"Deriving y & Solving y'' + 4y' - 5y = 0 with C" is a mathematical equation that represents a second-order linear differential equation. In simpler terms, it is a way to express a relationship between a function and its derivatives.
The 'y' in the equation represents the dependent variable, or the function that is being differentiated.
The constant 'C' is known as the arbitrary constant and it represents the family of solutions to the differential equation. It is used to account for any variations in the initial conditions or specific values of the function.
To solve "Deriving y & Solving y'' + 4y' - 5y = 0 with C", you would follow these steps:
1. Rewrite the equation in standard form: y'' + 4y' - 5y = 0
2. Find the roots of the characteristic equation: r^2 + 4r - 5 = 0
3. Use the roots to find the general solution: y = C1e^(-5x) + C2e^(x)
4. Use initial conditions or specific values to solve for the arbitrary constant 'C'.
Solving "Deriving y & Solving y'' + 4y' - 5y = 0 with C" allows us to find the function that satisfies the given relationship between the function and its derivatives. This is useful in many fields, such as physics, engineering, and economics, where differential equations are used to model and predict real-world phenomena.