# Differential Equation and Investment Growth

• Jun 23rd 2018, 08:38 PM
Vinod
Differential Equation and Investment Growth
Hello forumites,
A large corporation starts at time t=0 to continuously part of its receipts at a rate of P dollars per annum in a fund for future corporate expansion. Assume that the fund earns r percent interest per year compounded continuously. Thus , the rate of growth of the amount A in the fund is given by

$\frac{dA}{dt}=rA +P$ where A=0 when t=0 . I want to solve this differential equation for A as a function of t.

Using the above result, I want to find A for the following

a)P= 100000 dollars, r=6%, and t= 5 years b) P=250000 dollars, r=5%, and t=10 years.

I also want to find t if the corporation needs 800000 dollars and it can invest 75000 dollars per annum in a fund earning 8% interest compounded continuously.

Answer: I am working on these problems but still no results. It would be better if any forumite guide me.
• Jun 23rd 2018, 09:25 PM
MarkFL
Re: Differential Equation and Investment Growth
I would put the ODE into standard linear form:

$\displaystyle \frac{dA}{dt}+(-r)A=P$

Next, we need to compute the integrating factor:

$\displaystyle \mu(t)=\exp\left(-r\int\,dt\right)=\,?$
• Jun 24th 2018, 10:45 PM
Vinod
Re: Differential Equation and Investment Growth
Quote:

I would put the ODE into standard linear form:

$\displaystyle \frac{dA}{dt}+(-r)A=P$

Next, we need to compute the integrating factor:

$\displaystyle \mu(t)=\exp\left(-r\int\,dt\right)=\,?$

Hello,
So,$A=\frac{-P}{r}+C*e^{rt}$
So C=1666666.66666 and A=58,30,98 when P=100000 r=0.06 t=5 years.
and C=5000000 A=3243606.35 when P=250000 r=0.05 and t=10 years
• Jun 24th 2018, 11:06 PM
MarkFL
Re: Differential Equation and Investment Growth
Continuing where I left off...

$\displaystyle \mu(t)=\exp\left(-r\int\,dt\right)=e^{-rt}$

And thus the ODE becomes:

$\displaystyle \frac{d}{dt}\left(e^{-rt}A\right)=Pe^{-rt}$

Integrate:

$\displaystyle e^{-rt}A=-\frac{P}{r}e^{-rt}+c_1$

$\displaystyle A(t)=-\frac{P}{r}+c_1e^{rt}$

Now, we are given:

$\displaystyle A(0)=-\frac{P}{r}+c_1=0\implies c_1=\frac{P}{r}$

Hence, the solution to the IVP is:

$\displaystyle A(t)=-\frac{P}{r}+\frac{P}{r}e^{rt}=\frac{P}{r}\left(e^{ rt}-1\right)$

Can you proceed?
• Jun 24th 2018, 11:37 PM
Vinod
Re: Differential Equation and Investment Growth
Quote:

Continuing where I left off...

$\displaystyle \mu(t)=\exp\left(-r\int\,dt\right)=e^{-rt}$

And thus the ODE becomes:

$\displaystyle \frac{d}{dt}\left(e^{-rt}A\right)=Pe^{-rt}$

Integrate:

$\displaystyle e^{-rt}A=-\frac{P}{r}e^{-rt}+c_1$

$\displaystyle A(t)=-\frac{P}{r}+c_1e^{rt}$

Now, we are given:

$\displaystyle A(0)=-\frac{P}{r}+c_1=0\implies c_1=\frac{P}{r}$

Hence, the solution to the IVP is:

$\displaystyle A(t)=-\frac{P}{r}+\frac{P}{r}e^{rt}=\frac{P}{r}\left(e^{ rt}-1\right)$

Can you proceed?

Hello,
How to find t if the corporation needs 800000 dollars and it can invest 75000 dollars per annum in a fund earning 8% interest compounded continuously.
• Jun 25th 2018, 12:56 AM
Vinod
Re: Differential Equation and Investment Growth
Quote:

Hello,
How to find t if the corporation needs 800000 dollars and it can invest 75000 dollars per annum in a fund earning 8% interest compounded continuously.

I got the answer 8.0168 years in hp48 series calculator.
• Jun 25th 2018, 05:08 AM
DenisB
Re: Differential Equation and Investment Growth
Vinod, try something easier:
if interest is 8% compounded annually, what will be the value after 8 years?
• Jun 25th 2018, 05:34 AM
Vinod
Re: Differential Equation and Investment Growth
Quote:

Vinod, try something easier:
if interest is 8% compounded annually, what will be the value after 8 years?

Is it $797747.07? • Jun 25th 2018, 08:48 AM DenisB Re: Differential Equation and Investment Growth Right! OK: what if the rate compounds daily (use 1 year = 360 days)? • Jun 25th 2018, 10:38 PM Vinod Re: Differential Equation and Investment Growth Quote: Right! OK: what if the rate compounds daily (use 1 year = 360 days)? Hello, It would be$807253.35. If the rate compounded continuously, the amount after 8 years would be \$807280.79.
• Jun 25th 2018, 10:46 PM
dkugg
Re: Differential Equation and Investment Growth
We can model the growth of an initial deposit with respect to the interest rate with differential equations. If represents time, then the rate of change of the initial.
. Examples of ordinary differential equations:

dydx = u(y)

or

dydx = u(x, y)
• Jun 26th 2018, 05:04 AM
DenisB
Re: Differential Equation and Investment Growth
Quote:

Hello,
It would be 807253.35.
If the rate compounded continuously, the amount after 8 years would be 807280.79.

807280.79 - 807253.35 = 27.44
That's 27.44 / 8 = 3.43 average per year.

My point was to show you "continuously" is NOT that big a deal!!