Question

Solve the following differential equations by series and also by an elementary method and verify that your solutions agree. Note that the goal of these problems is not to get the answer (that's easy by computer or by hand) but to become familiar with the method of series solutions which we will be using later. Check your results by computer. $$y^{\prime}=3 x^{2} y$$

Short answer

The solution is \[ y = A e^{x^3} \]. Series and elementary solutions match, confirming correctness.

Step-by-step solution

Write the Differential Equation

Given the differential equation: \[ y' = 3x^2 y \]

- Solve Using Series Method

Assume a power series solution of the form \[ y = \sum_{n=0}^{\infty} a_n x^n \].Differentiate to obtain \[ y' = \sum_{n=1}^{\infty} n a_n x^{n-1} \].Substitute into the differential equation:\[ \sum_{n=1}^{\infty} n a_n x^{n-1} = 3x^2 \sum_{n=0}^{\infty} a_n x^n \].

- Matching Powers of x

Combine and reorder terms so we match powers of x:\[ \sum_{n=1}^{\infty} n a_n x^{n-1} = 3 \sum_{n=0}^{\infty} a_n x^{n+2} \].Let's reindex the right-hand series from j = n + 2, where j=n+3: \[ \sum_{j=2}^{\infty} j a_{j} x^{j-1} = 3 \sum_{j=0}^{\infty} a_j x^j \].After changing the limits and reindexing:\[ \sum_{n=0}^{\infty} (n + 1) a_{n + 1} x^n = 3 \sum_{n=2}^{\infty} a_{n - 2} x^n \].

- Equate Coefficients

By matching coefficients of x, we get:\[ (n+1) a_{n+1} = 3 a_{n-2} \].This is a recurrence relation which helps determine the coefficients.

- Initial Conditions

Assume initial condition \[ a_0 = A \], a constant.From \[ (n+1) a_{n+1} = 3 a_{n-2} \], we will find:\[ a_1 = 0 \] because y(0) is zero. Then, solving the recurrence, we get higher coefficients.

Recurrence - General Solution

The recurrence results in even indices having nonzero terms. For simplicity, let's solve a couple manually:\[ a_2 = \frac{3A}{2} \] \[ a_4= \frac{3^2 A}{4!} \]The series solution is:\[ y = A (1 + \frac{3x^2}{2!} + \frac{3^2 x^4}{4!} + ...) \].This series shows an exponential function behavior.

- Solve Using Elementary Method

Rewrite the equation as \[ \frac{dy}{y} = 3x^2 \ dx \].Integrate both sides to obtain:\[ \ln |y| = x^3 + C \].Thus, \[ y = e^{x^3+C} \]. Taking exponential and combining the constants\[ y = Ae^{x^3} \].

- Verify Solutions Agree

The exponential form can also be expanded into a series:\[ y = A e^{x^3} = A (1 + \frac{x^3}{1!} + \frac{x^6}{2!}...)\].This matches the series solution we derived earlier, confirming the agreement.

Key concepts

series method

The series method is a powerful technique for solving differential equations. It involves assuming that the solution can be expressed as an infinite sum of terms, typically in the form of a power series. This means that we look for solutions of the form \[ y = \sum_{n=0}^{\infty} a_n x^n \]. By differentiating this power series and substituting it into the differential equation, we obtain a new series that must match the terms on both sides of the equation. The key steps include:
- Substitute the power series into the differential equation
- Align terms with like powers of x
- Solve the resulting series for the coefficients. This step often leads to a recurrence relation, which helps determine each coefficient in the series.

power series

A power series is an infinite sum of terms in the form \[ y = \sum_{n=0}^{\infty} a_n x^n \], where each term involves the variable x raised to a power and multiplied by a coefficient a_n. Power series solutions are particularly useful for solving differential equations because they allow us to handle terms of varying powers of x in a systematic way. By plugging these series representations into differential equations, we can often simplify complex relationships into more manageable algebraic equations. The power series also provides a way to approximate the solution function around a given point, typically x=0 (the series center).

recurrence relation

A recurrence relation is an equation that expresses each term of a sequence as a function of its preceding terms. In the context of series solutions for differential equations, recurrence relations arise when equating coefficients of like powers of x. For instance, in our example, the matching of terms led to the relation \[ (n+1) a_{n+1} = 3 a_{n-2} \]. This relation allows us to compute higher-order coefficients from the known lower-order ones, systematically building the solution term by term. These relations are crucial for completely determining the series solution.

initial conditions

Initial conditions are the values specified for the solution of a differential equation at a particular point, usually for the function and its derivatives. These conditions are essential because they help determine the specific solution among the infinite family of potential solutions. For instance, in the given problem, the initial condition was assumed as \[ a_0 = A \]. This helps establish the values of the first few coefficients, which then propagate through the recurrence relation to define the entire series solution.

elementary method

Elementary methods refer to more straightforward, often more intuitive techniques for solving differential equations compared to the series method. For example, the given differential equation \[ y' = 3x^2 y \] can be solved by separation of variables. Rearrange the terms to isolate y on one side and x on the other: \[ \frac{dy}{y} = 3x^2 \ dx \]. Integrate both sides to obtain \[ \ln |y| = x^3 + C \], leading to the solution \[ y = e^{x^3 + C} \]. This further simplifies to \[ y = Ae^{x^3} \]. Verifying that this matches the series solution confirms the validity of the elementary method. Elementary methods provide a direct approach to finding solutions without initially assuming a series.