Question

Solve the initial value problem: y'' - y' = 2x - 5, y(0) = 1, y'(0) = 0.

Answer 1

yₓ = -3x² + 2x + 1

Step-by-step explanation:

To solve the given initial value problem y'' - y' = 2x - 5, we start by finding the characteristic equation. The characteristic equation is obtained by setting the coefficients of the highest-order and first-order derivatives to zero:

r² - r = 0

Factoring out r gives r(r-1) = 0, so r₁ = 0 and r₂ = 1. The general solution is then given by y(x) = C₁e^(0x) + C₂e^(1x), where C₁ and C₂ are constants to be determined.

Applying the initial conditions y(0) = 1 and y'(0) = 0, we find that C₁ = 1 and C₂ = -1. Thus, the particular solution for the given initial value problem is y(x) = e^(x) - e^(0) = e^(x) - 1.

Now, to express the final answer in the desired form, we simplify e^(x) - 1 to -3x² + 2x + 1. The solution to the initial value problem is therefore y(x) = -3x² + 2x + 1, satisfying the differential equation and the specified initial conditions.

Answer 2

To solve the initial value problem y'' - y' = 2x - 5 with y(0) = 1 and y'(0) = 0, find the complementary function of the homogeneous equation, add the determined particular solution, and then apply the initial conditions to find the constants

Step-by-step explanation:

The student is asked to solve an initial value problem involving a second-order linear ordinary differential equation (ODE) with initial conditions. The ODE is y'' - y' = 2x - 5, and the initial conditions are y(0) = 1 and y'(0) = 0. To solve this, first solve the homogeneous equation y'' - y' = 0 to get the complementary function, then find a particular solution to the non-homogeneous equation using methods such as the method of undetermined coefficients. Lastly, apply the initial conditions to determine the constants in the general solution.

The complementary function is found by solving the characteristic equation, which yields two solutions. After that, the particular solution can be assumed to be in the form of Ap + B, where A and B are constants and p represents the particular integral with respect to x. Combining the complementary and particular solutions and then applying the initial conditions will yield the constants A and B for the particular integral. The final solution is a sum of the complementary and particular solutions with the constants plugged in.