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Second-Order Linear Equations

Recognize homogeneous and nonhomogeneous linear differential equations.

Homogeneous Linear Equations

Consider the second-order differential equation

\[x{y}^{″}+2{x}^{2}{y}^{'}+5{x}^{3}y=0.\]

Notice that y and its derivatives appear in a relatively simple form. They are multiplied by functions of x, but are not raised to any powers themselves, nor are they multiplied together. As discussed in Introduction to Differential Equations, first-order equations with similar characteristics are said to be linear. The same is true of second-order equations. Also note that all the terms in this differential equation involve either y or one of its derivatives. There are no terms involving only functions of x. Equations like this, in which every term contains y or one of its derivatives, are called homogeneous.

Not all differential equations are homogeneous. Consider the differential equation

\[x{y}^{″}+2{x}^{2}{y}^{'}+5{x}^{3}y={x}^{2}.\]

The \({x}^{2}\) term on the right side of the equal sign does not contain y or any of its derivatives. Therefore, this differential equation is nonhomogeneous.

In linear differential equations, \(y\) and its derivatives can be raised only to the first power and they may not be multiplied by one another. Terms involving \({y}^{2}\) or \(\sqrt{{y}^{'}}\) make the equation nonlinear. Functions of \(y\) and its derivatives, such as \(\text{sin}\ y\) or \({e}^{{y}^{'}},\) are similarly prohibited in linear differential equations.

Note that equations may not always be given in standard form (the form shown in the definition). It can be helpful to rewrite them in that form to decide whether they are linear, or whether a linear equation is homogeneous.

Condensed — the full section is in OpenStax Calculus Volume 3.

Second-Order Equations with Constant Coefficients

Now that we have a better feel for linear differential equations, we are going to concentrate on solving second-order equations of the form

\[a{y}^{″}+b{y}^{'}+cy=0,\]

where \(a,\) \(b,\) and \(c\) are constants.

Since all the coefficients are constants, the solutions are probably going to be functions with derivatives that are constant multiples of themselves. We need all the terms to cancel out, and if taking a derivative introduces a term that is not a constant multiple of the original function, it is difficult to see how that term cancels out. Exponential functions have derivatives that are constant multiples of the original function, so let’s see what happens when we try a solution of the form \(y(x)={e}^{\lambda x},\) where \(\lambda\) (the lowercase Greek letter lambda) is some constant.

If \(y(x)={e}^{\lambda x},\) then \({y}^{'}(x)=\lambda {e}^{\lambda x}\) and \({y}^{″}={\lambda }^{2}{e}^{\lambda x}.\) Substituting these expressions into , we get

\[\begin{array}{ll}a{y}^{″}+b{y}^{'}+cy & =a({\lambda }^{2}{e}^{\lambda x})+b(\lambda {e}^{\lambda x})+c{e}^{\lambda x} \\ & ={e}^{\lambda x}(a{\lambda }^{2}+b\lambda +c).\end{array}\]

Since \({e}^{\lambda x}\) is never zero, this expression can be equal to zero for all x only if

\[a{\lambda }^{2}+b\lambda +c=0.\]

We call this the characteristic equation of the differential equation.

The characteristic equation is very important in finding solutions to differential equations of this form. We can solve the characteristic equation either by factoring or by using the quadratic formula

\[\lambda =\frac{\text{-}b\pm \sqrt{{b}^{2}-4ac}}{2a}.\]

If the characteristic equation has distinct real roots \({\lambda }_{1}\) and \({\lambda }_{2},\) then \({e}^{{\lambda }_{1}x}\) and \({e}^{{\lambda }_{2}x}\) are linearly independent solutions to , and the general solution is given by

\[y(x)={c}_{1}{e}^{{\lambda }_{1}x}+{c}_{2}{e}^{{\lambda }_{2}x},\]

where \({c}_{1}\) and \({c}_{2}\) are constants.

For example, the differential equation \({y}^{″}+9{y}^{'}+14y=0\) has the associated characteristic equation \({\lambda }^{2}+9\lambda +14=0.\) This factors into \((\lambda +2)(\lambda +7)=0,\) which has roots \({\lambda }_{1}=-2\) and \({\lambda }_{2}=-7.\) Therefore, the general solution to this differential equation is

\[y(x)={c}_{1}{e}^{-2x}+{c}_{2}{e}^{-7x}.\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Initial-Value Problems and Boundary-Value Problems

So far, we have been finding general solutions to differential equations. However, differential equations are often used to describe physical systems, and the person studying that physical system usually knows something about the state of that system at one or more points in time. For example, if a constant-coefficient differential equation is representing how far a motorcycle shock absorber is compressed, we might know that the rider is sitting still on his motorcycle at the start of a race, time \(t={t}_{0}.\) This means the system is at equilibrium, so \(y({t}_{0})=0,\) and the compression of the shock absorber is not changing, so \({y}^{'}({t}_{0})=0.\) With these two initial conditions and the general solution to the differential equation, we can find the specific solution to the differential equation that satisfies both initial conditions. This process is known as solving an initial-value problem. (Recall that we discussed initial-value problems in Introduction to Differential Equations.) Note that second-order equations have two arbitrary constants in the general solution, and therefore we require two initial conditions to find the solution to the initial-value problem.

Sometimes we know the condition of the system at two different times. For example, we might know \(y({t}_{0})={y}_{0}\) and \(y({t}_{1})={y}_{1}.\) These conditions are called boundary conditions, and finding the solution to the differential equation that satisfies the boundary conditions is called solving a boundary-value problem.

Mathematicians, scientists, and engineers are interested in understanding the conditions under which an initial-value problem or a boundary-value problem has a unique solution. Although a complete treatment of this topic is beyond the scope of this text, it is useful to know that, within the context of constant-coefficient, second-order equations, initial-value problems are guaranteed to have a unique solution as long as two initial conditions are provided. Boundary-value problems, however, are not as well behaved. Even when two boundary conditions are known, we may encounter boundary-value problems with unique solutions, many solutions, or no solution at all.

Condensed — the full section is in OpenStax Calculus Volume 3.

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 \({y}_{1}(x)\) and \({y}_{2}(x)\) are linearly independent solutions to a second-order, linear, homogeneous differential equation, then the general solution is given by
\[y(x)={c}_{1}{y}_{1}(x)+{c}_{2}{y}_{2}(x).\]
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	\({a}_{2}(x){y}^{″}+{a}_{1}(x){y}^{'}+{a}_{0}(x)y=r(x)\)
Second-order equation with constant coefficients	\(a{y}^{″}+b{y}^{'}+cy=0\)

Second-Order Linear Equations

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

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 c₁ and c₂. What do the solutions have in common?

Solve the initial-value problem.

Solve the boundary-value problem, if possible.

For each of the following differential equations:

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

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Classify each of the following equations as linear or nonlinear. If the equation is linear, determine further whether it is homogeneous or nonhomogeneous.
1. \({y}^{″}+3{x}^{4}{y}^{'}+{x}^{2}{y}^{2}={x}^{3}\)
2. \((\text{sin}\ x){y}^{″}+(\text{cos}\ x){y}^{'}+3y=0\)
3. \(4{t}^{2}{x}^{″}+3tx{x}^{'}+4x=0\)
4. \(5{y}^{″}+y=4{x}^{5}\)
5. \((\text{cos}\ x){y}^{″}-\text{sin}\ {y}^{'}+(\text{sin}\ x)y-\text{cos}\ x=0\)
6. \(8t{y}^{″}-6{t}^{2}{y}^{'}+4ty-3{t}^{2}=0\)
7. \(\text{sin}({x}^{2}){y}^{″}-(\text{cos}\ x){y}^{'}+{x}^{2}y={y}^{'}-3\)
8. \({y}^{″}+5x{y}^{'}-3y=\text{cos}\ y\)
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1. This equation is nonlinear because of the \({y}^{2}\) term.
2. This equation is linear. There is no term involving a power or function of \(y,\) and the coefficients are all functions of \(x\text{.}\) The equation is already written in standard form, and \(r(x)\) is identically zero, so the equation is homogeneous.
3. This equation is nonlinear. Note that, in this case, x is the dependent variable and t is the independent variable. The second term involves the product of \(x\) and \({x}^{'},\) so the equation is nonlinear.
4. This equation is linear. Since \(r(x)=4{x}^{5},\) the equation is nonhomogeneous.
5. This equation is nonlinear, because of the \(\text{sin}\ {y}^{'}\) term.
6. This equation is linear. Rewriting it in standard form gives
  \[8{t}^{2}{y}^{″}-6{t}^{2}{y}^{'}+4ty=3{t}^{2}.\]
  With the equation in standard form, we can see that \(r(t)=3{t}^{2},\) so the equation is nonhomogeneous.
7. This equation looks like it’s linear, but we should rewrite it in standard form to be sure. We get
  \[\text{sin}({x}^{2}){y}^{″}-(\text{cos}\ x+1){y}^{'}+{x}^{2}y=-3.\]
  This equation is, indeed, linear. With \(r(x)=-3,\) it is nonhomogeneous.
8. This equation is nonlinear because of the \(\text{cos}\ y\) term.
Classify each of the following equations as linear or nonlinear. If the equation is linear, determine further whether it is homogeneous or nonhomogeneous.
1. \({({y}^{″})}^{2}-{y}^{'}+8{x}^{3}y=0\)
2. \((\text{sin}\ t){y}^{″}+\text{cos}\ t-3t{y}^{'}=0\)
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1. Nonlinear
2. Linear, nonhomogeneous
Consider the linear, homogeneous differential equation
\[{x}^{2}{y}^{″}-x{y}^{'}-3y=0.\]
Looking at this equation, notice that the coefficient functions are polynomials, with higher powers of \(x\) associated with higher-order derivatives of \(y\text{.}\) Show that \(y={x}^{3}\) is a solution to this differential equation.

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Let \(y={x}^{3}.\) Then \({y}^{'}=3{x}^{2}\) and \({y}^{″}=6x\text{.}\) Substituting into the differential equation, we see that
\[\begin{array}{ll}{x}^{2}{y}^{″}-xy'-3y & ={x}^{2}(6x)-x(3{x}^{2})-3({x}^{3}) \\ & =6{x}^{3}-3{x}^{3}-3{x}^{3} \\ & =0.\end{array}\]
Show that \(y=2{x}^{2}\) is a solution to the differential equation
\[\frac{1}{2}{x}^{2}{y}^{″}-x{y}^{'}+y=0.\]
Consider the differential equation
\[{y}^{″}-4{y}^{'}-5y=0.\]
Given that \({e}^{\text{-}x}\) and \({e}^{5x}\) are solutions to this differential equation, show that \(4{e}^{\text{-}x}+{e}^{5x}\) is a solution.

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We have
\[y(x)=4{e}^{\text{-}x}+{e}^{5x},\ \text{so}\ {y}^{'}(x)=-4{e}^{\text{-}x}+5{e}^{5x}\ \text{and}\ {y}^{″}(x)=4{e}^{\text{-}x}+25{e}^{5x}.\]
Then
\[\begin{array}{ll}{y}^{″}-4{y}^{'}-5y & =(4{e}^{\text{-}x}+25{e}^{5x})-4(-4{e}^{\text{-}x}+5{e}^{5x})-5(4{e}^{\text{-}x}+{e}^{5x}) \\ & =4{e}^{\text{-}x}+25{e}^{5x}+16{e}^{\text{-}x}-20{e}^{5x}-20{e}^{\text{-}x}-5{e}^{5x} \\ & =0.\end{array}\]
Thus, \(y(x)=4{e}^{\text{-}x}+{e}^{5x}\) is a solution.
Consider the differential equation
\[{y}^{″}+5{y}^{'}+6y=0.\]
Given that \({e}^{-2x}\) and \({e}^{-3x}\) are solutions to this differential equation, show that \(3{e}^{-2x}+6{e}^{-3x}\) is a solution.
Determine whether the following pairs of functions are linearly dependent or linearly independent.
1. \({f}_{1}(x)={x}^{2},\) \({f}_{2}(x)=5{x}^{2}\)
2. \({f}_{1}(x)=\text{sin}\ x,\) \({f}_{2}(x)=\text{cos}\ x\)
3. \({f}_{1}(x)={e}^{3x},\) \({f}_{2}(x)={e}^{-3x}\)
4. \({f}_{1}(x)=3x,\) \({f}_{2}(x)=3x+1\)
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1. \({f}_{2}(x)=5{f}_{1}(x),\) so the functions are linearly dependent.
2. There is no constant C such that \({f}_{1}(x)=C{f}_{2}(x),\) so the functions are linearly independent.
3. There is no constant C such that \({f}_{1}(x)=C{f}_{2}(x),\) so the functions are linearly independent. Don’t get confused by the fact that the exponents are constant multiples of each other. With two exponential functions, unless the exponents are equal, the functions are linearly independent.
4. There is no constant C such that \({f}_{1}(x)=C{f}_{2}(x),\) so the functions are linearly independent.
Determine whether the following pairs of functions are linearly dependent or linearly independent: \({f}_{1}(x)={e}^{x},\) \({f}_{2}(x)=3{e}^{3x}.\)

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Linearly independent
If \({y}_{1}(t)={e}^{3t}\) and \({y}_{2}(t)={e}^{-3t}\) are solutions to \({y}^{″}-9y=0,\) what is the general solution?

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Note that \({y}_{1}\) and \({y}_{2}\) are not constant multiples of one another, so they are linearly independent. Then, the general solution to the differential equation is \(y(t)={c}_{1}{e}^{3t}+{c}_{2}{e}^{-3t}.\)
If \({y}_{1}(x)={e}^{3x}\) and \({y}_{2}(x)=x{e}^{3x}\) are solutions to \({y}^{″}-6{y}^{'}+9y=0,\) what is the general solution?

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\(y(x)={c}_{1}{e}^{3x}+{c}_{2}x{e}^{3x}\)
Find the general solution to the following differential equations. Give your answers as functions of x.
1. \({y}^{″}+3{y}^{'}-4y=0\)
2. \({y}^{″}+6{y}^{'}+13y=0\)
3. \({y}^{″}+2{y}^{'}+y=0\)
4. \({y}^{″}-5{y}^{'}=0\)
5. \({y}^{″}-16y=0\)
6. \({y}^{″}+16y=0\)
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Note that all these equations are already given in standard form (step 1).
1. The characteristic equation is \({\lambda }^{2}+3\lambda -4=0\) (step 2). This factors into \((\lambda +4)(\lambda -1)=0,\) so the roots of the characteristic equation are \({\lambda }_{1}=-4\) and \({\lambda }_{2}=1\) (step 3). Then the general solution to the differential equation is
  \[y(x)={c}_{1}{e}^{-4x}+{c}_{2}{e}^{x}\ \text{(step 4)}\text{.}\]
2. The characteristic equation is \({\lambda }^{2}+6\lambda +13=0\) (step 2). Applying the quadratic formula, we see this equation has complex conjugate roots \(-3\pm 2i\) (step 3). Then the general solution to the differential equation is
  \[y(t)={e}^{-3t}({c}_{1}\text{cos}\ 2t+{c}_{2}\text{sin}\ 2t)\ \text{(step 4)}\text{.}\]
3. The characteristic equation is \({\lambda }^{2}+2\lambda +1=0\) (step 2). This factors into \({(\lambda +1)}^{2}=0,\) so the characteristic equation has a repeated real root \(\lambda =-1\) (step 3). Then the general solution to the differential equation is
  \[y(t)={c}_{1}{e}^{\text{-}t}+{c}_{2}t{e}^{\text{-}t}\ \text{(step 4).}\]
4. The characteristic equation is \({\lambda }^{2}-5\lambda\) (step 2). This factors into \(\lambda (\lambda -5)=0,\) so the roots of the characteristic equation are \({\lambda }_{1}=0\) and \({\lambda }_{2}=5\) (step 3). Note that \({e}^{0x}={e}^{0}=1,\) so our first solution is just a constant. Then the general solution to the differential equation is
  \[y(x)={c}_{1}+{c}_{2}{e}^{5x}\ \text{(step 4)}\text{.}\]
5. The characteristic equation is \({\lambda }^{2}-16=0\) (step 2). This factors into \((\lambda +4)(\lambda -4)=0,\) so the roots of the characteristic equation are \({\lambda }_{1}=4\) and \({\lambda }_{2}=-4\) (step 3). Then the general solution to the differential equation is
  \[y(x)={c}_{1}{e}^{4x}+{c}_{2}{e}^{-4x}\ \text{(step 4)}\text{.}\]
6. The characteristic equation is \({\lambda }^{2}+16=0\) (step 2). This has complex conjugate roots \(\pm 4i\) (step 3). Note that \({e}^{0x}={e}^{0}=1,\) so the exponential term in our solution is just a constant. Then the general solution to the differential equation is
  \[y(t)={c}_{1}\text{cos}\ 4t+{c}_{2}\text{sin}\ 4t\ \text{(step 4)}\text{.}\]
Find the general solution to the following differential equations:
1. \({y}^{″}-2{y}^{'}+10y=0\)
2. \({y}^{″}+14{y}^{'}+49y=0\)
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1. \(y(x)={e}^{x}({c}_{1}\text{cos}\ 3x+{c}_{2}\text{sin}\ 3x)\)
2. \(y(x)={c}_{1}{e}^{-7x}+{c}_{2}x{e}^{-7x}\)
Solve the following initial-value problem: \({y}^{″}+3{y}^{'}-4y=0,\) \(y(0)=1,\) \({y}^{'}(0)=-9.\)

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We already solved this differential equation in a. and found the general solution to be
\[y(x)={c}_{1}{e}^{-4x}+{c}_{2}{e}^{x}.\]
Then
\[{y}^{'}(x)=-4{c}_{1}{e}^{-4x}+{c}_{2}{e}^{x}.\]
When \(x=0,\) we have \(y(0)={c}_{1}+{c}_{2}\) and \({y}^{'}(0)=-4{c}_{1}+{c}_{2}.\) Applying the initial conditions, we have
\[\begin{array}{lll}{c}_{1}+{c}_{2} & = & 1 \\ -4{c}_{1}+{c}_{2} & = & -9.\end{array}\]
Then \({c}_{1}=1-{c}_{2}.\) Substituting this expression into the second equation, we see that
\[\begin{array}{lll}-4(1-{c}_{2})+{c}_{2} & = & -9 \\ -4+4{c}_{2}+{c}_{2} & = & -9 \\ 5{c}_{2} & = & -5 \\ {c}_{2} & = & -1.\end{array}\]
So, \({c}_{1}=2\) and the solution to the initial-value problem is
\[y(x)=2{e}^{-4x}-{e}^{x}.\]
Solve the initial-value problem \({y}^{″}-3{y}^{'}-10y=0,\) \(y(0)=0,\) \({y}^{'}(0)=7.\)

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\(y(x)=\text{-}{e}^{-2x}+{e}^{5x}\)
Solve the following initial-value problem and graph the solution:
\[{y}^{″}+6{y}^{'}+13y=0,\ y(0)=0,\ {y}^{'}(0)=2\]

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We already solved this differential equation in b. and found the general solution to be
\[y(x)={e}^{-3x}({c}_{1}\text{cos}\ 2x+{c}_{2}\text{sin}\ 2x).\]
Then
\[{y}^{'}(x)={e}^{-3x}(-2{c}_{1}\text{sin}\ 2x+2{c}_{2}\text{cos}\ 2x)-3{e}^{-3x}({c}_{1}\text{cos}\ 2x+{c}_{2}\text{sin}\ 2x).\]
When \(x=0,\) we have \(y(0)={c}_{1}\) and \({y}^{'}(0)=2{c}_{2}-3{c}_{1}.\) Applying the initial conditions, we obtain
\[\begin{array}{lll}{c}_{1} & = & 0 \\ -3{c}_{1}+2{c}_{2} & = & 2.\end{array}\]
Therefore, \({c}_{1}=0,\) \({c}_{2}=1,\) and the solution to the initial value problem is shown in the following graph.
\[y={e}^{-3x}\text{sin}\ 2x\text{.}\]
Solve the following initial-value problem and graph the solution: \({y}^{″}-2{y}^{'}+10y=0,y(0)=2,{y}^{'}(0)=-1\)

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\(y(x)={e}^{x}(2\ \text{cos}\ 3x-\text{sin}\ 3x)\)
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.)
\[{y}^{″}+2{y}^{'}+y=0,y(0)=1,{y}^{'}(0)=0\]
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?

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In c. we found the general solution to this differential equation to be
\[y(t)={c}_{1}{e}^{\text{-}t}+{c}_{2}t{e}^{\text{-}t}.\]
Then
\[{y}^{'}(t)=\text{-}{c}_{1}{e}^{\text{-}t}+{c}_{2}(\text{-}t{e}^{\text{-}t}+{e}^{\text{-}t}).\]
When \(t=0,\) we have \(y(0)={c}_{1}\) and \({y}^{'}(0)=\text{-}{c}_{1}+{c}_{2}.\) Applying the initial conditions, we obtain
\[\begin{array}{lll}{c}_{1} & = & 1 \\ {\text{-}c}_{1}+{c}_{2} & = & 0.\end{array}\]
Thus, \({c}_{1}=1,\) \({c}_{2}=1,\) and the solution to the initial value problem is
\[y(t)={e}^{\text{-}t}+t{e}^{\text{-}t}.\]
This solution is represented in the following graph. At time \(t=2,\) the mass is at position \(y(2)={e}^{-2}+2{e}^{-2}=3{e}^{-2}\approx 0.406\) m below equilibrium.
To calculate the velocity at time \(t=1,\) we need to find the derivative. We have \(y(t)={e}^{\text{-}t}+t{e}^{\text{-}t},\) so
\[{y}^{'}(t)=\text{-}{e}^{\text{-}t}+{e}^{\text{-}t}-t{e}^{\text{-}t}=\text{-}t{e}^{\text{-}t}.\]
Then \({y}^{'}(1)=\text{-}{e}^{-1}\approx -0.3679.\) 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?
\[{y}^{″}+14{y}^{'}+49y=0,y(0)=0,{y}^{'}(0)=1\]

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

At time \(t=0.3,\) \(y(0.3)=0.3{e}^{(-7*0.3)}=0.3{e}^{-2.1}\approx 0.0367.\) The mass is 0.0367 ft below equilibrium. At time \(t=0.1,\) \({y}^{'}(0.1)=0.3{e}^{-0.7}\approx 0.1490.\) The mass is moving downward at a speed of 0.1490 ft/sec.
In f. we solved the differential equation \({y}^{″}+16y=0\) and found the general solution to be \(y(t)={c}_{1}\text{cos}\ 4t+{c}_{2}\text{sin}\ 4t\text{.}\) If possible, solve the boundary-value problem if the boundary conditions are the following:
1. \(y(0)=0,\) \(y(\frac{\pi }{4})=0\)
2. \(y(0)=1,\) \(y(\frac{\pi }{8})=0\)
3. \(y(\frac{\pi }{8})=0,\) \(y(\frac{3\pi }{8})=2\)
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We have
\[y(t)={c}_{1}\text{cos}\ 4t+{c}_{2}\text{sin}\ 4t\text{.}\]
1. Applying the first boundary condition given here, we get \(y(0)={c}_{1}=0.\) So the solution is of the form \(y(t)={c}_{2}\text{sin}\ 4t\text{.}\) When we apply the second boundary condition, though, we get \(y(\frac{\pi }{4})={c}_{2}\text{sin}\ (4(\frac{\pi }{4}))={c}_{2}\text{sin}\ \pi =0\) 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, \(y(t)={c}_{2}\text{sin}\ 4t\) is a solution for any value of \({c}_{2}.\)
2. Applying the first boundary condition given here, we get \(y(0)={c}_{1}=1.\) Applying the second boundary condition gives \(y(\frac{\pi }{8})={c}_{2}=0,\) so \({c}_{2}=0.\) In this case, we have a unique solution: \(y(t)=\text{cos}\ 4t\text{.}\)
3. Applying the first boundary condition given here, we get \(y(\frac{\pi }{8})={c}_{2}=0.\) However, applying the second boundary condition gives \(y(\frac{3\pi }{8})=\text{-}{c}_{2}=2,\) so \({c}_{2}=-2.\) We cannot have \({c}_{2}=0=-2,\) so this boundary value problem has no solution.
\({x}^{3}{y}^{″}+(x-1){y}^{'}-8y=0\)

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linear, homogenous
\((1+{y}^{2}){y}^{″}+x{y}^{'}-3y=\text{cos}\ x\)
\(x{y}^{″}+{e}^{y}{y}^{'}=x\)

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nonlinear
\({y}^{″}+\frac{4}{x}{y}^{'}-8xy=5{x}^{2}+1\)
\({y}^{″}+(\text{sin}\ x){y}^{'}-xy=4y\)

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linear, homogeneous
\({y}^{″}+(\frac{x+3}{y}){y}^{'}=0\)
[T]\({y}^{″}+2{y}^{'}-3y=0;\) \(y(x)={c}_{1}{e}^{x}+{c}_{2}{e}^{-3x}\)
[T]\({x}^{2}{y}^{″}-2y-3{x}^{2}+1=0;\) \(y(x)={c}_{1}{x}^{2}+{c}_{2}{x}^{-1}+{x}^{2}\text{ln}\ (x)+\frac{1}{2}\)
[T]\({y}^{″}+14{y}^{'}+49y=0;\) \(y(x)={c}_{1}{e}^{-7x}+{c}_{2}x{e}^{-7x}\)
[T]\(6{y}^{″}-49{y}^{'}+8y=0;\) \(y(x)={c}_{1}{e}^{x\text{/}6}+{c}_{2}{e}^{8x}\)
Find the general solution to the linear differential equation.
\({y}^{″}-3{y}^{'}-10y=0\)

ເປີດເຜີຍຄຳຕອບ

\(y={c}_{1}{e}^{5x}+{c}_{2}{e}^{-2x}\)
\({y}^{″}-7{y}^{'}+12y=0\)
\({y}^{″}+4{y}^{'}+4y=0\)

ເປີດເຜີຍຄຳຕອບ

\(y={c}_{1}{e}^{-2x}+{c}_{2}x{e}^{-2x}\)
\(4{y}^{″}-12{y}^{'}+9y=0\)
\(2{y}^{″}-3{y}^{'}-5y=0\)

ເປີດເຜີຍຄຳຕອບ

\(y={c}_{1}{e}^{5x\text{/}2}+{c}_{2}{e}^{\text{-}x}\)
\(3{y}^{″}-14{y}^{'}+8y=0\)
\({y}^{″}+{y}^{'}+y=0\)

ເປີດເຜີຍຄຳຕອບ

\(y={e}^{\text{-}x\text{/}2}({c}_{1}\text{cos}\ \frac{\sqrt{3}x}{2}+{c}_{2}\text{sin}\ \frac{\sqrt{3}x}{2})\)
\(5{y}^{″}+2{y}^{'}+4y=0\)
\({y}^{″}-121y=0\)

ເປີດເຜີຍຄຳຕອບ

\(y={c}_{1}{e}^{-11x}+{c}_{2}{e}^{11x}\)
\(8{y}^{″}+14{y}^{'}-15y=0\)
\({y}^{″}+81y=0\)

ເປີດເຜີຍຄຳຕອບ

\(y={c}_{1}\text{cos}\ 9x+{c}_{2}\text{sin}\ 9x\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\lambda

lambda (eigenvalue)
The factor by which an eigenvector is stretched: Av = λv.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

\frac{\partial u}{\partial t},\ \nabla^2 u

partial derivative in time, Laplacian
Rate of change in time; sum of second partials (the diffusion operator).

How to: Second-Order Linear Equations

Recognize homogeneous and nonhomogeneous linear differential equations.
Determine the characteristic equation of a homogeneous linear equation.
Use the roots of the characteristic equation to find the solution to a homogeneous linear equation.
Solve initial-value and boundary-value problems involving linear differential equations.
This equation is nonlinear because of the
This equation is linear. There is no term involving a power or function of
This equation is nonlinear. Note that, in this case,
This equation is linear. Since

Questions people ask

What is a differential equation?

An equation whose unknown is a function, relating it to its own derivatives. "The rate of growth is proportional to the population" is y′ = ky, and solving it means finding y as a function of time.

Why does the solution have arbitrary constants?

Integrating loses information: many functions share the same derivative. An n-th order equation has n constants, fixed by n initial or boundary conditions.

ພະຍາຍາມເອງ

Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

ເພີ່ມເຕີມໃນ Differential Equations

Separable equations First-order linear equations Second-order, constant coefficients Nonhomogeneous equations Modelling with differential equations

Homogeneous Linear Equations

Second-Order Equations with Constant Coefficients

Initial-Value Problems and Boundary-Value Problems

Key Concepts

Key Equations

Second-Order Linear Equations

Practice (40)

Symbols used here

How to: Second-Order Linear Equations

Questions people ask

ພະຍາຍາມ​ເອງ

ເພີ່ມເຕີມໃນ Differential Equations

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