Let y(t) represent your retirement account balance, in dollars, after t years. Each year the...


Let {eq}y(t) {/eq} represent your retirement account balance, in dollars, after {eq}t {/eq} years. Each year the account earns {eq}3\% {/eq} interest, and you deposit {eq}6\% {/eq} of your annual income. Your current annual income is {eq}\$36000 {/eq}, but it is growing at a continuous rate of {eq}4\% {/eq} per year. Write the differential equation modeling this situation.

First-Order Linear Differential Equations:

An equation which can be written in the form {eq}y'(t)+a(t)y(t)=b(t) {/eq} is called a first-order linear differential equation. First-order differential equations can be solved by applying the method of integrating factors. To apply this method, multiply the equation by a function {eq}e^{\int a(t) \, dt} {/eq} (which is called an integrating factor of the equation). This will put the equation into a form where both sides of the equation can be integrated.

Answer and Explanation:

Let {eq}I(t) {/eq} denote your income at time {eq}t {/eq} in dollars per year. Then {eq}I(0)=36000 {/eq} and {eq}I(t) {/eq} grows exponentially at a rate of 4% per year, and so {eq}I(t)=36000e^{0.04t} {/eq}.

If you deposit 6% of your income per year in your retirement account, then your deposits increase {eq}y(t) {/eq} at a rate of {eq}0.06I(t) {/eq}. The retirement account also earns 3% interest, which increases {eq}y(t) {/eq} at a rate of {eq}0.03y(t) {/eq}. So the rate of increase in your retirement account is:

{eq}\begin{align*} y'(t)&=0.03y(t)+0.06I(t)\\ &=0.03y(t)+0.06(36000e^{0.04t})&&\text{(by the above expression for }I(t)\text{)}\\ &=0.03y(t)+2160e^{0.04t} \, . \end{align*} {/eq}

So {eq}y(t) {/eq} obeys the first-order linear differential equation {eq}\boxed{y'(t)=0.03y(t)+2160e^{0.04t} \, .} {/eq}

Learn more about this topic:

Exponential Growth: Definition & Examples

from High School Algebra I: Help and Review

Chapter 6 / Lesson 10

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