17–20. Verifying solutions of initial value problems Verify that the given function y is a solution of the initial value problem that follows it.
y(t) = 8t⁶ - 3; ty'(t) - 6y(t) = 18, y(1) = 5

Verified step by step guidance

1. Start by identifying the given function and the initial value problem. The function is \(y(t) = 8t^{6} - 3\), and the differential equation is \(t y'(t) - 6 y(t) = 18\) with the initial condition \(y(1) = 5\).
2. Find the derivative of the function \(y(t)\) with respect to \(t\). Use the power rule to differentiate \(y(t) = 8t^{6} - 3\), which gives \(y'(t) = 48t^{5}\).
3. Substitute \(y(t)\) and \(y'(t)\) into the differential equation \(t y'(t) - 6 y(t) = 18\). This means replacing \(y'(t)\) with \$48t^{5}\( and \)y(t)\( with \)8t^{6} - 3$.
4. Simplify the left-hand side expression after substitution: calculate \(t imes 48t^{5} - 6 imes (8t^{6} - 3)\) and simplify the terms to check if it equals 18 for all \(t\).
5. Verify the initial condition by substituting \(t = 1\) into the original function \(y(t) = 8t^{6} - 3\) and check if \(y(1) = 5\) holds true.

Key Concepts

Initial Value Problem (IVP)

An initial value problem consists of a differential equation along with a specified value of the unknown function at a particular point. Solving an IVP means finding a function that satisfies both the differential equation and the initial condition, ensuring the solution is unique and fits the given starting value.

Verification of a Solution

To verify a solution to a differential equation, substitute the proposed function and its derivative into the equation to check if the equality holds. Additionally, confirm that the function satisfies the initial condition by evaluating it at the given point.

Differentiation of Polynomial Functions

Differentiation involves finding the derivative of a function, which represents its rate of change. For polynomial functions like y(t) = 8t⁶ - 3, apply the power rule by multiplying the coefficient by the exponent and reducing the exponent by one to find y'(t).