# Section3.8Linear Systems in Higher Dimensions¶ permalink

Suppose that we have two masses on a table, $m_1$ and $m_2\text{,}$ connected by three springs with the outside springs connected to two walls (Figure 3.65), and the masses are free to move horizontally. We will assume that the springs are uniform and all have the same spring constant $k\text{.}$ The horizontal displacements of the springs are denoted by $x_1(t)$ and $x_2(t)$ for the masses $m_1$ and $m_2\text{,}$ respectively. Assuming that there is no damping, the only forces acting on mass $m_1$ at time $t$ are those of left and middle springs. The force from the left spring will be $-kx_1$ while the force from middle spring will be $k(x_2 - x_1)\text{.}$ By Newton's Second Law of motion, we have \begin{equation*} m_1 x_1'' = -kx_1 + k(x_2 - x_1). \end{equation*} Similarly, the only forces acting on the second mass, $m_2\text{,}$ will come from middle and right springs. Again using Newton's Second Law, \begin{equation*} m_2 x_2'' = -k(x_2 - x_1) - kx_2. \end{equation*}

We now have a system of two second-order linear equations \begin{align*} m_1 x_1'' & = -2kx_1 + kx_2\\ m_2 x_2'' & = kx_1 - 2kx_2. \end{align*} If we define $x_3$ and $x_4$ by $x_3 = x_1'$ and $x_4 = x_2'\text{,}$ we now have first-order linear system of four equations, \begin{align*} x_1' & = x_3\\ x_2' & = x_4\\ x_3' & = -2 \frac{k}{m_1} x_1 + \frac{k}{m_1} x_2\\ x_4' & = \frac{k}{m_2} x_1 - 2\frac{k}{m_2} x_2 \end{align*} We can represent this system in the matrix form ${\mathbf x}' = A {\mathbf x}\text{,}$ where \begin{equation*} A = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -2k/m_1 & k/m_1 & 0 & 0 \\ k / m_2 & - 2k/m_2 & 0 & 0 \end{pmatrix}. \end{equation*} We will learn how to analyze and solve such systems in the next two sections.

We can write the system \begin{align*} x_1' & = a_{11} x_1 + \cdots a_{1n} x_n\\ x_2' & = a_{21} x_1 + \cdots a_{2n} x_n\\ & \vdots\\ x_n' & = a_{n1} x_1 + \cdots a_{nn} x_n \end{align*} in matrix form ${\mathbf x}' = A {\mathbf x}\text{,}$ where \begin{equation*} A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix} \qquad\text{and}\qquad {\mathbf x} = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}. \end{equation*}

The strategy for finding solutions to the system ${\mathbf x}' = A {\mathbf x}$ is the same as for systems of two equations. If $\lambda$ is an eigenvalue of $A$ with eigenvector ${\mathbf v}\text{,}$ \begin{equation*} {\mathbf x}(t) = e^{\lambda t} {\mathbf v} \end{equation*} is a solution for our system. Indeed, \begin{equation*} {\mathbf x}'(t) = \lambda e^{\lambda t} {\mathbf v} = e^{\lambda t} (\lambda {\mathbf v}) = e^{\lambda t} A {\mathbf v} = A(e^{\lambda t} {\mathbf v}) = A {\mathbf x}(t). \end{equation*}

##### Example3.66

The system \begin{align*} x' & = -5x - 8y - 2z\\ y' & = 5x + 12y + 4z\\ z' & = -11x - 19y - 5z \end{align*} can be rewritten as \begin{equation*} {\mathbf x}' = \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} = \begin{pmatrix} -5 & -8 & -2 \\ 5 & 12 & 4 \\ -11 & -19 & -5 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = A {\mathbf x}. \end{equation*} We can compute the eigenvalues of $A$ by finding the roots of its characteristic polynomial \begin{equation*} \det( A - \lambda I) = -\lambda^3 - 2\lambda^2 - \lambda + 2 = (\lambda - 1)(\lambda + 1)(\lambda - 2). \end{equation*} Thus, the eigenvalues of $A$ are $\lambda = 2\text{,}$ $\lambda = 1\text{,}$ and $\lambda = -1\text{.}$ To find an eigenvector for $\lambda = 2\text{,}$ we must find a nontrivial solution for system of equations $(A - 2I){\mathbf x} = {\mathbf 0}\text{,}$ \begin{align*} -7x - 8y - 2z \amp = 0\\ -5x + 10y + 4z \amp = 0\\ -11x - 19y - 7z \amp = 0. \end{align*} It is easy to check that $(2, -3, 5)$ is a solution. Similarly, we can determine that $(5, -7, 13)$ is an eigenvector for $\lambda = 1$ and $(1, -1, 2)$ is an eigenvector for $\lambda = -1\text{.}$ Thus, we have found three solutions for the system ${\mathbf x}' = A {\mathbf x}\text{,}$ \begin{equation*} \mathbf x_1(t) = e^{2t} \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix}, \mathbf x_2(t) = e^{t} \begin{pmatrix} 5 \\ -7 \\ 13 \end{pmatrix}, \quad \text{and} \quad \mathbf x_3(t) = e^{-t} \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}. \end{equation*}

The Principle of Superposition also holds for higher-order systems. If ${\mathbf x}_1(t)$ and ${\mathbf x}_2(t)$ are solutions for ${\mathbf x}' = A {\mathbf x}\text{,}$ then \begin{equation*} c_1 {\mathbf x}_1(t) + c_2 {\mathbf x}_2(t) \end{equation*} is a solution for the system, since \begin{align*} \frac{d}{dt}[c_1 {\mathbf x}_1(t) + c_2 {\mathbf x}_2(t)] & = c_1 \frac{d}{dt}{\mathbf x}_1(t) + c_2 \frac{d}{dt}{\mathbf x}_2(t)\\ & = c_1 A {\mathbf x}_1(t) + c_2 A {\mathbf x}_2(t)\\ & = A[c_1 {\mathbf x}_1(t) + c_2 {\mathbf x}_2(t)]. \end{align*} Consequently, \begin{equation*} {\mathbf x}(t) = c_1 e^{2t} \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} + e^{t} \begin{pmatrix} 5 \\ -7 \\ 13 \end{pmatrix} + c_3 e^{-t} \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \end{equation*} is a solution for our system. This is, in fact, the general solution for the system.

Although we shall not cover the notions of linear independence, canonical matrices, and change of coordinates for ${\mathbb R}^n\text{,}$ the same ideas that we used for systems first-order linear differential equations in ${\mathbb R}^2$ carry over to ${\mathbb R}^n\text{.}$The necessary linear algebra is covered in any good linear algebra course. Also, see Chapters 5 and 6 in [12].

##### Example3.67

Let us see how the linear algebra works in the previous example. If we form the matrix \begin{equation*} T = \begin{pmatrix} 2 & 5 & 1 \\ -3 & -7 & -1 \\ 5 & 13 & 2 \end{pmatrix} \end{equation*} from the eigenvectors of $A\text{,}$ we can convert the system ${\mathbf x}' = A {\mathbf x}$ to the system \begin{align*} {\mathbf y}' \amp = (T^{-1} A T) {\mathbf y}\\ \amp \begin{pmatrix} 1 & -3 & -2\\ -1 & 1 & 1 \\ 4 & 1 & -1 \end{pmatrix} \begin{pmatrix} -5 & -8 & -2 \\ 5 & 12 & 4 \\ -11 & -19 & -5 \end{pmatrix} \begin{pmatrix} 2 & 5 & 1 \\ -3 & -7 & -1 \\ 5 & 13 & 2 \end{pmatrix} {\mathbf y},\\ \amp = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} {\mathbf y}, \end{align*} which we can immediately solve: \begin{equation*} {\mathbf y}(t) = c_1 e^{2t} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + c_2 e^{t} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + c_3 e^{-t} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}. \end{equation*} Multiplying our solution by $T$ yields the general solution \begin{equation*} {\mathbf x}(t) = T {\mathbf y} = c_1 e^{2t} \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} + e^{t} \begin{pmatrix} 5 \\ -7 \\ 13 \end{pmatrix} + c_3 e^{-t} \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}, \end{equation*} the solution to the system $\mathbf x' A \mathbf x\text{.}$

# Subsection3.8.2The Geometry of Solutions¶ permalink

In Section 3.6, we classified all of the geometry of the solutions for planar systems using the trace-determinant plane. The geometry for linear systems in three variables is a bit more complicated. For a system \begin{align*} x' \amp = a_1 x + a_1 y + a_3 z\\ y' \amp = b_1 x + b_2 y + b_3 z\\ z' \amp = c_1 x + c_2 y + c_3 z, \end{align*} our solutions curves live in $\mathbb R^3\text{,}$ and there is simply a lot more room to move around in three dimensions than in two dimensions. The origin is still an equilibrium solution for a system of linear differential equations in three variables. The origin is a stable equilibrium solution if any solution $\mathbf x(t)$ approaches $\mathbf 0 = (0, 0, 0)$ as $t \to \infty\text{;}$ otherwise, $\mathbf 0$ is an unstable equilibrium solution. In the case of planar systems, an unstable solution is a nodal saddle, a nodal source, a spiral source, or a source with a single unstable line. In the case of $\mathbb R^3\text{,}$ we could have a stable line of solutions and an unstable plane of solutions. In this case, all solutions of the system with initial condition lying on the stable line would approach the origin as $t \to \infty\text{,}$ but all solutions with initial conditions that are a nonzero point on the unstable plane would move away from the origin.

##### Example3.68

In Example 3.66, we had solutions \begin{equation*} \mathbf x_1(t) = e^{2t} \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix}, \mathbf x_2(t) = e^{t} \begin{pmatrix} 5 \\ -7 \\ 13 \end{pmatrix}, \quad \text{and} \quad \mathbf x_3(t) = e^{-t} \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}. \end{equation*} The straight line through the origin and the point $(1, -1, 2)$ is a stable line. That is, for any initial condition ${\mathbf x}(0) = (x_0, y_0, z_0)$ lying on this line, our solution will tend toward the origin as $t \to \infty\text{.}$ On the other hand, the plane spanned by $(2, -3, 5)$ and $(5, -7, 13)$ is unstable plane. Solutions on this plane move away from the origin as $t \to \infty\text{.}$ Of course, $(0, 0, 0)$ is an equlibrium solution for our system. We say that the origin is a saddle in this example (Figure 3.69).

##### Example3.70

For the system \begin{equation*} {\mathbf x}' = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} {\mathbf x} \end{equation*} we have a very different type of unstable equilibrium solution. The eigenvalues of this matrix are $\lambda = \pm i$ and $\lambda = -1\text{.}$ Thus, a solution satisfying the initial condition ${\mathbf x}(0) = (x_0, y_0, z_0)$ is given by \begin{equation*} {\mathbf x}(t) = x_0 \begin{pmatrix} \cos t \\ - \sin t \\ 0 \end{pmatrix} + y_0 \begin{pmatrix} \sin t \\ \cos t \\ 0 \end{pmatrix} + z_0 e^{-t} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}. \end{equation*} If $z_0 = 0\text{,}$ then our initial condition is in the $xy$-plane and all of the solutions lie on circles centered at the origin. On the other hand, if $x_0 = 0$ and $y_0 = 0\text{,}$ we have a stable line of solutions lying along the $z$-axis. In fact, each solution that does not lie on the stable line lies on a cylinder in ${\mathbb R}^3$ given by $x^2 + y^2 = r^2$ for some constant $r \gt 0\text{.}$ These solutions spiral towards the circular solution of radius $r$ in the $xy$-plane if $z_0 \neq 0$ (Figure 3.71).

##### Example3.72

For an example of a stable plane and an unstable line, let us consider the system \begin{equation*} {\mathbf x}' = \begin{pmatrix} -1 & 1 & 0 \\ -1 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix} {\mathbf x} = A {\mathbf x}. \end{equation*} The characteristic equation of the matrix $A$ is \begin{equation*} \lambda^3 + \lambda^2 - 2 = (\lambda - 1)(\lambda^2 + 2\lambda + 2) = 0. \end{equation*} Thus, the eigenvalues of $A$ are $\lambda = 1$ and $\lambda = -1 \pm i\text{.}$ Solving \begin{equation*} [A - (-1 + i) ]{\mathbf x} = {\mathbf 0} \end{equation*} gives us an eigenvector $(1, i, 0)$ for $\lambda = -1 + i\text{.}$ Since \begin{equation*} \begin{pmatrix} 1 \\ i \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + i \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \end{equation*} we will let ${\mathbf v}_1 = (1, 0, 0)$ and ${\mathbf v}_2 = (0, 1, 0)\text{.}$ Since ${\mathbf v}_3 = (0, 0, 1)$ is an eigenvector for $\lambda = 1\text{,}$ our system has solution \begin{equation*} {\mathbf x}(t) = x_0 e^{-t} \begin{pmatrix} \cos t \\ - \sin t \\ 0 \end{pmatrix} + y_0 e^{-t} \begin{pmatrix} \sin t \\ \cos t \\ 0 \end{pmatrix} + z_0 e^t \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \end{equation*} where $\mathbf x(0) = (x_0, y_0, z_0)\text{.}$ If $z_0 = 0\text{,}$ our initial condition lies in the $xy$-plane and solution curves sprial in towards the origin. Thus, we have a stable plane. On the other hand, if $x_0 = 0$ and $y_0 = 0$ but $z_0 \neq 0\text{,}$ then our solution approaches $\pm \infty$ as $t \to \infty\text{.}$ In this case, the $z$-axis is an unstable line (Figure 3.73).

For an example of a stable equilibrium solution at the origin, consider the system \begin{equation*} {\mathbf x}' = \begin{pmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{pmatrix} {\mathbf x}, \end{equation*} where $\lambda_3 \lt \lambda_2 \lt \lambda_1 \lt 0\text{.}$ For an initial condition $(x_0, y_0, z_0)$ with at least one coordinate nonzero, the corresponding solution tends towards the origin tangentially to the $x$-axis as $t \to \infty$ (Figure 3.74).

Changing the system that in Example3.72 to be \begin{equation*} {\mathbf x}' = \begin{pmatrix} -1 & 1 & 0 \\ -1 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix} {\mathbf x}, \end{equation*} we obtain the solution satisfying the initial condition ${\mathbf x}(0) = (x_0, y_0, z_0)$ to be \begin{equation*} {\mathbf x}(t) = x_0 e^{-t} \begin{pmatrix} \cos t \\ - \sin t \\ 0 \end{pmatrix} + y_0 e^{-t} \begin{pmatrix} \sin t \\ \cos t \\ 0 \end{pmatrix} + z_0 e^{-t} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \end{equation*} where $\mathbf x(0) = (x_0, y_0, z_0)\text{.}$ In this case, all solutions will approach the origin as $t \to \infty\text{.}$

# Subsection3.8.3The Double Spring-Mass Systems Revisited¶ permalink

Let us return to our spring-mass system ${\mathbf x}' = A {\mathbf x}\text{,}$ where \begin{equation*} A = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -2k/m_1 & k/m_1 & 0 & 0 \\ k / m_2 & - 2k/m_2 & 0 & 0 \end{pmatrix}. \end{equation*} To keep matter simple, we will assume that $m_1 = m_2 = k = 1\text{.}$ Thus, our matrix now becomes \begin{equation*} A = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -2 & 1 & 0 & 0 \\ 1 & - 2 & 0 & 0 \end{pmatrix}. \end{equation*} The characteristic polynomial of $A$ is \begin{equation*} \det(A - \lambda I) = \lambda^4 + 4\lambda^2 + 3 = (\lambda^2 + 1)(\lambda^2 + 3). \end{equation*} Thus the eigenvalues of $A$ are $\lambda = \pm i$ and $\lambda = \pm i \sqrt{3}\text{.}$ We can find eigenvectors \begin{equation*} {\mathbf v}_1 = \begin{pmatrix} i / \sqrt{3} \\ - i / \sqrt{3} \\ -1 \\ 1 \end{pmatrix}, {\mathbf v}_2 = \begin{pmatrix} -i / \sqrt{3} \\ i / \sqrt{3} \\ -1 \\ 1 \end{pmatrix}, {\mathbf v}_3 = \begin{pmatrix} -i \\ - i \\ 1 \\ 1 \end{pmatrix}, {\mathbf v}_4 = \begin{pmatrix} i \\ i \\ 1 \\ 1 \end{pmatrix}, \end{equation*} corresponding to the eigenvalues $\lambda_1 = i \sqrt{3}\text{,}$ $\lambda_2 = - i \sqrt{3}\text{,}$ $\lambda_3 = i\text{,}$ $\lambda_4 = - i\text{,}$ respectively. Consequently, the general solution to our system is \begin{equation*} {\mathbf x}(t) = c_1 e^{ i \sqrt{3} t} {\mathbf v}_1 + c_1 e^{ -i \sqrt{3} t} {\mathbf v}_2 + c_1 e^{ i t} {\mathbf v}_3 + c_1 e^{ -i t} {\mathbf v}_4; \end{equation*} however, this form of the solution is not very useful. By examining real and imaginary parts of $e^{ i \sqrt{3} t} {\mathbf v}_1$ and $c_1 e^{ i t} {\mathbf v}_3\text{,}$ we can rewrite the solution as \begin{equation*} {\mathbf x}(t) = c_1 \begin{pmatrix} \cos \sqrt{3} t \\ - \cos \sqrt{3}t \\ - \sqrt{3} \sin \sqrt{3}t \\ \sqrt{3} \sin \sqrt{3}t \end{pmatrix} + c_2 \begin{pmatrix} - \sin \sqrt{3} t \\ \sin \sqrt{3}t \\ -\sqrt{3} \cos \sqrt{3} t \\ \sqrt{3} \cos \sqrt{3} t \end{pmatrix} + c_3 \begin{pmatrix} \cos t \\ \cos t \\ -\sin t \\ - \sin t \end{pmatrix} + c_4 \begin{pmatrix} \sin t \\ \sin t \\ \cos t \\ \cos t \end{pmatrix}. \end{equation*} If we have the following initial conditions, \begin{align*} x_1(0) & = 0\\ x_2(0) & = 0\\ x_1'(0) = x_3(0) & = 2\\ x_2'(0) = x_4(0) & = 2, \end{align*} we can determine $c_1 = c_2 = c_3 = 0$ and $c_4 = 2\text{.}$ Thus, the solution to our initial value problem is \begin{equation*} {\mathbf x}(t) = \begin{pmatrix} 2 \sin t \\ 2 \sin t \\ 2 \cos t \\ 2 \cos t \end{pmatrix}, \end{equation*} and the two masses will oscillate with a frequency of one and an amplitude of two. We leave the details as an exercise.

• \begin{align*} x_1' & = a_{11} x_1 + \cdots a_{1n} x_n\\ x_2' & = a_{21} x_1 + \cdots a_{2n} x_n\\ & \vdots\\ x_n' & = a_{n1} x_1 + \cdots a_{nn} x_n \end{align*} can be written in matrix form ${\mathbf x}' = A {\mathbf x}\text{,}$ where \begin{equation*} A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix} \qquad\text{and}\qquad {\mathbf x} = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}. \end{equation*}
• As in the case of ${\mathbb R}^2\text{,}$ we can solve the system ${\mathbf x}' = A {\mathbf x}$ by finding eigenvalues and eigenvectors for $A\text{.}$
• The Principle of Superposition holds for higher-order systems. If ${\mathbf x}_1(t)$ and ${\mathbf x}_2(t)$ are solutions for ${\mathbf x}' = A {\mathbf x}\text{,}$ then \begin{equation*} c_1 {\mathbf x}_1(t) + c_2 {\mathbf x}_2(t) \end{equation*} is a solution for the system.
• The geometry for system in ${\mathbb R}^3$ is more complicated than the planar case. However, the solutions are usually characterized by stable lines or stable planes.

##### 1

Solve the system \begin{align*} x' & = -5x + 15y - 4z\\ y' & = -x + 5y - z\\ z' & =4 x -6y + 3z\\ x(0) & = -1\\ y(0) & = 2\\ z(0) & = 0 \end{align*}

##### 2

Find the general solution of the system \begin{align*} x_1' & = 6x_1 + 9x_2 + x_3 + 4 x_4\\ x_2' & = -x_1 + x_2 + x_3\\ x_3' & = 3x_1 + 8x_2 + 4x_3 + 4 x_4\\ x_4' & = -3x_1 - 8x_2 -7 x_3 - 5 x_4 \end{align*}

##### 3

Let $a x'' + b x' + cx = 0\text{,}$ where $a \neq 0$ and $b^2 - 4ac = 0\text{.}$

1. Show that $x_1(t) = e^{-bt/2a}$ is a solution to $a x'' + b x' + cx = 0\text{.}$
2. Assume that \begin{equation*} y = v(t) x_1(t) = v(t) e^{-bt/2a} \end{equation*} is a solution to $a x'' + b x' + cx = 0$ and show that $v(t) = c_1 + c_2 t\text{.}$ Thus, \begin{equation*} x(t) = c_1 e^{-bt/2a} + c_2 t e^{-bt/2a} \end{equation*} is a general solution for $a x'' + b x' + cx = 0\text{.}$
##### 4

Consider the equation \begin{equation*} y'' - (2 \alpha -1) y' + \alpha (\alpha - 1) y =0. \end{equation*} Determine all values of $\alpha\text{,}$ if any, for which all solutions tend toward zero as $t \to \infty\text{.}$ Also, determine the values of $\alpha\text{,}$ if any, for which all nonzero solutions become unbounded as $t \to \infty\text{.}$

##### 5

Solve each of the following initial value problems.

1. \begin{align*} y'' - 2y' + 5y & = 0\\ y(\pi/2) & = 0\\ y'(\pi/2) & = 2 \end{align*}
2. \begin{align*} 9y'' - 12y' + 4y & = 0\\ y(0) & = 2\\ y'(0) & = -1 \end{align*}
3. \begin{align*} y'' + 8y' -9y & = 0\\ y(1) & = 1\\ y'(1) & = 0 \end{align*}
4. \begin{align*} y'' + 2ay' + (a^2 + 1)y & = 0\\ y(0) & = 1\\ y'(0) & = 0 \end{align*}
##### 6

Reduction of Order. Suppose that $x_1(t)$ is a solution (not identically zero) to the equation \begin{equation*} x'' + p(t) x' + q(t) x = 0. \end{equation*}

1. Assume that $x(t) = v(t) x_1(t)$ is a solution to $x'' + p(t) x' + q(t) x = 0$ and derive the equation $$x_1 v'' +(2x_1' + px_1)v' = 0.\label{equation-exercise-linear08-reduction-of-order-1}\tag{3.12}$$
2. Let $u = v'$ and show that ((3.12)) is a first-order linear differential equation in $u\text{.}$
3. Show that $x_1(t) = 1/t$ is a solution to $$2 t^2 x'' + 3t x' - x = 0\label{equation-exercise-linear08-reduction-of-order-2}\tag{3.13}$$ for $t \gt 0$ and find a second linearly independent solution to ((3.13)).
##### 7

Euler Equations. An important class of second-order linear differential equations are equations of the form \begin{equation*} t^2 y'' + \alpha t y' + \beta y = 0, \end{equation*} where $t \gt 0$ and $\alpha$ and $\beta$ are real constants. An equation of this form is called an Euler equation.

1. Show that the substitution $x = \ln t$ transforms an Euler equation into an equation of constant coefficients. {\em Hint}: Show that \begin{align*} \frac{dy}{dt} & = \frac{dx}{dt} \frac{dy}{dx}\\ \frac{d^2y}{dt^2} & = \left( \frac{dx}{dt} \right)^2 \frac{d^2y}{dx^2} + \frac{d^2x}{dt^2} \frac{dy}{dx}. \end{align*}
2. Solve the equation \begin{equation*} t^2 y'' + 4ty' + 2y = 0. \end{equation*}