Which of the following forcing terms have characteristic form? For the ones that do, give the characteristic and the degree.
\(u\)
\(te^{-\frac{t}{2}}\)
\((v^2 + v + 1)e^v\)
\(te^t\sin t + (t^2 + 1)e^t\cos t\)
\(\frac{\sin(t)}{t}\,e^{-3t}\)
\(z\cos(z)\)
\(\frac{\cos(x)}{e^x} + e^{2x}\sin(x)\)
\(t\tan(t) + t\sec(t)\)
\(\cos(2w)\)
\(\cos(3y)\cos(4y)\)
For #2–#5, find a general solution to the differential equation by using undetermined coefficients.
\(v'' + 3v' + 2v = 200e^{3u}\)
\(w'' - 4w' + 5w = e^{2x}\)
\(\ddot{y} + 5\dot{y} + 6y = e^{-3t}\)
\(y''+ 2y' - 3y = 16e^t\cos(2t)\)
For #6–#9, find a general solution to the differential equation by using key identity evaluations.
\(y'' - 6y' + 8y = \cos(2t)\)
\(y'' - 4y' + 13y = 5e^{3t}\)
\(w'' + 9w = 2\sin(3z)\)
\(x'' + x = t\cos(t)\)
For #10–#20, solve the following initial value problems. Some of them will be easier with undetermined coefficients and some will be easier with key identity evaluations, but the choice of which method to use is left up to you.
\(y'' - y' = 8\), \(y(0) = 2\), \(y'(0) = 8\)
\(v'' - 4v = 15e^{3u}\), \(v(0) = 5\), \(v'(0) = 21\)
\(y'' - 2y' + 2y = 100\cos(2t)\), \(y(0) = 0\), \(y'(0) = -25\)
\(w'' - 3w' + 2w = 4ze^z\cos(z)\), \(w(0) = 0\), \(w'(0) = 0\)
\(y'' - 5y' + 6y = 2\sin(2t)\sin(3t)\), \(y(0) = \frac{1176}{9860}\), \(y'(0) = \frac{264}{9860}\) [Hint. These numbers were chosen so that some cancellation would occur.]
\(w'' - 3w' + w = e^t + e^{2t}\), \(w(0) = -2\), \(w'(0) = 0\)
\(y'' - 2y' + 10y = 3 e^t + 37\cos(3t)\), \(y(0) = y'(0) = 1\)
\(y'' - 8y' + 17y = 100x^2e^x\), \(y(0) = 5\), \(y'(0) = 0\)
\(y'' + 2y' + 5y = 4e^{-t}\sin(2t)\), \(y(0) = \frac{1}{2}\), \(y'(0) = \frac{3}{2}\)
\(y'' - 2y' + y = (3u+1)e^u\), \(y(0) = 4\), \(y'(0) = 0\)
\(y'' + y = 3t^2\cos(2t) - 6t^2\sin(2t)\), \(y(0) = 3\), \(y'(0) = 0\)
For #21–#23, find the Green function for the following differential equations.
Don’t try to produce a general solution.
\(v'' + 2v' + 5v = \log(u)\)
\(y'' - 4y' + 4y = \frac{1}{t}\)
\(\ddot{y} - 6\dot{y} - 7y = \sec(2t)\)
For #24–#27, find general solutions to the differential equations using Green functions.
Some of them may involve antiderivatives that you can’t completely evaluate; leaving things in terms of integrals is okay.
\(\ddot{y} = \frac{1}{t}\) [Note. Yes you can solve this by just integrating twice. This is an easy differential equation to solve, even with Green functions.]
\(\ddot{w} + 16w = \sec(4x)\)
\[y'' - 3y' = \frac{1}{e^{3t} + 1}\]
\[\ddot{y} - 4y = \frac{1}{e^{2t} + e^{-2t}}\]
Remark: While we’ve focused so far on forcing terms that are continuous, it’s possible to create solutions to differential equations with discontinuous forcing using the methods we’ve learned. Often physical models of real-world situations will include functions with jumps in them; this could represent the almost-instantaneous effect of flipping a switch that causes a voltage to pass through a current, for instance. Since the solution to the differential equation is supposed to represent something physical (like a position or a voltage), we assume the solution is going to be continuous. This suggests the following method of solution:(a) solve the differential equation using the forcing term as it looks at \(t=0\), using the initial conditions, then(b) find the value of the differential equation when the forcing term jumps, and finally (c) use that value as an initial condition to solve a second differential equation, starting at the jump point, using the second case of the forcing term.
Use the above framework to solve the following two differential equations.
\[y'' + y = \left\{ \begin{array}{ll} 3\sin(2t) & 0 \leq t < \pi \\ \sin(t) & t \geq \pi \end{array} \right. \text{ where }\;\;y(0) = 1,\, y'(0) = -2\]
\[z'' - z = \left\{ \begin{array}{ll} 1 & 0 \leq u < 1 \\ e^{-2u} & u \geq 1 \end{array} \right. \text{ where }\;\; z(0) = z'(0) = 0\] Suggestion. This could possibly get a little ugly; try using three decimal places to estimate \(e\).]
Consider the equation
\[v'' - 6v' + 25v = e^{x^2}.\]
a) Compute the Green function \(g(x)\) associated with the differential equation.
b) Find a particular solution \(V_p(x)\) in terms of definite integrals.
Find the Green function associated with the differential operator \(\Lop = \Dop^2 - 16\).
Give a particular solution to the following equation:
\[w'' + 8w' + 25w = 10e^{-t},\] using two of the methods discussed in this section. How do they compare to the third method discussed?
\(\bf{A~bit~of~thinking~outside~the~box}\)
(a) Consider the equation \(z^{(7)} - 2z^{(6)} + z^{(5)} - 2z^{(4)} = 0\). Given that the characteristic polynomial associated with this equation is of the form \(z^4(z - 2)(z^2 + 1)\), write down a general solution to this homogeneous, constant coefficient, linear seventh-order differential equation.
(b) Now consider the equation \(z^{(7)} - 2z^{(6)} + z^{(5)} - 2z^{(4)} = 4x + \cos(x) + e^{16x}\). Write down a guess for a particular solution \(Z_p(x)\) to this non-homogeneous equation. There is no need to solve for the undetermined coefficients!
Write down the form of the particular solution to
\[\ddot{y} + p(x)\dot{y} + q(x)y = g(x)\] for the following \(g(x)'s\):
a) \(g(x) = e^{7x} + 6\);
b) \(g(x) = 10e^{-5x} + e^{-5x} \cos(6x) - \sin(6x)\);
c) \(g(x) = x^2 \cos(x) - 5x\sin(x)\).
\(\bf{Hint}\) Don’t worry about the form of \(p(x)\) or \(g(x)\) for now, but concentrate on the form of the composite characteristic form instead.
Consider the following differential equation
\[a \ddot{w} + b\dot{w} + cw = g(u),\] where \(a, b,\) and \(c\) are positive. If \(W_1(u)\) and \(W_2(u)\) are solutions of the differential equation, prove that \(W_1(u) - W_2(u) \rightarrow 0\) as \( u\rightarrow \infty\). Does the result hold when \(b = 0\)?
Here we will show an alternative method towards solving the differential equation
\[y'' + by' + cy = (\Dop^2 + b\Dop + c)y = f(x),\] where \(b\) and \(c\) are constants, and \(\Dop\) is the differentiation operator with respect to \(x\). Let \(z_1\) and \(z_2\) be the zeros of the characteristic polynomial of the corresponding homogeneous equation. Note that these roots could be distinct real, real and equal or complex conjugates of each other.
(a) Show that the differential equation \(y'' + by' + cy = (\Dop^2 + b\Dop + c)y = f(x), \) can be rewritten as \((\Dop - z_1) (\Dop - z_2)y = f(x)\) and find out what \(z_1\) and \(z_2\) are in terms of \(b\) and \(c\).
(b) Let \(v = (\Dop - z_2)y\). Show that the solution of \(y'' + by' + cy = (\Dop^2 + b\Dop + c)y = f(x), \) can be found by solving the following system of first-order non-homogeneous differential equations: \[(\Dop - z_1)v(x) = f(x),\] \[(\Dop - z_2)y(x) = v(x).\]
In the light of the previous problem, use the method outlined above to solve the following differential equation:
\[w'' -3w' - 4w = 3e^{2u}.\]
\(\bf{Note:}\) We have other methods for solving this differential equation as well, but here we would like to illustrate how annihilating the second-order operator yields a system of first-order equations.
Find a second order homogeneous constant coefficient equation whose general solution is of the form \(v(u) = c_1 \cos(5u) + c_2 \sin(5u) - e^u \sin(2u).\)
Suppose that \(w_1 = 2u \sin(3u)\) is a solution of the following equation \[\ddot{w} + 2\dot{w} + 2w = f_1(u),\] and \(w_2 = \cos(6u) - e^{-u}\cos(u)\) is a solution of the following equation \[\ddot{w} + 2\dot{w} + 2w = f_1(u).\] What is the general solution of the following equation? \[\ddot{w} + 2\dot{w} + 2w = 5 f_1(u) - 2 f_2(u)\]
Find a second order homogeneous linear constant coefficient differential equation whose general solution is of the from \(y(x) = c_1 e^{-x} + c_2 x e^{-x} + x^3 - 5x\).
Suppose the equation \(w'' - 4w' - 5w = f(x)\) has \(w = 3x^3\) has one of its solutions.
(a) Which one of the following functions is also a solution?
\(\bf{Note:}\) More than one answer might be possible.
\(w = e^{-u} + 3x^3;\)
\(w = 2080 e^{5x} + 3x^3;\)
\(w = 2080 e^{5x} + 6x^3;\)
\(w = e^{-x} + e^{5x} + 3x^3;\)
\(w = 20e^{-x} + 14e^{5x} + x^3.\)
(b) What is the general solution of the equation?
(c) Find \(f(x)\).
(d) Knowing that \(w(0) = 3\) and that \(w'(0) = 3\), solve the initial value problem.
\(\bf{More~on~Annihilation~Operators}\)
As we’ve seen before, an annihilation operator for a given function \(f\) is a polynomial \(P(\Dop)\) for the differential operator \(\Dop\) such that \(P(\Dop)f = 0\). In other words, f is the solution to the homogeneous equation characterized by \(P(\Dop)f = 0\). For example, \(f(t) = e^t\) satisfied the first-order differential equation \(f' - f = 0\) and the corresponding annihilation operator for \(f\) is \(\Dop - 1\), in that \((\Dop - 1)f = 0\). For another example, \(f(t) = 2t\), the corresponding annihilation operator is \(\Dop^2\) because \(\Dop^2(f) = 0\).
Just to get ourselves thinking in the annihilation operator mode one more time, find the corresponding annihilation operator for the following functions and write down the corresponding homogeneous differential equation:
a) \(\sin(x)\);
b) \(v e^{v}\);
c) \(2 e^{3x} + \cos(x)\).
Consider the following differential equation \(w'' -5w' + 6w = e^{2v}\). Write down a general solution to the differential equation using the method of annihilators and starting from the general solution, name exactly which is the particular solution. \(\bf{Note:}\) This problem can be tackled using either of the three methods discussed in this section, but here we are illustrating the use of annihilators in achieving the same goal!
Consider the following differential equation \(v''' - v'= x + 1\). Write down a general solution to the differential equation using the method of annihilators and starting from the general solution, name exactly which is the particular solution.
\(\bf{Note:}\) Once again, this problem can also be tackled using either of the three methods discussed in this section, but here we are illustrating the use of annihilators in achieving the same goal!
Using the method of annihilators, determine the general solution to the following constant coefficient differential equations:
a) \((\Dop - 4) (\Dop + 1)w = 16 u e^{3u}\);
b) \((\Dop - 2) v = 3\cos(u) + 4\sin(u)\).
Find the differential equations (in operator form) solved by the following functions:
a) \(v(x) = 14 + x^2 - 21 \cos(x);\)
b) \(w(t) = 10t^2e^{4t};\)
c) \(v(u) = - e^{3u} + u e^{3u} - 4ue^{-u}\cos{3u}.\)