Question

find the fundamental set of solutions specified by Theorem 3.2.5 for the given differential equation and initial point. $$ y^{\prime \prime}+4 y^{\prime}+3 y=0, \quad t_{0}=1 $$

Short answer

Based on the given differential equation and the step-by-step solution provided, the fundamental set of solutions for the differential equation is: $$ \{e^{-3t}, e^{-t}\} $$

Step-by-step solution

Write down the characteristic equation

The characteristic equation of the given differential equation is: $$ r^2 + 4r + 3 = 0 $$

Find the roots of the characteristic equation

We can solve the quadratic equation by factoring or using the quadratic formula, but in this case, the characteristic equation can be factored as: $$ (r + 3)(r + 1) = 0 $$ Hence, the roots are \(r_1 = -3\) and \(r_2 = -1\).

Write down the general solution

Since we have found two distinct real roots, the general solution of the differential equation is given by: $$ y(t) = C_1 e^{-3t} + C_2 e^{-t} $$ Now, we have the general solution, and we can find the fundamental set of solutions by applying the initial condition.

Apply the initial condition

The given initial point is \((t_0, y_0) = (1, 0)\). Substitute \(t_0=1\) into the general solution: $$ 0 = C_1 e^{-3(1)} + C_2 e^{-1(1)} $$ However, finding the constants \(C_1\) and \(C_2\) in this case is not necessary, because we only need the fundamental set of solutions. Notice that we do not have a value for \(y_0\) which would allow us to find a unique solution by solving for \(C_1\) and \(C_2\).

Write down the fundamental set of solutions

Since we have a second-order homogeneous linear differential equation, the fundamental set of solutions consists of the two linearly independent solutions given by the general solution. Hence, the fundamental set of solutions can be written as: $$ \{e^{-3t}, e^{-t}\} $$