Question

a. Find a power series for the solution of the following differential equations. b. Identify the function represented by the power series. $$y^{\prime}(t)-y(t)=0, y(0)=2$$

Short answer

The power series solution is $$y(t) = 2 + 2t + t^2 + \frac{t^3}{3} + ...$$, and the function it represents is $$y(t) = 2e^t$$.

Step-by-step solution

Form a power series solution

Attempt to form a power series solution of the form: $$y(t) = \sum_{n=0}^{\infty} a_nt^n$$, where $$a_n$$ is a constant coefficient. Differentiating this power series with respect to t: $$y'(t) = \sum_{n=1}^{\infty} na_nt^{n-1}$$ Now, substitute the power series of $$y(t)$$ and $$y'(t)$$ into the differential equation: $$\sum_{n=1}^{\infty} na_nt^{n-1} - \sum_{n=0}^{\infty} a_nt^n = 0$$

Use initial condition and match the coefficients

Using the given initial condition, $$y(0) = 2$$,we have: $$y(0) = \sum_{n=0}^{\infty} a_n0^n = a_0 = 2$$ This gives us our first coefficient, $$a_0 = 2$$. To match the coefficients in the differential equation, consider changing the index in the first sum: $$\sum_{n=0}^{\infty} (n+1)a_{n+1}t^n - \sum_{n=0}^{\infty} a_nt^n = 0$$ The coefficients of the corresponding powers of t must match, then we can write: $$(n+1)a_{n+1} - a_n = 0$$ From this equation, we can find the other coefficients: $$a_{n+1} = \frac{a_n}{n+1}$$

Find the coefficients recursively

Using the recurrence relation found above, we can find all the coefficients: For $$n=0$$: $$a_1 = \frac{a_0}{1} = \frac{2}{1} = 2$$ For $$n=1$$: $$a_2 = \frac{a_1}{2} = \frac{2}{2} = 1$$ For $$n=2$$:$$a_3 = \frac{a_2}{3} = \frac{1}{3}$$ and so on.

Write the power series solution

Now that we have the coefficients, write the power series solution as follows: $$y(t) = \sum_{n=0}^{\infty} a_nt^n = 2 + 2t + t^2 + \frac{t^3}{3} +...$$

Identify the function represented by the power series

Observe the obtained power series. It resembles the Maclaurin series of the function $$y(t) = 2e^t$$, which is: $$2e^t= 2 + 2t + t^2 + \frac{t^3}{3} +...$$ Hence, the function represented by the power series is $$y(t) = 2e^t$$. To conclude, the power series solution for the given differential equation is $$y(t) = 2 + 2t + t^2+ \frac{t^3}{3} + ...$$, and the function that this power series represents is $$y(t) = 2e^t$$.

Key concepts

Power Series

A power series is an infinite series of the form \( y(t) = \sum_{n=0}^{\infty} a_n t^n \), where \( a_n \) are coefficients and \( t^n \) represents each term raised to a power. These series are used to represent functions as an infinite sum, allowing for flexible approximations of complex functions by simplifying them into polynomial terms.
Power series can converge to represent certain functions within specific intervals. An important property of power series is that they can be easily differentiated term-by-term, allowing the use of calculus on series representations.
In the context of solving differential equations, power series allow us to express solutions in a manageable form, especially when exact solutions are hard to find or do not exist in simple forms. By using power series, we can often revert back to familiar algebraic or trigonometric functions after identifying coefficient patterns.

Maclaurin Series

The Maclaurin series is a special type of power series centered at zero. It is a way to express functions as a sum of polynomial terms evaluated at \( t = 0 \). The general formula for a Maclaurin series is given by:

• \( f(t) = f(0) + f'(0)t + \frac{f''(0)t^2}{2!} + \frac{f'''(0)t^3}{3!} + \ldots \)

This series simplifies the computation involved when evaluating functions around the origin, making it a handy tool in approximating functions near \( t = 0 \).
For example, the function \( e^t \) can be represented by its Maclaurin series as \( e^t = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \ldots \). In the original exercise, the power series for \( y(t) = 2e^t \) follows the same pattern with a multiplier of 2, showing that the resulting power series is indeed a Maclaurin series scaled by a factor of two.

Recurrence Relation

A recurrence relation is an equation that recursively defines a sequence of values: each term is defined as a function of its preceding terms. In the realm of power series solutions for differential equations, recurrence relations help identify coefficients based on a relationship between the terms.
In the given problem, the recurrence relation \( (n+1)a_{n+1} - a_n = 0 \) arises when matching coefficients in the differentiated power series. Using this relation, we can solve for each \( a_n \).
This relation can be rewritten as \( a_{n+1} = \frac{a_n}{n+1} \), starting with the initial coefficient \( a_0 = 2 \). By following the recurrence relation, the coefficients \( a_1 = 2 \), \( a_2 = 1 \), \( a_3 = \frac{1}{3} \), and so forth are determined systematically.

Initial Conditions

Initial conditions are particular values or 'starting points' that allow us to find specific solutions to differential equations. These conditions specify certain values for the function or its derivatives at a specific point, which serve as background information to fully determine the unique solution for given problems.
In the exercise, the initial condition is \( y(0) = 2 \). This means that when \( t = 0 \), the function \( y(t) \) should equal 2. The initial condition helps establish the first coefficient in the power series nested within the differential equation-solving process.
By setting \( y(0) = 2 \), we directly find that \( a_0 = 2 \). This starting point is crucial as it forms the base case for calculating subsequent coefficients through the recurrence relation, ensuring that the full power series solution aligns with the specified initial conditions.