7.10. Practice Problems

Solve y′′+5y′+6y =3e−x

This problem requires solving a second-order linear non-homogeneous differential equation. The equation consists of a homogeneous part (y′′ + 5y′ + 6y = 0) and a non-homogeneous part with a forcing function (3e−x). Step 1 will involve finding the complementary function by solving the characteristic equation associated with the homogeneous part. Step 2 will involve guessing a particular integral that fits the form of the non-homogeneous term (3e−x). After substituting this guess into the original equation and solving for unknown coefficients, we will combine the results to obtain the general solution.

Consider a scenario where you're trying to model how an object behaves when an external force (like a spring in damping motion) is applied. The homogeneous part represents the natural behavior of the system without external force, while the forcing function represents how the system responds to that external push.

Solve y′′−4y = x²

For this problem, we also have a second-order linear non-homogeneous differential equation. The first step is to find the complementary function by solving the homogeneous equation (y′′ - 4y = 0). The characteristic polynomial will be r² - 4 = 0, which yields roots r = 2 and r = -2. This leads to the complementary function y_c = C₁e^(2x) + C₂e^(-2x). Next, we guess a particular solution for the non-homogeneous term (x²) by assuming a trial solution that fits the polynomial form. Once substituted into the original equation and simplified, we solve for coefficients to find the particular integral. Combining both solutions gives the general solution.

Think of this as modeling the displacement of a beam under a specific load – here, the load can be represented as the polynomial term x². The first part of the problem shows how the beam moves without any load, allowing us to see its natural deflection before understanding how additional loads like x² modify its behavior.

Solve y′′+y = xsin(x)

This differential equation features a forced term involving a product of functions (xsin(x)). The complementary function is first found by solving the associated homogeneous equation (y′′ + y = 0). We find y_c = C₁cos(x) + C₂sin(x). Next, for the particular solution, we guess a trial function reflecting the non-homogeneous term. Since both x and sin(x) are present, we could assume a form such as x(Acos(x) + Bsin(x)). After determining the derivatives and substituting into the original equation, we solve for coefficients A and B by matching against the non-homogeneous term.

Imagine swinging on a swing while someone is pushing you periodically. Your natural swinging motion represents the homogeneous solution, while the multiple pushes during your swing can be modeled by the xsin(x) term. This analogy helps visualize how we can analyze both the natural behavior and responses due to external influences.

Solve y′′+2y′+y = e−x Hint: The complementary function here will include e−x, so use modification.

In this problem, the method starts similarly by deriving the complementary function from the homogeneous part (y′′ + 2y′ + y = 0), leading us to a characteristic equation with roots that provide a solution of the form y_c = C₁e^(-x) + C₂xe^(-x). Notably, we have to modify the form of our trial solution for the particular integral because e^(-x) appears in y_c. Therefore, our trial guess must be x²e^(-x) (to eliminate duplication). This approach avoids overlap with the complementary function when we substitute this into the original equation to solve for coefficients and find the complete general solution.

This can relate to how a damped oscillator behaves under a steady external force. You have the natural decay of the oscillator's movement represented by the e^(-x) form, while the external push must be modeled correctly to ensure we capture the whole behavior of the system, ensuring we handle overlaps effectively.

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