For this first group of problems: (a) Show the given \(Y_P\) is a solution to the non-homogeneous equation (b) Find a general solution to the equation.
\(\ddot{w}-\dot{w}-2w = 2e^u\) with \(W_P(u) = -e^u\).
\(y'' + 4y = \sin(t)\) with \(Y_P(t) = \frac{1}{3}\sin(t).\)
\(w'' +w = 2\cos(x)\) with \(W_P(x)= x\sin(x).\)
\(y''+2y'+y = 3e^{2t}\) with \(Y_P(t) = \frac{1}{3}e^{2t}.\)
\(x'' + 2x' + 2x = e^{-t}\) with \(X_P(t) = e^{-t}.\)
\(y'' - 6y'+9y = 2e^{3t}\) with \(Y_P(t) = t^2e^{3t}.\)
\(\ddot{w} - 2\dot{w} + w = -e^z/z^2\) with \(W_P(z)= e^z\log(z).\)
\(t^2y'' - (t^2 + 2t)y' + (t+2)y = 2t^3\) with \(Y_P(t) = -2t^2\) (For part (b) note that \(y_1(t) = t\) and \(y_2(t) = te^t\) are both solutions to the homogeneous equation).
In the next group of problems show the solution given is a solution to the equation and then use it to find the solution to the given initial value problem.
\(y''+4y'+4y=9e^x\) with \(Y_P(x) = e^x\) where \(y(0) = 3\) and \(y'(0) = 4.\)
\(y''-4y = 5e^{3t}\) with \(Y_P(t) = e^{3t}\) where \(y(0) = 3\) and \(y'(0) = 0.\)
\(w''-2w'-3w= 3e^{2u}\) with \(W_P(u) = e^{3u}-e^{2u}\) where \(w(0) = 2\) and \(w'(0) = 1\).
\(y''-2y'+6y = -29\cos(t)\) with \(Y_p(t) = 2\sin(t)-5\cos(t)\) where \(y(0)=1\)
and \(y'(0) = 2\).
\(z''-z= 4we^w\) with \(Z_P(w) = w^2e^w-we^w\) where \(z(0) = 2\) and \(z'(0) = -1\).
\(y^{(4)} -3y'' -4y = 2t\) with \(Y_P(t) = -t/2\) where \(y(0) = 10\), \(y'(0) = 3/2\), \(y''(0) = 0\), and \(y'''(0) = -2\).
\(x''' - x'' - x'+x = 3e^{2u}\) with \(X_P(u) = e^{2u}\) where \(x(0) = 1\), \(x'(0) = 2\) and \(x''(0) = 4\).
\(t^2y''-ty'+y = t^2\) with \(Y_P(t)=t^2\) and \(y_1(t) = t\) is a solution to the homogeneous equation, and where \(y(1) = 1\) and \(y'(1) = 0\) (Note: you will have to first find a second independent solution to the homogeneous equation).
Use reduction of order to find general solutions to the following non-homgeneous equations. That is given a solution \(y_1\) to the corresponding homogeneous equation set \(y_2 = y_1v\), and reduce the second order non-homogeneous equations to a first order equation.
\(z^2w''-2zw'+2w= 4z^2\) where \(w_1(z) = z\).
\(x^2\ddot{w}+7x\dot{w}+5w = x\) where \(w_1(x) = \frac{1}{x}\).
In the following please justify your responses
The recipe for finding a general solution to a non-homogeneous equation is to first find a single solution to the non-homogeneous equation \(Y_P\) and then add it to the general solution for the corresponding homogeneous equation \(Y_H\).
(a) Justify why \(Y_P(t) + Y_H(t)\) will be a solution for any homogeneous solution \(Y_H\).
(b) Suppose that we also have an initial condition. Can we solve the initial value problem by adding a solution to the non-homogeneous equation and adding it to the solution to the corresponding homogeneous initial value problem? Why or why not?
This problem touches on the method of annihilators. The next section depends on having the non-homogeneous part be a solution to some homogeneous equation. It gives us a way to find a non-homogeneous solution when the non-homogeneous part has the special property that it is annihilated by a differential operator.
(a) Suppose that \(y''-y = g(t)\) is such that \(g(t)\) is a solution to the homogeneous equation \(y'' -y = 0\). Justify why a solution to \(y''-y = g(t)\) is also a solution to the homogeneous equation \(y^{(4)} - 2y'' + y=0\).
(b) More generally suppose that \(\Lop_1\) and \(\Lop_2\) are differential operators such that \(\Lop_1Y_P = g\) and \(\Lop_2g = 0\). Justify why \(Y_P\) is a solution to \(\Lop_2 \Lop_1 y=0\).
(c) If \(Y\) is the general solution to \(\Lop_2 \Lop_1 y=0\), and \(z\) is the general solution to \(\Lop_1 y=0\), and \(g\) is a solution to \(\Lop_2 y=0\), justify why we can find coefficients such that \(Y-z\) is a solution to \(\Lop_1 y=g\).
This is a two-part problem.
a) What should \(h(x)\) be so that \(y(x)\) is a solution to \[y'' - 2y' + y = h(x),\] when \(i) y(x) = \cos(2x)?\) \(ii) y(x) = \sin(2x)?\)
b) Using the results obtained from \(a)\), find a particular solution \(y_p(x)\) and a general soluton to each of the following equations: \(i) y'' - 2 y' + y = \cos(2x)\) \(ii) y'' - 2 y' + y = \sin(2x)\).
(\(\bf{Hint:}\) Here you will have to make an educated guess for a particular solution. Think about a linear combination of solutions from i) and ii) and solve for the unknown coefficients of that linear combination.)
\(\bf{Remark:}\) This foreshadows the appearance of the method of “Undetermined Coefficients” from Chapter 6.