Question

Substitute \(y=a+b t+c t^{2}\) into \(y^{\prime}+y=1+t^{2}\) to find a particular solution.

Short answer

The particular solution is \(y = 1 + t^2\).

Step-by-step solution

Differentiate the Substitution Function

Start with the given function: \(y = a + b t + c t^2\). Differentiate it with respect to \(t\). This gives:\[y' = \frac{d}{dt}(a + b t + c t^2) = b + 2ct.\]

Substitute into the Differential Equation

Substitute both \(y\) and \(y'\) from the previous step into the differential equation \(y' + y = 1 + t^2\).\[(b + 2ct) + (a + b t + c t^2) = 1 + t^2.\]This simplifies to:\[a + 2b t + c t^2 = 1 + t^2.\]

Equate Coefficients

Since the equation \(a + 2b t + c t^2 = 1 + t^2\) must hold for all \(t\), equate the coefficients of like powers of \(t\):- Constant term: \(a = 1\).- \(t\) term: \(b = 0\).- \(t^2\) term: \(c = 1\).

Write the Particular Solution

Substitute the found coefficients \(a = 1\), \(b = 0\), and \(c = 1\) into the original \(y\) expression. The particular solution is:\[y = 1 + 0 \cdot t + 1 \cdot t^2 = 1 + t^2.\]

Key concepts

Particular Solution

A particular solution in the context of differential equations refers to a specific solution that satisfies both the differential equation and initial conditions, if any are provided. In problems like this, the goal is to find one such solution among possibly infinitely many other solutions.

When the differential equation is given without initial conditions, the particular solution is primarily used because it's tied to the specific form of the equation. For instance, with our function involving terms like \(a + bt + ct^2\), the particular solution helped us match the form \(1 + t^2\) on the right-hand side of the equation. This ensures that the left side matches for any value of \(t\).

Particular solutions are critical in scenarios where the equation model or scenario provides a "target" outcome, which in this example was \(1 + t^2\). While the general solution covers all potential answers, the particular solution narrows down to the one result that matches specific system behavior.

Polynomial Substitution

Polynomial substitution is a technique used in solving differential equations, where we guess a solution and plug it into the equation to simplify or solve. In this exercise, we substitute the polynomial \(y = a + bt + ct^2\) into our differential equation.

The polynomial form here involves:

• Constant term: \(a\)
• Linear term: \(bt\)
• Quadratic term: \(ct^2\)

This form is substituted because it mirrors the terms on the right side of the equation \(1 + t^2\). This seems strategic as it allows us to directly compare and equate each term according to powers of \(t\).

By doing so, we align both sides of the equation, making it easier to see how coefficients must be selected, leading us directly to a particular solution. Polynomial substitution is a commonly used method when the differential equation's form suggests a polynomial structure.

Equating Coefficients

Equating coefficients is a technique used to solve for unknowns in polynomial expressions by matching coefficients of corresponding terms on both sides. In the equation \(a + 2bt + ct^2 = 1 + t^2\), each term's power must match.

The process involves:

• Matching constant terms: \(a\) on the left with \(1\) on the right, giving us \(a = 1\).
• Matching \(t\) terms: \(2b\) on the left with no corresponding term on the right, resulting in \(b = 0\).
• Matching \(t^2\) terms: \(c\) on the left with \(1\) on the right, leading to \(c = 1\).

Equating coefficients efficiently breaks down the problem by dealing with each element individually. By addressing each piece separately, we can more easily find the values of unknown coefficients. This step-by-step itemization is instrumental in simplifying complex equations down to solvable problems, quickly leading us to the correct solution of the differential equation.