# 7.1 Second-order linear equations  (Page 8/15)

 Page 8 / 15

Solve the following initial-value problem and graph the solution: $y\text{″}-2{y}^{\prime }+10y=0,y\left(0\right)=2,{y}^{\prime }\left(0\right)=-1$

$y\left(x\right)={e}^{x}\left(2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}3x-\text{sin}\phantom{\rule{0.1em}{0ex}}3x\right)$

## Initial-value problem representing a spring-mass system

The following initial-value problem models the position of an object with mass attached to a spring. Spring-mass systems are examined in detail in Applications . The solution to the differential equation gives the position of the mass with respect to a neutral (equilibrium) position (in meters) at any given time. (Note that for spring-mass systems of this type, it is customary to define the downward direction as positive.)

$y\text{″}+2{y}^{\prime }+y=0,y\left(0\right)=1,{y}^{\prime }\left(0\right)=0$

Solve the initial-value problem and graph the solution. What is the position of the mass at time $t=2$ sec? How fast is the mass moving at time $t=1$ sec? In what direction?

In [link] c. we found the general solution to this differential equation to be

$y\left(t\right)={c}_{1}{e}^{\text{−}t}+{c}_{2}t{e}^{\text{−}t}.$

Then

${y}^{\prime }\left(t\right)=\text{−}{c}_{1}{e}^{\text{−}t}+{c}_{2}\left(\text{−}t{e}^{\text{−}t}+{e}^{\text{−}t}\right).$

When $t=0,$ we have $y\left(0\right)={c}_{1}$ and ${y}^{\prime }\left(0\right)=\text{−}{c}_{1}+{c}_{2}.$ Applying the initial conditions, we obtain

$\begin{array}{ccc}\hfill {c}_{1}& =\hfill & 1\hfill \\ \hfill {\text{−}c}_{1}+{c}_{2}& =\hfill & 0.\hfill \end{array}$

Thus, ${c}_{1}=1,$ ${c}_{2}=1,$ and the solution to the initial value problem is

$y\left(t\right)={e}^{\text{−}t}+t{e}^{\text{−}t}.$

This solution is represented in the following graph. At time $t=2,$ the mass is at position $y\left(2\right)={e}^{-2}+2{e}^{-2}=3{e}^{-2}\approx 0.406$ m below equilibrium.

To calculate the velocity at time $t=1,$ we need to find the derivative. We have $y\left(t\right)={e}^{\text{−}t}+t{e}^{\text{−}t},$ so

${y}^{\prime }\left(t\right)=\text{−}{e}^{\text{−}t}+{e}^{\text{−}t}-t{e}^{\text{−}t}=\text{−}t{e}^{\text{−}t}.$

Then ${y}^{\prime }\left(1\right)=\text{−}{e}^{-1}\approx -0.3679.$ At time $t=1,$ the mass is moving upward at 0.3679 m/sec.

Suppose the following initial-value problem models the position (in feet) of a mass in a spring-mass system at any given time. Solve the initial-value problem and graph the solution. What is the position of the mass at time $t=0.3$ sec? How fast is it moving at time $t=0.1$ sec? In what direction?

$y\text{″}+14{y}^{\prime }+49y=0,y\left(0\right)=0,{y}^{\prime }\left(0\right)=1$

$y\left(t\right)=t{e}^{-7t}$

At time $t=0.3,$ $y\left(0.3\right)=0.3{e}^{\left(-7*0.3\right)}=0.3{e}^{-2.1}\approx 0.0367.$ The mass is 0.0367 ft below equilibrium. At time $t=0.1,$ ${y}^{\prime }\left(0.1\right)=0.3{e}^{-0.7}\approx 0.1490.$ The mass is moving downward at a speed of 0.1490 ft/sec.

## Solving a boundary-value problem

In [link] f. we solved the differential equation $y\text{″}+16y=0$ and found the general solution to be $y\left(t\right)={c}_{1}\text{cos}\phantom{\rule{0.1em}{0ex}}4t+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}4t\text{.}$ If possible, solve the boundary-value problem if the boundary conditions are the following:

1. $y\left(0\right)=0,$ $y\left(\frac{\pi }{4}\right)=0$
2. $y\left(0\right)=1,$ $y\left(\frac{\pi }{8}\right)=0$
3. $y\left(\frac{\pi }{8}\right)=0,$ $y\left(\frac{3\pi }{8}\right)=2$

We have

$y\left(x\right)={c}_{1}\text{cos}\phantom{\rule{0.1em}{0ex}}4t+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}4t\text{.}$
1. Applying the first boundary condition given here, we get $y\left(0\right)={c}_{1}=0.$ So the solution is of the form $y\left(t\right)={c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}4t\text{.}$ When we apply the second boundary condition, though, we get $y\left(\frac{\pi }{4}\right)={c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(4\left(\frac{\pi }{4}\right)\right)={c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}\pi =0$ for all values of ${c}_{2}.$ The boundary conditions are not sufficient to determine a value for ${c}_{2},$ so this boundary-value problem has infinitely many solutions. Thus, $y\left(t\right)={c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}4t$ is a solution for any value of ${c}_{2}.$
2. Applying the first boundary condition given here, we get $y\left(0\right)={c}_{1}=1.$ Applying the second boundary condition gives $y\left(\frac{\pi }{8}\right)={c}_{2}=0,$ so ${c}_{2}=0.$ In this case, we have a unique solution: $y\left(t\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}4t\text{.}$
3. Applying the first boundary condition given here, we get $y\left(\frac{\pi }{8}\right)={c}_{2}=0.$ However, applying the second boundary condition gives $y\left(\frac{3\pi }{8}\right)=\text{−}{c}_{2}=2,$ so ${c}_{2}=-2.$ We cannot have ${c}_{2}=0=-2,$ so this boundary value problem has no solution.

## Key concepts

• Second-order differential equations can be classified as linear or nonlinear, homogeneous or nonhomogeneous.
• To find a general solution for a homogeneous second-order differential equation, we must find two linearly independent solutions. If ${y}_{1}\left(x\right)$ and ${y}_{2}\left(x\right)$ are linearly independent solutions to a second-order, linear, homogeneous differential equation, then the general solution is given by
$y\left(x\right)={c}_{1}{y}_{1}\left(x\right)+{c}_{2}{y}_{2}\left(x\right).$
• To solve homogeneous second-order differential equations with constant coefficients, find the roots of the characteristic equation. The form of the general solution varies depending on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots.
• Initial conditions or boundary conditions can then be used to find the specific solution to a differential equation that satisfies those conditions, except when there is no solution or infinitely many solutions.

## Key equations

• Linear second-order differential equation
${a}_{2}\left(x\right)y\text{″}+{a}_{1}\left(x\right){y}^{\prime }+{a}_{0}\left(x\right)y=r\left(x\right)$
• Second-order equation with constant coefficients
$ay\text{″}+b{y}^{\prime }+cy=0$

Classify each of the following equations as linear or nonlinear. If the equation is linear, determine whether it is homogeneous or nonhomogeneous.

${x}^{3}y\text{″}+\left(x-1\right){y}^{\prime }-8y=0$

linear, homogenous

$\left(1+{y}^{2}\right)y\text{″}+x{y}^{\prime }-3y=\text{cos}\phantom{\rule{0.1em}{0ex}}x$

$xy\text{″}+{e}^{y}{y}^{\prime }=x$

nonlinear

$y\text{″}+\frac{4}{x}{y}^{\prime }-8xy=5{x}^{2}+1$

$y\text{″}+\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right){y}^{\prime }-xy=4y$

linear, homogeneous

$y\text{″}+\left(\frac{x+3}{y}\right){y}^{\prime }=0$

For each of the following problems, verify that the given function is a solution to the differential equation. Use a graphing utility to graph the particular solutions for several values of c 1 and c 2 . What do the solutions have in common?

[T] $y\text{″}+2{y}^{\prime }-3y=0;$ $y\left(x\right)={c}_{1}{e}^{x}+{c}_{2}{e}^{-3x}$

[T] ${x}^{2}y\text{″}-2y-3{x}^{2}+1=0;$ $y\left(x\right)={c}_{1}{x}^{2}+{c}_{2}{x}^{-1}+{x}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(x\right)+\frac{1}{2}$

[T] $y\text{″}+14{y}^{\prime }+49y=0;$ $y\left(x\right)={c}_{1}{e}^{-7x}+{c}_{2}x{e}^{-7x}$

[T] $6y\text{″}-49{y}^{\prime }+8y=0;$ $y\left(x\right)={c}_{1}{e}^{x\text{/}6}+{c}_{2}{e}^{8x}$

Find the general solution to the linear differential equation.

$y\text{″}-3{y}^{\prime }-10y=0$

$y={c}_{1}{e}^{5x}+{c}_{2}{e}^{-2x}$

$y\text{″}-7{y}^{\prime }+12y=0$

$y\text{″}+4{y}^{\prime }+4y=0$

$y={c}_{1}{e}^{-2x}+{c}_{2}x{e}^{-2x}$

$4y\text{″}-12{y}^{\prime }+9y=0$

$2y\text{″}-3{y}^{\prime }-5y=0$

$y={c}_{1}{e}^{5x\text{/}2}+{c}_{2}{e}^{\text{−}x}$

$3y\text{″}-14{y}^{\prime }+8y=0$

$y\text{″}+{y}^{\prime }+y=0$

$y={e}^{\text{−}x\text{/}2}\left({c}_{1}\text{cos}\phantom{\rule{0.1em}{0ex}}\frac{\sqrt{3}x}{2}+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}\frac{\sqrt{3}x}{2}\right)$

$5y\text{″}+2{y}^{\prime }+4y=0$

$y\text{″}-121y=0$

$y={c}_{1}{e}^{-11x}+{c}_{2}{e}^{11x}$

$8y\text{″}+14{y}^{\prime }-15y=0$

$y\text{″}+81y=0$

$y={c}_{1}\text{cos}\phantom{\rule{0.1em}{0ex}}9x+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}9x$

$y\text{″}-{y}^{\prime }+11y=0$

$2y\text{″}=0$

$y={c}_{1}+{c}_{2}x$

$y\text{″}-6{y}^{\prime }+9y=0$

$3y\text{″}-2{y}^{\prime }-7y=0$

$y={c}_{1}{e}^{\left(\left(1+\sqrt{22}\right)\text{/}3\right)x}+{c}_{2}{e}^{\left(\left(1-\sqrt{22}\right)\text{/}3\right)x}$

$4y\text{″}-10{y}^{\prime }=0$

$36\frac{{d}^{2}y}{d{x}^{2}}+12\frac{dy}{dx}+y=0$

$y={c}_{1}{e}^{\text{−}x\text{/}6}+{c}_{2}x{e}^{\text{−}x\text{/}6}$

$25\frac{{d}^{2}y}{d{x}^{2}}-80\frac{dy}{dx}+64y=0$

$\frac{{d}^{2}y}{d{x}^{2}}-9\frac{dy}{dx}=0$

$y={c}_{1}+{c}_{2}{e}^{9x}$

$4\frac{{d}^{2}y}{d{x}^{2}}+8y=0$

Solve the initial-value problem.

$y\text{″}+5{y}^{\prime }+6y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=0,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(0\right)=-2$

$y=-2{e}^{-2x}+2{e}^{-3x}$

$y\text{″}+2{y}^{\prime }-8y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=5,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(0\right)=4$

$y\text{″}+4y=0,\phantom{\rule{4em}{0ex}}y\left(0\right)=3,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(0\right)=10$

$y=3\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(2x\right)+5\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(2x\right)$

$y\text{″}-18{y}^{\prime }+81y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=1,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(0\right)=5$

$y\text{″}-{y}^{\prime }-30y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=1,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(0\right)=-16$

$y=\text{−}{e}^{6x}+2{e}^{-5x}$

$4y\text{″}+4{y}^{\prime }-8y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=2,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(0\right)=1$

$25y\text{″}+10{y}^{\prime }+y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=2,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(0\right)=1$

$y=2{e}^{\text{−}x\text{/}5}+\frac{7}{5}x{e}^{\text{−}x\text{/}5}$

$y\text{″}+y=0,\phantom{\rule{4em}{0ex}}y\left(\pi \right)=1,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime }\left(\pi \right)=-5$

Solve the boundary-value problem, if possible.

$y\text{″}+{y}^{\prime }-42y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=0,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(1\right)=2$

$y=\left(\frac{2}{{e}^{6}-{e}^{-7}}\right){e}^{6x}-\left(\frac{2}{{e}^{6}-{e}^{-7}}\right){e}^{-7x}$

$9y\text{″}+y=0,\phantom{\rule{4em}{0ex}}y\left(\frac{3\pi }{2}\right)=6,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(0\right)=-8$

$y\text{″}+10{y}^{\prime }+34y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=6,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(\pi \right)=2$

No solutions exist.

$y\text{″}+7{y}^{\prime }-60y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=4,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(2\right)=0$

$y\text{″}-4{y}^{\prime }+4y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=2,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(1\right)=-1$

$y=2{e}^{2x}-\frac{2{e}^{2}+1}{{e}^{2}}x{e}^{2x}$

$y\text{″}-5{y}^{\prime }=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=3,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(-1\right)=2$

$y\text{″}+9y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=4,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(\frac{\pi }{3}\right)=-4$

$y=4\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}3x+{c}_{2}\text{sin}\phantom{\rule{0.1em}{0ex}}3x,\phantom{\rule{0.2em}{0ex}}\text{infinitely many solutions}$

$4y\text{″}+25y=0,\phantom{\rule{2em}{0ex}}y\left(0\right)=2,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}y\left(2\pi \right)=-2$

Find a differential equation with a general solution that is $y={c}_{1}{e}^{x\text{/}5}+{c}_{2}{e}^{-4x}.$

$5y\text{″}+19{y}^{\prime }-4y=0$

Find a differential equation with a general solution that is $y={c}_{1}{e}^{x}+{c}_{2}{e}^{-4x\text{/}3}.$

For each of the following differential equations:

1. Solve the initial value problem.
2. [T] Use a graphing utility to graph the particular solution.

$y\text{″}+64y=0;\phantom{\rule{2em}{0ex}}y\left(0\right)=3,\phantom{\rule{1em}{0ex}}{y}^{\prime }\left(0\right)=16$

a. $y=3\phantom{\rule{0.1em}{0ex}}\text{cos}\left(8x\right)+2\phantom{\rule{0.1em}{0ex}}\text{sin}\left(8x\right)$
b.

$y\text{″}-2{y}^{\prime }+10y=0\phantom{\rule{2em}{0ex}}y\left(0\right)=1,\phantom{\rule{1em}{0ex}}{y}^{\prime }\left(0\right)=13$

$y\text{″}+5{y}^{\prime }+15y=0\phantom{\rule{2em}{0ex}}y\left(0\right)=-2,\phantom{\rule{1em}{0ex}}{y}^{\prime }\left(0\right)=7$

a. $y={e}^{\left(-5\text{/}2\right)x}\left[-2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(\frac{\sqrt{35}}{2}x\right)+\frac{4\sqrt{35}}{35}\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(\frac{\sqrt{35}}{2}x\right)\right]$
b.

(Principle of superposition) Prove that if ${y}_{1}\left(x\right)$ and ${y}_{2}\left(x\right)$ are solutions to a linear homogeneous differential equation, $y\text{″}+p\left(x\right){y}^{\prime }+q\left(x\right)y=0,$ then the function $y\left(x\right)={c}_{1}{y}_{1}\left(x\right)+{c}_{2}{y}_{2}\left(x\right),$ where ${c}_{1}$ and ${c}_{2}$ are constants, is also a solution.

Prove that if a, b, and c are positive constants, then all solutions to the second-order linear differential equation $ay\text{″}+b{y}^{\prime }+cy=0$ approach zero as $x\to \infty \text{.}$ ( Hint: Consider three cases: two distinct roots, repeated real roots, and complex conjugate roots.)

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