Question

Problems In each of Problems 1 through 3, transform the given equation into a system of first-order equations. s 1. u" +0.5u' + 2u = 0) 2. tu" + tu' + (12 – 0.25) u = 0 3. u(4) – 1 = 0

Answer 1

To transform the given equation into a system of first-order equations, we introduce new variables. The equation tc' + tv + (12 - 0.25)c = 0 gives us a system of first-order equations: c' = v and v' = -(12 - 0.25)c - tv. The initial condition for this system is c(4) - 1 = 0.

Step-by-step explanation:

In order to transform the given equation into a system of first-order equations, we need to introduce new variables. Let's define two new variables: v = u' and c = u

The given equation 2 can be rewritten as tu" + tu' + (12 - 0.25)u = 0. Substituting v = u' and c = u, we obtain tc' + tv + (12 - 0.25)c = 0. This gives us a system of first-order equations: c' = v and v' = -(12 - 0.25)c - tv.

For the given equation 3 u(4) - 1 = 0, we can express it as c(4) - 1 = 0, where c(4) denotes the value of c at t = 4. This represents an initial condition for the system of first-order equations we obtained in the previous step.

Answer 2

1. The system of first-order equations is u' = v and v' = -0.5v - 2u.

2. The system of first-order equations is u' = v and v' = -0.25u - u'/t.

3. The system of first-order equations is u' = 0 and u = 1.

Step-by-step explanation:

In problem 1, the given second-order differential equation u" + 0.5u' + 2u = 0 is transformed into a system of first-order equations. We introduce a new variable, v = u', and express the original equation as a system of two equations: u' = v and v' = -0.5v - 2u. This substitution allows us to break down the higher-order equation into a set of simpler first-order equations. The first equation captures the relationship between u and v, while the second equation represents the derivatives of u and v in terms of themselves and each other.

For problem 2, we adopt a similar strategy. Introducing v = u', we rewrite the given equation tu" + tu' + (12 – 0.25)u = 0 as the system u' = v and v' = -0.25u - u'/t. This transformation simplifies the problem by expressing the second-order differential equation in terms of two first-order equations.

In problem 3, the equation u(4) – 1 = 0 is already a first-order equation as it involves only the first derivative. Thus, the system becomes u' = 0 and u = 1. This represents a constant function, as the derivative is zero, and the initial condition is satisfied.