For #1–#7, find a general solution to the differential equation by using variation of parameters.
(#1—#3 can be done by other methods as well, but stick to variation of parameters for them.)
\(v'' - v = e^u\)
\(y'' + y = \tan(t)\)
\(tx'' + x' = 1\)
(The homogeneous solutions are \(1\) and \(\ln(t)\).)
\(x^2y'' - 6xy' + 10y = x^6\cos(2x)\)
(The homogeneous solutions are \(x^2\) and \(x^5\).)
\(xw'' - (1+x)w' + w = x^2e^x\)
(The homogeneous solutions are \(e^x\) and \(1+x\).)
\(\displaystyle x^3y'' - xy' + y = \frac{1}{x^2}e^{-\frac{1}{x}}\)
(The homogeneous solutions are \(x\) and \(xe^{-\frac{1}{x}}\).)
\(\displaystyle z^2w'' + zw' + \left( z^2 - \frac{1}{4}\right)w = z\sqrt{z}\)
(The homogeneous solutions are \(\frac{\cos(z)}{\sqrt{z}}\) and \(\frac{\sin(z)}{\sqrt{z}}\).)
For #8–#14, find a general solution to the differential equation by integrating against a Green function. (Some can be done by other methods as well, we admit.)
\(y'' + y = t + 1\)
\(xw'' + w' = \frac{1}{x}\)
(The homogeneous solutions are \(1\) and \(\log(x)\). Look for a solution on \((0, \infty)\).)
\((t+1)y'' + 2ty' - 4y = (t+1)^3\)
(The homogeneous solutions are \(e^{-2t}\) and \(2t^2 + 2t + 1\). Look for a solution on \((-1, \infty)\))
\(6u^2\ddot{v} + u\dot{v} + v = u+1\)
(The homogeneous solutions are \(\sqrt{u}\) and \(\sqrt[3]{u}\). Find a solution on \((0, \infty)\).)
\(\displaystyle y'' + \frac{2}{t}\,y' - y = \frac{\cos(2t)}{t}\)
(The homogeneous solutions are \(\frac{1}{t}\,e^t\) and \(\frac{1}{t}\,e^{-t}\). Find a solution on \((0, \infty)\).)
\(\cos(x)\ddot{z} + \sin(x)\dot{z} = \cos^2(x)\sin(x)\)
(The homogeneous solutions are \(1\) and \(\sin(x)\). Look for a solution on \((-\frac{\pi}{2}, \frac{\pi}{2})\).)
\(\displaystyle y'' + \frac{4}{t} y' + \frac{2}{t^2} y = \log(t)\)
(The homogeneous solutions are \(\frac{1}{t}\) and \(\frac{1}{t^2}\). Look for a solution on \((0, \infty)\).] )
Suppose the differential operator \(\Lop\) has constant coefficients, say \[\Lop y = D^ny + a_1D^{n-1}y + a_2D^{n-2}y + \cdots + a_{n-2}D^2y + a_{n-1}D y + a_ny\text{,}\] where all the \(a_i\) are real numbers. Let \(g(x)\) be the Green function for this operator, as defined in the previous section (that is, a solution to the initial value problem \(\Lop g = 0\), \(g(0) = 0\), \(g'(0) = 1\)). Let \(G(x, s)\) be the Green function as defined in this section. Show that \(G(x, s) = g(x-s)\).
For #16–#17, find a general solution to the differential equation by using variation of parameters.
\(y''' - y' = \cos(2t)\)
\(xz''' - z'' = \log(x)\)
(The homogeneous solutions are \(1\), \(x\), and \(x^3\).)
For #18–#19, find a general solution to the differential equation by integrating against a Green function.
\(y''' - 19y' + 30y = e^{2t}\)
(Note \(z^3 - 19z + 30 = (z-2)(z-3)(z+5)\))
\(\displaystyle w''' - \frac{2}{t^2}\,w' = t\)
( The homogeneous solutions are \(1\), \(t^3\), \(\log(t)\). Find a solution on \((0, \infty)\).)
\(\bf{A~comparison~between~different~solving~techniques}\)
This is a good review of the solving techniques discussed so far for second-order, linear non-homogeneous differential equations.
a) Find a general solution to the following equation: \[w'' + 9w = 0.\]
b) Can you think of multiple ways of solving the following differential equation \(w'' + 9w = e^x?\)
c) Can you think of multiple ways of solving the folllowing differential equation \(w'' + 9w = \sec(3x)?\)
\(\bf{Remark}\) There should be at least 3 ways to approach part c) :).
The functions \(v^2\) and \(v\) are solutions to the following homogeneous equation
\[v^2 w'' - 2v w' + 2 w = 0,\] for \(v > 0\). (No need to check that \(v^2\) and \(v\) are indeed solutions!)
a) Compute the Wronskian of the two functions and evaluate it at \( v = 5\).
b) Solve the initial value problem \[v^2 w'' - 2v w' + 2 w = v^3 e^v,~~w(1) = w'(1) = 0,~~~v~>~0,\] using both solving technques discussed in this section. You should be able to evaluate all the definite integrals.
Write down a general solution to the non-homogeneous second-order linear equation \(x'' + x = \sec(t)\), for \(t \in (-\frac{\pi}{2}, \frac{\pi}{2})\), using the two methods discussed in this section. Which one is lengthier/easier to solve with?
Here we will illustrate how the nonhomogeneity in an initial value problem can be separated in two initial value problems.
Show that the solution to the differential equation \(\Lop[v] = v'' + p(x) v' + q(x) v = f(x), v(x_0) = v_0, v'(x_0) = v'_0\) can be written as \(v(x) = y(x) + w(x)\), where \(y\) and \(w\) are solutions to the following initial value problems: \[\Lop[y] = 0,~~y(x_0) = v_0,~~ y'(x_0) = v'_0~~ ,\] \[\Lop[w] = f(x),~~ w(x_0) = 0,~~ w'(x_0) = 0.\]
\(\bf{Note:}\) If a set of fundamental solutions is known for the differential equation \(\Lop[y] = 0\), then \(y(x)\) can be relatively easy to find. Moreover, finding \(w(x)\) could be done by the method of variation of parameters or Green’s function method.
The method of reduction of order can also be used with non-homogeneous non-constant coefficient linear differential equations. Specifically, consider the following equation \(w'' + p(u)w' + q(u)w = g(u)\), and suppose that one solution to the associated homogeneous equation is known, \(w_1(u)\). Now let \(w(u) = v(u) w_1(u)\) and show that if \(w(u)\) is a solution to \(w'' + p(u)w' + q(u)w = g(u)\), then \(v(u)\) must be a solution to \(w_1(u) v'' + [2w'_1(u) + p(u)w_1(u)]v' = g(u)\).
\(\bf{Remark:}\) The equation \(w_1(u) v'' + [2w'_1(u) + p(u)w_1(u)]v' = g(u)\) is a first order linear equation in \(v'\). After solving this equation, integrating the result to obtain \(v(u)\) and multiplying that by \(w_1(u)\), we obtain the general solution \(w(u) = v(u) w_1(u)\) to the original differential equation \(w'' + p(u)w' + q(u)w = g(u)\).
Given that \(y_1(x) = e^x, ~~y_2(x) = x e^x,\) and \(y_3(x) = e^{-x}\) are solutions to the homogeneous equation associated with \(y''' - y '' - y' + y = f(x)\), determine a particular solution to the differential equation in terms of the definite integrals. Can you write down the general solution as well?
Find a formula involving integrals for a particular solution of the integral equation \[w''' - w'' + w' - w = f(u).\]