]

Answer 3

I am having trouble with the first part...
If we substitute
\[ \Psi(t)=\frac{ 1 }{ a }(1-e^{a(t-T)}) \]
into the differential equation, I don't get 0

Answer 4

What if we let (STEP 1) be
\[0=\gamma \frac{ d \Psi(t) }{ dt }+\left[ \left( 1-\gamma \right)\left( r+\frac{ 1 }{ 2 }\frac{ \lambda^2 }{ \gamma } \right)-\rho \right]*\Psi(t)+\gamma\]

and

\[a= \frac{ \rho-(1-\gamma)\left( r+\frac{ 1 }{ 2 }\frac{ \lambda^2 }{ \gamma } \right) }{ \gamma }\]

Answer 5

a is the negative of the coefficient in front of Psi divided by gamma ?

Answer 6

are you trying to verify some statement (e.g. in a book) or solve a problem ?

Answer 7

2 sec.

Answer 8

I am trying to verify a statement in a book because have to solve the same problem just wist a other boundary condition.

Answer 9

Attached picture

Answer 10

Here is one way to do it.

Answer 11

Great thank you.

What if we change the boundary condition to \[\Psi(T)=He^{-\rho T}\frac{ 1 }{ 1- \gamma }W^{1- \gamma}\]
Can i just change the initial condition?

Answer 12

I see that I messed up the signs at the very bottom
Here is a correction

Answer 13

Yes, I think you use equation (8) with t= T and set that equal to the right-hand side of your new initial condition. But it does look hairy.

Answer 14

to find the constant C. then use that constant C in equation (8) to get the final answer.

Answer 15

Thank you @phi. It was a big help.