Question

\(y^{\prime \prime}=t y^{\prime}+y+1\), with \(y(0)=1\), and \(y^{\prime}(0)=0\).

Short answer

Power series or iterative solution can solve the equation, starting with initial conditions.

Step-by-step solution

Identify the Type of Differential Equation

We are given a second-order linear differential equation with variable coefficients: \[ y'' = t y' + y + 1 \]where \( y(0) = 1 \) and \( y'(0) = 0 \). This is a non-homogeneous linear differential equation.

Substitute Known Values

Use the initial conditions to find specific solutions or verify any derived general solutions. Since we are dealing with variable coefficients, let's identify any possibilities for substitution.

Attempt Power Series Expansion

To solve this equation, assume a power series solution:\[ y(t) = \sum_{n=0}^{ } a_n t^n \]Then, differentiate it to find expressions for \( y' \) and \( y'' \) in terms of series.

Calculate Derivatives

Calculate the derivatives:\[ y'(t) = \sum_{n=1}^{ } n a_n t^{n-1} \]\[ y''(t) = \sum_{n=2}^{ } n(n-1) a_n t^{n-2} \]Substitute these into the given differential equation.

Substitute Derivatives into the Differential Equation

Substitute the power series expressions for \( y'' \), \( y' \), and \( y \) into:\[\sum_{n=2}^{ } n(n-1) a_n t^{n-2} = t \sum_{n=1}^{ } n a_n t^{n-1} + \sum_{n=0}^{ } a_n t^n + 1\]Align and simplify terms based on the coefficients of powers of \( t \).

Match Coefficients for Power Series

Equate the coefficients of like powers of \( t \) from both sides of the equation. This will generate a sequence of equations in terms of \( a_n \). Solve these iteratively using the initial conditions.

Solve for Unknown Coefficients

Substitute \( y(0)=1 \) and \( y'(0)=0 \) to determine specific coefficients (such as \( a_0 \) and \( a_1 \)). Use these to solve subsequent terms in the power series for \( y(t) \).

Construct the Solution

Using the determined coefficients, construct the explicit function for \( y(t) \). Verify it satisfies both the differential equation and initial conditions.

Key concepts

Power Series Expansion

A power series is a way of expressing a function as an infinite sum of terms calculated from its derivatives at a single point. When it comes to differential equations, a power series expansion is incredibly useful, especially when dealing with variable coefficients.
In our example, we assume the solution takes the form of a power series:

• \[ y(t) = \sum_{n=0}^{\infty} a_n t^n \]

This expression allows us to expand each function within the differential equation into its series form. By differentiating the power series, we get expressions for both \(y'\) and \(y''\).
This step is crucial as it helps transform the differential equation into an algebraic equation related to the coefficients \(a_n\).
The primary goal here is to align series and solve for unknown coefficients by matching terms of the same power in \(t\). This method is a cornerstone for solving differential equations with variable coefficients.

Initial Conditions

Initial conditions are specific values used to find the particular solution to a differential equation. They specify the value of the function and its derivatives at a particular point, usually \(t=0\).

For our differential equation, we have:

• \( y(0) = 1 \)
• \( y'(0) = 0 \)

These initial values allow us to uniquely determine the coefficients \(a_0\) and \(a_1\) of our power series expansion. Substituting these conditions ensures that our series solution satisfies these specific values at \(t = 0\).
Without initial conditions, the solution would remain in its most general form, containing arbitrary constants. With them, we tailor our solution to fit the particular scenario described by the original problem.

Non-Homogeneous Linear Differential Equation

A differential equation is termed as non-homogeneous if it contains a term that does not depend on the function or its derivatives. In our case:

• The equation \[ y'' = t y' + y + 1 \] has a constant \(+1\), making it non-homogeneous.

The form of such an equation usually is \(Ly = g(t)\), where \(L\) is a linear differential operator, and \(g(t)\) is a non-zero function.
To solve it, one typically finds both the complementary solution to the homogeneous equation \(Ly = 0\) and a particular solution to the non-homogeneous part.
The complete solution is the sum of these two solutions.
In our problem, the presence of the variable \(t\) in the coefficients and the non-zero constant term requires a more advanced technique, such as a power series expansion, to find the particular solution.

Variable Coefficients

The coefficients of a differential equation are functions of the independent variable rather than constants. In our differential equation:

• \( y'' = t y' + y + 1 \)

The presence of the variable \(t\) as a multiplier of \(y'\) classifies it as a variable coefficient differential equation.
These complications often rule out straightforward solutions like the characteristic equation, which is used for constant coefficients.
The variable coefficients make this equation particularly interesting because it can describe more complex, real-world phenomena.

Dealing with variable coefficients requires techniques like substitution or, as we've explored, a power series approach. By expanding in a power series and matching coefficients for each power of \(t\), we can find solutions even in the presence of these varying coefficients.