1.5.3. Example

Solve: d²y/dx² - 5 dy/dx + 6y = 0

This chunk presents the differential equation that we need to solve. It is a second-order linear homogeneous differential equation, indicated by the presence of derivatives of y up to the second order (d²y/dx²) and no explicit external driving force (non-homogeneous term). To solve this type of equation, we need to find a solution that satisfies this equation for all values of x.

Think of this equation as a representation of how a spring might behave when it's stretched or compressed. The terms in the equation describe the forces acting on the spring, with the goal of understanding how it will return to its original position after being disturbed.

• AE: m² - 5m + 6 = 0 • Solve for roots m₁, m₂

To solve the differential equation, we first form the Auxiliary Equation (AE) from the given differential equation. This transforms our problem of differential equations into an algebraic form. The AE helps us determine the roots m₁ and m₂ that will contribute to our general solution. We can solve this quadratic equation using the quadratic formula or by factoring.

Imagine trying to find the points at which a parabolic shape (like a water fountain’s arc) crosses the ground (the x-axis). The solutions or roots to this quadratic equation tell us where those intersections (real values of x) occur.

• m = 2, 3

By solving the Auxiliary Equation, we find that the roots are m = 2 and m = 3. These roots are the key to forming our general solution. Each root corresponds to a particular solution that we can combine for the complete solution to the original differential equation.

Consider a race where two cars, one fast and one slower, leave the starting line at the same time. The points (roots) where they meet on the track can be seen as critical junctions that help us predict how the race (the solution) will progress.

• General solution: y = C₁e²ˣ + C₂e³ˣ

With the found roots m₁ and m₂, we can now state the general solution to the differential equation. The general solution combines both exponential functions based on the roots we found, with C₁ and C₂ being arbitrary constants that represent the influence of initial conditions on the solution.

Imagine creating a custom blend of tea by mixing two different types of tea leaves. C₁ and C₂ represent the proportions of each type in your blend. Depending on how much of each you add (initial conditions), the flavor (the shape of the solution) changes, but both types of leaves (the solution components) remain crucial.

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