# Solving Compound Interest using Ordinary Differential Equation

I'm reading a book about finance and it says that if an investor makes a deposit of P dollars into a cash account that pays interest rate r 100% per year, compounded continuously, the evolution of account balance as a function of time t (measured in years) satisfies the Ordinary Differential Equation:

y'(t) = r y(t)

Questions:

• What I learnt in high school is that for compound interest the account balance as a function of time t is calculated by the equation $P(1+r)^{t}$. Hence, its derivative should be $P(1+r)^{t}\ln [P(1+r)]$, which doesn't always satisfy the equation y'(t) = r y(t), i.e. $P(1+r)^{t}\ln [P(1+r)]$ does not always equal to $r\ P(1+r)^{t}$. Am I thinking about it in the wrong way? Where does the Ordinary Differential Equation comes from?

• The book further explains that the amount of change in the account balance is equal to the (interest rate) * (previous balance) * (elapsed time) with an initial condition y(0) = P, why is it so?

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The formula from school you give $A = P(1+r)^t$ is for interest compounded once a year, after $t$ years. The book is talking about continuous compounding. I might have time to add a longer answer later, but for now you may want to look at en.wikipedia.org/wiki/Compound_interest#Continuous_compounding and en.wikipedia.org/wiki/Definitions_of_the_exponential_function –  in_wolframAlpha_we_trust Feb 7 '13 at 11:33
@in_wolfram_we_trust: Thanks for pointing it out. I wasn't aware of the existence of continuous compounding :-) Now I understand that the equation for continuous compounding can be derive by solving the Ordinary Differential Equation and y(0) = P, is it one of the ways they derive the continuous compounding's formula? But still, where does the ODE comes from? Is it.... a mere observation? –  user61350 Feb 8 '13 at 10:19
the equation $y'=ry$ states that the change in y (which is $y'$) equals interest rate (which is r) multiplied by y. But $r*y$ is the amount by which y changes. You see that? Ex.g. Lets say interest rate is 10%, r=0.1, and our investment is 50 bucks, y=50. So when compounded the change of our investments, $y'$, is going to equal to r*y=5. So, our return will be 5 bucks. To check 50*1.1=55. However, notice that I am using constants for y whereas in your book they refer to fucntions of time $y(t)$. This ODE is mere reasoning. Change in deposits,y', equals the interest rate share of your deposits –  Koba Apr 11 '13 at 19:19

Eg. if the interest rate is 10% simple per year, initial investment is \\$500, then after 3 months i.e. 3/12 years, the interest gained is $$\frac{10\%}{\text{year}} * \frac{3}{12} \text{years} * \500 = 2.5\% * 500.$$