# Thread: Word problem with DE

1. ## Word problem with DE

Ok, I simply cannot get the answer that they have in the solutions manual, nor do they explain how they got the answer, so I really have no clue what I'm doing wrong.

The instructions for the problem are: Investment. A brokerage firm opens a new real estate investment plan for which the earnings are equivalent to continuous compounding at the rate of $\displaystyle r$. The firm estimates that deposits from investors will create a net cash flow of $\displaystyle Pt$ dollars, where $\displaystyle t$ is the time in years. The rate of change in the total investment $\displaystyle A$ is modeled by

$\displaystyle \frac{dA}{dt}=rA+Pt$

(a) Solve the differential equation and find the total investment $\displaystyle A$ as a function of $\displaystyle t$. Assume that $\displaystyle A = 0$ when $\displaystyle t = 0$.

(b) Find the total investment A after 10 years given that P = $500,000 and r = 9%. Ok, so I rearranged the DE to be$\displaystyle A' - rA = Pt$and then determined that$\displaystyle P(t) = -r$and$\displaystyle Q(t)=Pt$. From there, I found$\displaystyle IF=e^{\int-rdt}=e^{-rt}$. I then mutiplied each side by the IF to get$\displaystyle e^{-rt}A'-rAe^{-rt}=Pte^{-rt}$. From there I doublechecked that the IF was correct, it was, and then I did$\displaystyle \int \frac{dA}{dt} (e^{-rt} * A) = \int Pte^{-rt}$The integral on the left cancels with the derivative so$\displaystyle e^{-rt}A=\int Pte^{-rt}$I then pulled P out as a constant to get$\displaystyle e^{-rt}A=P\int te^{-rt}$I then did parts, using$\displaystyle u=t$,$\displaystyle du=dt$,$\displaystyle V=e^{-rt}$to get$\displaystyle e^{-rt}A=P(te^{-rt} - \int e^{-rt}dt)$From there I used$\displaystyle u=-rt$and$\displaystyle du=-rdt$and since I only have a$\displaystyle dt$I did$\displaystyle -\frac{1}{r}du=dt$Solving the final integral I came up with$\displaystyle e^{-rt}A=P(te^{-rt}+\frac{1}{r}e^{-rt}+C)$Dividing both sides by$\displaystyle e^{-rt}$to get$\displaystyle A$by itself, I got$\displaystyle A=P(t+\frac{1}{r}+Ce^{rt})$Which is different from what the solutions manual got, which was$\displaystyle A=\frac{P}{r^2}(-rt-1+Ce^{rt})$What am I doing wrong? 2. Originally Posted by emttim84 Ok, I simply cannot get the answer that they have in the solutions manual, nor do they explain how they got the answer, so I really have no clue what I'm doing wrong. The instructions for the problem are: Investment. A brokerage firm opens a new real estate investment plan for which the earnings are equivalent to continuous compounding at the rate of$\displaystyle r$. The firm estimates that deposits from investors will create a net cash flow of$\displaystyle Pt$dollars, where$\displaystyle t$is the time in years. The rate of change in the total investment$\displaystyle A$is modeled by$\displaystyle \frac{dA}{dt}=rA+Pt$(a) Solve the differential equation and find the total investment$\displaystyle A$as a function of$\displaystyle t$. Assume that$\displaystyle A = 0$when$\displaystyle t = 0$. (b) Find the total investment A after 10 years given that P =$500,000 and r = 9%.

Ok, so I rearranged the DE to be $\displaystyle A' - rA = Pt$ and then determined that $\displaystyle P(t) = -r$ and $\displaystyle Q(t)=Pt$. From there, I found $\displaystyle IF=e^{\int-rdt}=e^{-rt}$.

I then mutiplied each side by the IF to get $\displaystyle e^{-rt}A'-rAe^{-rt}=Pte^{-rt}$.

From there I doublechecked that the IF was correct, it was, and then I did $\displaystyle \int \frac{dA}{dt} (e^{-rt} * A) = \int Pte^{-rt}$

The integral on the left cancels with the derivative so

$\displaystyle e^{-rt}A=\int Pte^{-rt}$

I then pulled P out as a constant to get $\displaystyle e^{-rt}A=P\int te^{-rt}$

I then did parts, using $\displaystyle u=t$, $\displaystyle du=dt$, $\displaystyle V=e^{-rt}$ to get

$\displaystyle e^{-rt}A=P(te^{-rt} - \int e^{-rt}dt)$

From there I used $\displaystyle u=-rt$ and $\displaystyle du=-rdt$ and since I only have a $\displaystyle dt$ I did $\displaystyle -\frac{1}{r}du=dt$

Solving the final integral I came up with

$\displaystyle e^{-rt}A=P(te^{-rt}+\frac{1}{r}e^{-rt}+C)$

Dividing both sides by $\displaystyle e^{-rt}$ to get $\displaystyle A$ by itself, I got

$\displaystyle A=P(t+\frac{1}{r}+Ce^{rt})$

Which is different from what the solutions manual got, which was

$\displaystyle A=\frac{P}{r^2}(-rt-1+Ce^{rt})$

What am I doing wrong?
Is P a function or a variable or a constant? if it is a constant this is a simple seperable differential equation

3. your integration by parts is incorrect
the correct integration is

$\displaystyle \int Pte^{-rt}=P\left[ -\frac{t}{r}e^{-rt}-\frac{1}{r^2}e^{-rt}+C \right]$

4. Originally Posted by TheEmptySet
your integration by parts is incorrect
the correct integration is

$\displaystyle \int Pte^{-rt}=P\left[ -\frac{t}{r}e^{-rt}-\frac{1}{r^2}e^{-rt}+C \right]$
Right, I figured the integration is incorrect, but where is the error so I can see how to get the correct answer, and not just have the correct answer?