Match the graphs of solutions in Figure 11.89 with the differential equations below. (a) x^''+4

step 1 The equation is in the case B squared minus 4C to be less than 0. So this implies that your gene

Match the graphs of solutions in Figure 11.89 with the...

Key Concepts

Stability of Solutions

The stability of a differential equation's solutions is determined by the real parts of the roots of its characteristic equation. If the real parts are negative, solutions decay over time (stable); if positive, they grow exponentially (unstable); and if zero, they neither grow nor decay, often corresponding to sustained oscillations in undamped systems.

Damping

Damping refers to the presence of a first derivative term in the differential equation, which can either reduce or amplify oscillations over time. A positive damping coefficient generally results in the decay of oscillations (energy dissipation), whereas a negative coefficient can lead to growing oscillations (anti-damping), affecting the stability of the system.

Oscillatory Behavior

When the roots of the characteristic equation are complex conjugates, the solutions exhibit oscillatory behavior. This means the system will have solutions that involve sine and cosine functions, often corresponding to periodic motion in physical systems such as undamped or underdamped oscillators.

Second Order Linear Differential Equations

These are equations involving the second derivative of a function, typically expressed in the form a·x'' + b·x' + c·x = 0, where the coefficients a, b, and c are constants. The solutions to these equations describe dynamic systems and their behavior over time, such as oscillations or exponential growth/decay.

Characteristic Equation

For a second order linear differential equation with constant coefficients, the characteristic equation is obtained by assuming a solution of the form x = e^(rt). This leads to a quadratic equation in r (r^2 + (b/a)r + (c/a) = 0), and the nature of its roots (real distinct, real repeated, or complex conjugate) determines the qualitative behavior of the solution.

Transcript

00:01 The equation is in the case b squared minus 4c to be less than 0.

00:10 So this implies that your general solution will be yt to be equal to c 1e to the power alpha t cause v3 plus c2a to the power alpha t sign.

00:32 B t, beta t.

00:35 So the characteristic equation, you have r squared plus 4 equal to 0.

00:42 So this implies that the roots or the solutions of r are going to be r1, r2 is going to be plus or minus 2.

00:54 So you have alpha to be equal to 0, then beta to be equal to 2.

01:00 So therefore the general solution to your general solution then solution is going to be you have x t which is equal to c 1 cost you have see one course 2 t plus c2 sign 2 t so then this implies that the graph of equation so the graph of equation x t is the graph shown i v then for the second part b so b as well the case for b is b squared minus 4 c to be greater than zero so the solution for that is going to be y to be equal to c1 f c1 e to the power r1 t plus c2 to the power r2 t and the characteristic equation we have is r to the power 2 minus 4 equal to 0 so then the roots of r 2 is going to be plus or minus 2.

03:04 So this implies that our general solution, which is our xt, is going to be c1.

03:19 We have c1, e to the power negative 2t, plus c2, c2, c2, e to the power 2t.

03:38 So this represents the graph i -i then for the third part so for the third part c you have the case where you have b squared minus 4c to be less than zero.

04:21 So that your general solution is of the form y t which is c1 e to the power alpha t cause beta t plus c2 e to the power alpha t sign beta t in a characteristic equation it's r squared minus 0 .2r plus 1 .01 to be equal to 0...

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