Question

. Given an ordinary differential equation y" - 4y' + 3y = 1. a. Show that y₁ (t) = et and y₂ (t): =e³t are the solutions for the differential equation above.

Answer 1

By taking the first and second derivatives of y₁(t) = e⁴t and y₂(t) = e³t, and substituting into the differential equation y" - 4y' + 3y = 1, we can show that both functions satisfy the equation, proving they are solutions.

Step-by-step explanation:

To show that y₁(t) = e⁴t and y₂(t) = e³t are solutions to the differential equation y" - 4y' + 3y = 1, we must verify that they satisfy the equation when substituted. Taking the first and second derivatives of y₁(t) gives us y₁'(t) = e⁴t and y₁"(t) = 4e⁴t. Substituting these into the equation yields (4e⁴t) - 4(e⁴t) + 3(e⁴t) which simplifies to 1, the right-hand side of the equation, thus y₁(t) is indeed a solution. Repeating the process for y₂(t), with its first and second derivatives y₂'(t) = 3e³t and y₂"(t) = 9e³t, and substituting, we get (9e³t) - 4(3e³t) + 3(e³t), which again simplifies to 1, confirming y₂(t) is also a solution.