A second order differential equation is said to be *linear* if it can be written as

[label{eq:5.1.1} y''+p(x)y'+q(x)y=f(x).]

We call the function (f) on the right a *forcing function*, since in physical applications it is often related to a force acting on some system modeled by the differential equation. We say that Equation
ef{eq:5.1.1} is *homogeneous* if (fequiv0) or *nonhomogeneous* if (f
otequiv0). Since these definitions are like the corresponding definitions in Section 2.1 for the linear first order equation

[label{eq:5.1.2} y'+p(x)y=f(x),]

it is natural to expect similarities between methods of solving Equation ef{eq:5.1.1} and Equation ef{eq:5.1.2}. However, solving Equation ef{eq:5.1.1} is more difficult than solving Equation ef{eq:5.1.2}. For example, while Theorem (PageIndex{1}) gives a formula for the general solution of Equation ef{eq:5.1.2} in the case where (fequiv0) and Theorem (PageIndex{2}) gives a formula for the case where (f otequiv0), there are no formulas for the general solution of Equation ef{eq:5.1.1} in either case. Therefore we must be content to solve linear second order equations of special forms.

In Section 2.1 we considered the homogeneous equation (y'+p(x)y=0) first, and then used a nontrivial solution of this equation to find the general solution of the nonhomogeneous equation (y'+p(x)y=f(x)). Although the progression from the homogeneous to the nonhomogeneous case isn’t that simple for the linear second order equation, it is still necessary to solve the homogeneous equation

[label{eq:5.1.3} y''+p(x)y'+q(x)y=0]

in order to solve the nonhomogeneous equation Equation ef{eq:5.1.1}. This section is devoted to Equation ef{eq:5.1.3}.

The next theorem gives sufficient conditions for existence and uniqueness of solutions of initial value problems for Equation ef{eq:5.1.3}. We omit the proof.

Theorem (PageIndex{1})

Suppose (p) and (q) are continuous on an open interval ((a,b),) let (x_0) be any point in ((a,b),) and let (k_0) and (k_1) be arbitrary real numbers(.) Then the initial value problem

[y''+p(x)y'+q(x)y=0, y(x_0)=k_0, y'(x_0)=k_1 onumber ]

has a unique solution on ((a,b).)

Since (yequiv0) is obviously a solution of Equation
ef{eq:5.1.3} we call it the *trivial* solution. Any other solution is *nontrivial*. Under the assumptions of Theorem (PageIndex{1}), the only solution of the initial value problem

[y''+p(x)y'+q(x)y=0, y(x_0)=0, y'(x_0)=0 onumber ]

on ((a,b)) is the trivial solution (*Exercise 5.1.24*).

The next three examples illustrate concepts that we’ll develop later in this section. You shouldn’t be concerned with how to *find* the given solutions of the equations in these examples. This will be explained in later sections.

Example (PageIndex{1})

The coefficients of (y') and (y) in

[label{eq:5.1.4} y''-y=0]

are the constant functions (pequiv0) and (qequiv-1), which are continuous on ((-infty,infty)). Therefore Theorem (PageIndex{1}) implies that every initial value problem for Equation ef{eq:5.1.4} has a unique solution on ((-infty,infty)).

- Verify that (y_1=e^x) and (y_2=e^{-x}) are solutions of Equation ef{eq:5.1.4} on ((-infty,infty)).
- Verify that if (c_1) and (c_2) are arbitrary constants, (y=c_1e^x+c_2e^{-x}) is a solution of Equation ef{eq:5.1.4} on ((-infty,infty)).
- Solve the initial value problem [label{eq:5.1.5} y''-y=0,quad y(0)=1,quad y'(0)=3.]

**Solution**:

a. If (y_1=e^x) then (y_1'=e^x) and (y_1''=e^x=y_1), so (y_1''-y_1=0). If (y_2=e^{-x}), then (y_2'=-e^{-x}) and (y_2''=e^{-x}=y_2), so (y_2''-y_2=0).

b. If [label{eq:5.1.6} y=c_1e^x+c_2e^{-x}] then [label{eq:5.1.7} y'=c_1e^x-c_2e^{-x}] and [y''=c_1e^x+c_2e^{-x}, onumber ]

so [egin{aligned} y''-y&=(c_1e^x+c_2e^{-x})-(c_1e^x+c_2e^{-x}) &=c_1(e^x-e^x)+c_2(e^{-x}-e^{-x})=0end{aligned} onumber ] for all (x). Therefore (y=c_1e^x+c_2e^{-x}) is a solution of Equation ef{eq:5.1.4} on ((-infty,infty)).

c.

We can solve Equation ef{eq:5.1.5} by choosing (c_1) and (c_2) in Equation ef{eq:5.1.6} so that (y(0)=1) and (y'(0)=3). Setting (x=0) in Equation ef{eq:5.1.6} and Equation ef{eq:5.1.7} shows that this is equivalent to

[egin{aligned} c_1+c_2&=1 c_1-c_2&=3.end{aligned} onumber ]

Solving these equations yields (c_1=2) and (c_2=-1). Therefore (y=2e^x-e^{-x}) is the unique solution of Equation ef{eq:5.1.5} on ((-infty,infty)).

Example (PageIndex{2})

Let (omega) be a positive constant. The coefficients of (y') and (y) in

[label{eq:5.1.8} y''+omega^2y=0]

are the constant functions (pequiv0) and (qequivomega^2), which are continuous on ((-infty,infty)). Therefore Theorem (PageIndex{1}) implies that every initial value problem for Equation ef{eq:5.1.8} has a unique solution on ((-infty,infty)).

- Verify that (y_1=cosomega x) and (y_2=sinomega x) are solutions of Equation ef{eq:5.1.8} on ((-infty,infty)).
- Verify that if (c_1) and (c_2) are arbitrary constants then (y=c_1cosomega x+c_2sinomega x) is a solution of Equation ef{eq:5.1.8} on ((-infty,infty)).
- Solve the initial value problem [label{eq:5.1.9} y''+omega^2y=0,quad y(0)=1,quad y'(0)=3.]

**Solution**:

a. If (y_1=cosomega x) then (y_1'=-omegasinomega x) and (y_1''=-omega^2cosomega x=-omega^2y_1), so (y_1''+omega^2y_1=0). If (y_2=sinomega x) then, (y_2'=omegacosomega x) and (y_2''=-omega^2sinomega x=-omega^2y_2), so (y_2''+omega^2y_2=0).

b. If [label{eq:5.1.10} y=c_1cosomega x+c_2sinomega x] then [label{eq:5.1.11} y'=omega(-c_1sinomega x+c_2cosomega x)] and [y''=-omega^2(c_1cosomega x+c_2sinomega x), onumber ] so [egin{aligned} y''+omega^2y&= -omega^2(c_1cosomega x+c_2sinomega x) +omega^2(c_1cosomega x+c_2sinomega x) &=c_1omega^2(-cosomega x+cosomega x)+ c_2omega^2(-sinomega x+sinomega x)=0end{aligned} onumber ] for all (x). Therefore (y=c_1cosomega x+c_2sinomega x) is a solution of Equation ef{eq:5.1.8} on ((-infty,infty)).

c. To solve Equation ef{eq:5.1.9}, we must choosing (c_1) and (c_2) in Equation ef{eq:5.1.10} so that (y(0)=1) and (y'(0)=3). Setting (x=0) in Equation ef{eq:5.1.10} and Equation ef{eq:5.1.11} shows that (c_1=1) and (c_2=3/omega). Therefore

[y=cosomega x+{3overomega}sinomega x onumber ]

is the unique solution of Equation ef{eq:5.1.9} on ((-infty,infty)).

Theorem (PageIndex{1}) implies that if (k_0) and (k_1) are arbitrary real numbers then the initial value problem

[label{eq:5.1.12} P_0(x)y''+P_1(x)y'+P_2(x)y=0,quad y(x_0)=k_0,quad y'(x_0)=k_1]

has a unique solution on an interval ((a,b)) that contains (x_0), provided that (P_0), (P_1), and (P_2) are continuous and (P_0) has no zeros on ((a,b)). To see this, we rewrite the differential equation in Equation ef{eq:5.1.12} as

[y''+{P_1(x)over P_0(x)}y'+{P_2(x)over P_0(x)}y=0 onumber ]

and apply Theorem (PageIndex{1}) with (p=P_1/P_0) and (q=P_2/P_0).

Example (PageIndex{3})

The equation

[label{eq:5.1.13} x^2y''+xy'-4y=0]

has the form of the differential equation in Equation ef{eq:5.1.12}, with (P_0(x)=x^2), (P_1(x)=x), and (P_2(x)=-4), which are are all continuous on ((-infty,infty)). However, since (P(0)=0) we must consider solutions of Equation ef{eq:5.1.13} on ((-infty,0)) and ((0,infty)). Since (P_0) has no zeros on these intervals, Theorem (PageIndex{1}) implies that the initial value problem

[x^2y''+xy'-4y=0,quad y(x_0)=k_0,quad y'(x_0)=k_1 onumber ]

has a unique solution on ((0,infty)) if (x_0>0), or on ((-infty,0)) if (x_0<0).

- Verify that (y_1=x^2) is a solution of Equation ef{eq:5.1.13} on ((-infty,infty)) and (y_2=1/x^2) is a solution of Equation ef{eq:5.1.13} on ((-infty,0)) and ((0,infty)).
- Verify that if (c_1) and (c_2) are any constants then (y=c_1x^2+c_2/x^2) is a solution of Equation ef{eq:5.1.13} on ((-infty,0)) and ((0,infty)).
- Solve the initial value problem [label{eq:5.1.14} x^2y''+xy'-4y=0,quad y(1)=2,quad y'(1)=0.]
- Solve the initial value problem [label{eq:5.1.15} x^2y''+xy'-4y=0,quad y(-1)=2,quad y'(-1)=0.]

**Solution**:

a. If (y_1=x^2) then (y_1'=2x) and (y_1''=2), so [x^2y_1''+xy_1'-4y_1=x^2(2)+x(2x)-4x^2=0 onumber ] for (x) in ((-infty,infty)). If (y_2=1/x^2), then (y_2'=-2/x^3) and (y_2''=6/x^4), so [x^2y_2''+xy_2'-4y_2=x^2left(6over x^4 ight)-xleft(2over x^3 ight)-{4over x^2}=0 onumber ] for (x) in ((-infty,0)) or ((0,infty)).

b. If [label{eq:5.1.16} y=c_1x^2+{c_2over x^2}] then [label{eq:5.1.17} y'=2c_1x-{2c_2over x^3}] and [y''=2c_1+{6c_2over x^4}, onumber ] so [egin{aligned} x^{2}y''+xy'-4y&=x^{2}left(2c_{1}+frac{6c_{2}}{x^{4}} ight)+xleft(2c_{1}x-frac{2c_{2}}{x^{3}} ight)-4left(c_{1}x^{2}+frac{c_{2}}{x^{2}} ight) &=c_{1}(2x^{2}+2x^{2}-4x^{2})+c_{2}left(frac{6}{x^{2}}-frac{2}{x^{2}}-frac{4}{x^{2}} ight) &=c_{1}cdot 0+c_{2}cdot 0 = 0 end{aligned} onumber ] for (x) in ((-infty,0)) or ((0,infty)).

c. To solve Equation ef{eq:5.1.14}, we choose (c_1) and (c_2) in Equation ef{eq:5.1.16} so that (y(1)=2) and (y'(1)=0). Setting (x=1) in Equation ef{eq:5.1.16} and Equation ef{eq:5.1.17} shows that this is equivalent to

[egin{aligned} phantom{2}c_1+phantom{2}c_2&=2 2c_1-2c_2&=0.end{aligned} onumber ]

Solving these equations yields (c_1=1) and (c_2=1). Therefore (y=x^2+1/x^2) is the unique solution of Equation ef{eq:5.1.14} on ((0,infty)).

d. We can solve Equation ef{eq:5.1.15} by choosing (c_1) and (c_2) in Equation ef{eq:5.1.16} so that (y(-1)=2) and (y'(-1)=0). Setting (x=-1) in Equation ef{eq:5.1.16} and Equation ef{eq:5.1.17} shows that this is equivalent to

[egin{aligned} phantom{-2}c_1+phantom{2}c_2&=2 -2c_1+2c_2&=0.end{aligned} onumber ]

Solving these equations yields (c_1=1) and (c_2=1). Therefore (y=x^2+1/x^2) is the unique solution of Equation ef{eq:5.1.15} on ((-infty,0)).

Although the *formulas* for the solutions of Equation
ef{eq:5.1.14} and Equation
ef{eq:5.1.15} are both (y=x^2+1/x^2), you should not conclude that these two initial value problems have the same solution. Remember that a solution of an initial value problem is defined *on an interval that contains the initial point*; therefore, the solution of Equation
ef{eq:5.1.14} is (y=x^2+1/x^2) *on the interval* ((0,infty)), which contains the initial point (x_0=1), while the solution of Equation
ef{eq:5.1.15} is (y=x^2+1/x^2) *on the interval* ((-infty,0)), which contains the initial point (x_0=-1).

## The General Solution of a Homogeneous Linear Second Order Equation

If (y_1) and (y_2) are defined on an interval ((a,b)) and (c_1) and (c_2) are constants, then

[y=c_1y_1+c_2y_2 onumber ]

is a *linear combination of (y_1)* and (y_2). For example, (y=2cos x+7 sin x) is a linear combination of (y_1= cos x) and (y_2=sin x), with (c_1=2) and (c_2=7).

The next theorem states a fact that we’ve already verified in Examples (PageIndex{1}), (PageIndex{2}), (PageIndex{3}).

Theorem (PageIndex{2})

If (y_1) and (y_2) are solutions of the homogeneous equation

[label{eq:5.1.18} y''+p(x)y'+q(x)y=0]

on ((a,b),) then any linear combination

[label{eq:5.1.19} y=c_1y_1+c_2y_2]

of (y_1) and (y_2) is also a solution of (eqref{eq:5.1.18}) on ((a,b).)

**Proof**If [y=c_1y_1+c_2y_2 onumber ] then [y'=c_1y_1'+c_2y_2'quad ext{ and} quad y''=c_1y_1''+c_2y_2''. onumber ]

Therefore[egin{aligned} y''+p(x)y'+q(x)y&=(c_1y_1''+c_2y_2'')+p(x)(c_1y_1'+c_2y_2') +q(x)(c_1y_1+c_2y_2) &=c_1left(y_1''+p(x)y_1'+q(x)y_1 ight) +c_2left(y_2''+p(x)y_2'+q(x)y_2 ight) &=c_1cdot0+c_2cdot0=0,end{aligned} onumber ]

since (y_1) and (y_2) are solutions of Equation ef{eq:5.1.18}.We say that ({y_1,y_2}) is a

*fundamental set of solutions of (eqref{eq:5.1.18})*on ((a,b)) if every solution of Equation ef{eq:5.1.18} on ((a,b)) can be written as a linear combination of (y_1) and (y_2) as in Equation ef{eq:5.1.19}. In this case we say that Equation ef{eq:5.1.19} is*general solution of (eqref{eq:5.1.18})*on ((a,b)).

## Linear Independence

We need a way to determine whether a given set ({y_1,y_2}) of solutions of Equation ef{eq:5.1.18} is a fundamental set. The next definition will enable us to state necessary and sufficient conditions for this.

We say that two functions (y_1) and (y_2) defined on an interval ((a,b)) are *linearly independent on* ((a,b)) if neither is a constant multiple of the other on ((a,b)). (In particular, this means that neither can be the trivial solution of Equation
ef{eq:5.1.18}, since, for example, if (y_1equiv0) we could write (y_1=0y_2).) We’ll also say that the set ({y_1,y_2}) *is linearly independent on* ((a,b)).

Theorem (PageIndex{3})

Suppose (p) and (q) are continuous on ((a,b).) Then a set ({y_1,y_2}) of solutions of

[label{eq:5.1.20} y''+p(x)y'+q(x)y=0]

on ((a,b)) is a fundamental set if and only if ({y_1,y_2}) is linearly independent on ((a,b).)

**Proof**We’ll present the proof of Theorem (PageIndex{3}) in steps worth regarding as theorems in their own right. However, let’s first interpret Theorem (PageIndex{3}) in terms of Examples (PageIndex{1}), (PageIndex{2}), (PageIndex{3}).

Example (PageIndex{4})

Since (e^x/e^{-x}=e^{2x}) is nonconstant, Theorem (PageIndex{3}) implies that (y=c_1e^x+c_2e^{-x}) is the general solution of (y''-y=0) on ((-infty,infty)).

Since (cosomega x/sinomega x=cotomega x) is nonconstant, Theorem (PageIndex{3}) implies that (y=c_1cosomega x+c_2sinomega x) is the general solution of (y''+omega^2y=0) on ((-infty,infty)).

Since (x^2/x^{-2}=x^4) is nonconstant, Theorem (PageIndex{3}) implies that (y=c_1x^2+c_2/x^2) is the general solution of (x^2y''+xy'-4y=0) on ((-infty,0)) and ((0,infty)).

## The Wronskian and Abel's Formula

To motivate a result that we need in order to prove Theorem (PageIndex{3}), let’s see what is required to prove that ({y_1,y_2}) is a fundamental set of solutions of Equation ef{eq:5.1.20} on ((a,b)). Let (x_0) be an arbitrary point in ((a,b)), and suppose (y) is an arbitrary solution of Equation ef{eq:5.1.20} on ((a,b)). Then (y) is the unique solution of the initial value problem

[label{eq:5.1.21} y''+p(x)y'+q(x)y=0,quad y(x_0)=k_0,quad y'(x_0)=k_1;]

that is, (k_0) and (k_1) are the numbers obtained by evaluating (y) and (y') at (x_0). Moreover, (k_0) and (k_1) can be any real numbers, since Theorem (PageIndex{1}) implies that Equation ef{eq:5.1.21} has a solution no matter how (k_0) and (k_1) are chosen. Therefore ({y_1,y_2}) is a fundamental set of solutions of Equation ef{eq:5.1.20} on ((a,b)) if and only if it is possible to write the solution of an arbitrary initial value problem Equation ef{eq:5.1.21} as (y=c_1y_1+c_2y_2). This is equivalent to requiring that the system

[label{eq:5.1.22} egin{array}{rcl} c_1y_1(x_0)+c_2y_2(x_0)&=k_0 c_1y_1'(x_0)+c_2y_2'(x_0)&=k_1 end{array}]

has a solution ((c_1,c_2)) for every choice of ((k_0,k_1)). Let’s try to solve Equation ef{eq:5.1.22}.

Multiplying the first equation in Equation ef{eq:5.1.22} by (y_2'(x_0)) and the second by (y_2(x_0)) yields

[egin{aligned} c_1y_1(x_0)y_2'(x_0)+c_2y_2(x_0)y_2'(x_0)&= y_2'(x_0)k_0 c_1y_1'(x_0)y_2(x_0)+c_2y_2'(x_0)y_2(x_0)&= y_2(x_0)k_1,end{aligned}]

and subtracting the second equation here from the first yields

[label{eq:5.1.23} left(y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0) ight)c_1= y_2'(x_0)k_0-y_2(x_0)k_1.]

Multiplying the first equation in Equation ef{eq:5.1.22} by (y_1'(x_0)) and the second by (y_1(x_0)) yields

[egin{aligned} c_1y_1(x_0)y_1'(x_0)+c_2y_2(x_0)y_1'(x_0)&= y_1'(x_0)k_0 c_1y_1'(x_0)y_1(x_0)+c_2y_2'(x_0)y_1(x_0)&= y_1(x_0)k_1,end{aligned}]

and subtracting the first equation here from the second yields

[label{eq:5.1.24} left(y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0) ight)c_2= y_1(x_0)k_1-y_1'(x_0)k_0.]

If

[y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)=0, onumber ]

it is impossible to satisfy Equation ef{eq:5.1.23} and Equation ef{eq:5.1.24} (and therefore Equation ef{eq:5.1.22} ) unless (k_0) and (k_1) happen to satisfy

[egin{aligned} y_1(x_0)k_1-y_1'(x_0)k_0&=0 y_2'(x_0)k_0-y_2(x_0)k_1&=0.end{aligned}]

On the other hand, if

[label{eq:5.1.25} y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0) e0]

we can divide Equation ef{eq:5.1.23} and Equation ef{eq:5.1.24} through by the quantity on the left to obtain

[label{eq:5.1.26} egin{array}{rcl} c_1&={y_2'(x_0)k_0-y_2(x_0)k_1over y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)} c_2&={y_1(x_0)k_1-y_1'(x_0)k_0over y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)}, end{array}]

no matter how (k_0) and (k_1) are chosen. This motivates us to consider conditions on (y_1) and (y_2) that imply Equation ef{eq:5.1.25}.

Theorem (PageIndex{4})

Suppose (p) and (q) are continuous on ((a,b),) let (y_1) and (y_2) be solutions of

[label{eq:5.1.27} y''+p(x)y'+q(x)y=0]

on ((a,b)), and define

[label{eq:5.1.28} W=y_1y_2'-y_1'y_2.]

Let (x_0) be any point in ((a,b).) Then

[label{eq:5.1.29} W(x)=W(x_0) e^{-int^x_{x_0}p(t):dt}, quad a

Therefore either (W) has no zeros in ((a,b)) or (Wequiv0) on ((a,b).)

**Proof**Differentiating Equation ef{eq:5.1.28} yields

[label{eq:5.1.30} W'=y'_1y'_2+y_1y''_2-y'_1y'_2-y''_1y_2= y_1y''_2-y''_1y_2.]

Since (y_1) and (y_2) both satisfy Equation ef{eq:5.1.27},[y''_1 =-py'_1-qy_1quad ext{and} quad y''_2 =-py'_2-qy_2. onumber ]

Substituting these into Equation ef{eq:5.1.30} yields[egin{aligned} W'&= -y_1igl(py'_2+qy_2igr) +y_2igl(py'_1+qy_1igr) &= -p(y_1y'_2-y_2y'_1)-q(y_1y_2-y_2y_1) &= -p(y_1y'_2-y_2y'_1)=-pW.end{aligned} onumber ]

Therefore (W'+p(x)W=0); that is, (W) is the solution of the initial value problem[y'+p(x)y=0,quad y(x_0)=W(x_0). onumber ]

We leave it to you to verify by separation of variables that this implies Equation ef{eq:5.1.29}. If (W(x_0) e0), Equation ef{eq:5.1.29} implies that (W) has no zeros in ((a,b)), since an exponential is never zero. On the other hand, if (W(x_0)=0), Equation ef{eq:5.1.29} implies that (W(x)=0) for all (x) in ((a,b)).

The function (W) defined in Equation
ef{eq:5.1.28} is the *Wronskian of ({y_1,y_2})*. Formula Equation
ef{eq:5.1.29} is *Abel’s formula*.

The Wronskian of ({y_1,y_2}) is usually written as the determinant

[W=left| egin{array}{cc} y_1 & y_2 y'_1 & y'_2 end{array} ight|. onumber ]

The expressions in Equation ef{eq:5.1.26} for (c_1) and (c_2) can be written in terms of determinants as

[c_1={1over W(x_0)} left| egin{array}{cc} k_0 & y_2(x_0) k_1 & y'_2(x_0) end{array} ight| quad ext{and} quad c_2={1over W(x_0)} left| egin{array}{cc} y_1(x_0) & k_0 y'_1(x_0) &k_1 end{array} ight|. onumber ]

If you’ve taken linear algebra you may recognize this as *Cramer’s rule*.

Example (PageIndex{5})

Verify Abel’s formula for the following differential equations and the corresponding solutions, from Examples (PageIndex{1}), (PageIndex{2}), (PageIndex{3}).

- (y''-y=0;quad y_1=e^x,; y_2=e^{-x})
- (y''+omega^2y=0;quad y_1=cosomega x,; y_2=sinomega x)
- (x^2y''+xy'-4y=0;quad y_1=x^2,; y_2=1/x^2)

**Solution**:

a. Since (pequiv0), we can verify Abel’s formula by showing that (W) is constant, which is true, since

[W(x)=left| egin{array}{rr} e^x & e^{-x} e^x & -e^{-x} end{array} ight|=e^x(-e^{-x})-e^xe^{-x}=-2 onumber ]

for all (x).

b. Again, since (pequiv0), we can verify Abel’s formula by showing that (W) is constant, which is true, since

[egin{aligned} W(x)&={left| egin{array}{cc} cosomega x & sinomega x -omegasinomega x &omegacosomega x end{array} ight|} &=cosomega x (omegacosomega x)-(-omegasinomega x)sinomega x &=omega(cos^2omega x+sin^2omega x)=omegaend{aligned} onumber ]

for all (x).

c. Computing the Wronskian of (y_1=x^2) and (y_2=1/x^2) directly yields

[label{eq:5.1.31} W=left| egin{array}{cc} x^2 & 1/x^2 2x & -2/x^3 end{array} ight|=x^2left(-{2over x^3} ight)-2xleft(1over x^2 ight)=-{4over x}.]

To verify Abel’s formula we rewrite the differential equation as

[y''+{1over x}y'-{4over x^2}y=0 onumber ]

to see that (p(x)=1/x). If (x_0) and (x) are either both in ((-infty,0)) or both in ((0,infty)) then

[int_{x_0}^x p(t),dt=int_{x_0}^x {dtover t}=lnleft(xover x_0 ight), onumber ]

so Abel’s formula becomes

[egin{aligned} W(x)&=W(x_0)e^{-ln(x/x_0)}=W(x_0){x_0over x} &=-left(4over x_0 ight)left(x_0over x ight)quad ext{from} eqref{eq:5.1.31} &=-{4over x},end{aligned} onumber ]

which is consistent with Equation ef{eq:5.1.31}.

The next theorem will enable us to complete the proof of Theorem (PageIndex{3}).

Theorem (PageIndex{5})

Suppose (p) and (q) are continuous on an open interval ((a,b),) let (y_1) and (y_2) be solutions of

[label{eq:5.1.32} y''+p(x)y'+q(x)y=0]

on ((a,b),) and let (W=y_1y_2'-y_1'y_2.) Then (y_1) and (y_2) are linearly independent on ((a,b)) if and only if (W) has no zeros on ((a,b).)

**Proof**We first show that if (W(x_0)=0) for some (x_0) in ((a,b)), then (y_1) and (y_2) are linearly dependent on ((a,b)). Let (I) be a subinterval of ((a,b)) on which (y_1) has no zeros. (If there’s no such subinterval, (y_1equiv0) on ((a,b)), so (y_1) and (y_2) are linearly independent, and we are finished with this part of the proof.) Then (y_2/y_1) is defined on (I), and

[label{eq:5.1.33} left(y_2over y_1 ight)'={y_1y_2'-y_1'y_2over y_1^2}={Wover y_1^2}.]

However, if (W(x_0)=0), Theorem (PageIndex{4}) implies that (Wequiv0) on ((a,b)). Therefore Equation ef{eq:5.1.33} implies that ((y_2/y_1)'equiv0), so (y_2/y_1=c) (constant) on (I). This shows that (y_2(x)=cy_1(x)) for all (x) in (I). However, we want to show that (y_2=cy_1(x)) for all (x) in ((a,b)). Let (Y=y_2-cy_1). Then (Y) is a solution of Equation ef{eq:5.1.32} on ((a,b)) such that (Yequiv0) on (I), and therefore (Y'equiv0) on (I). Consequently, if (x_0) is chosen arbitrarily in (I) then (Y) is a solution of the initial value problem

[y''+p(x)y'+q(x)y=0,quad y(x_0)=0,quad y'(x_0)=0, onumber ]

which implies that (Yequiv0) on ((a,b)), by the paragraph following Theorem (PageIndex{1}). (See also

*Exercise 5.1.24*). Hence, (y_2-cy_1equiv0) on ((a,b)), which implies that (y_1) and (y_2) are not linearly independent on ((a,b)).

Now suppose (W) has no zeros on ((a,b)). Then (y_1) can’t be identically zero on ((a,b)) (why not?), and therefore there is a subinterval (I) of ((a,b)) on which (y_1) has no zeros. Since Equation ef{eq:5.1.33} implies that (y_2/y_1) is nonconstant on (I), (y_2) isn’t a constant multiple of (y_1) on ((a,b)). A similar argument shows that (y_1) isn’t a constant multiple of (y_2) on ((a,b)), since

[left(y_1over y_2 ight)'={y_1'y_2-y_1y_2'over y_2^2}=-{Wover y_2^2} onumber ]

on any subinterval of ((a,b)) where (y_2) has no zeros.

We can now complete the proof of Theorem (PageIndex{3}). From Theorem (PageIndex{5}), two solutions (y_1) and (y_2) of Equation ef{eq:5.1.32} are linearly independent on ((a,b)) if and only if (W) has no zeros on ((a,b)). From Theorem (PageIndex{4}) and the motivating comments preceding it, ({y_1,y_2}) is a fundamental set of solutions of Equation ef{eq:5.1.32} if and only if (W) has no zeros on ((a,b)). Therefore ({y_1,y_2}) is a fundamental set for Equation ef{eq:5.1.32} on ((a,b)) if and only if ({y_1,y_2}) is linearly independent on ((a,b)).

The next theorem summarizes the relationships among the concepts discussed in this section.

Theorem (PageIndex{6})

Suppose (p) and (q) are continuous on an open interval ((a,b)) and let (y_1) and (y_2) be solutions of

[label{eq:5.1.34} y''+p(x)y'+q(x)y=0]

on ((a,b).) Then the following statements are equivalent(;) that is(,) they are either all true or all false(.)

- The general solution of (eqref{eq:5.1.34}) on ((a,b)) is (y=c_1y_1+c_2y_2).
- ({y_1,y_2}) is a fundamental set of solutions of (eqref{eq:5.1.34}) on ((a,b).)
- ({y_1,y_2}) is linearly independent on ((a,b).)
- The Wronskian of ({y_1,y_2}) is nonzero at some point in ((a,b).)
- The Wronskian of ({y_1,y_2}) is nonzero at all points in ((a,b).)

We can apply this theorem to an equation written as

[P_0(x)y''+P_1(x)y'+P_2(x)y=0 onumber ]

on an interval ((a,b)) where (P_0), (P_1), and (P_2) are continuous and (P_0) has no zeros.dd proof here and it will automatically be hidden

Theorem (PageIndex{7})

Suppose (c) is in ((a,b)) and (alpha) and (eta) are real numbers, not both zero. Under the assumptions of Theorem (PageIndex{7}), suppose (y_{1}) and (y_{2}) are solutions of Equation ef{eq:5.1.34} such that

[label{eq:5.1.35} alpha y_{1}(c)+eta y_{1}'(c)=0quad ext{and}quad alpha y_{2}(c)+eta y_{2}'(c)=0.]

Then ({y_{1},y_{2}}) isn’t linearly independent on ((a,b).)

**Proof**Since (alpha) and (eta) are not both zero, Equation ef{eq:5.1.35} implies that

[left|egin{array}{ccccccc} y_{1}(c)&y_{1}'(c)y_{2}(c)& y_{2}'(c) end{array} ight|=0, quad ext{so}quad left|egin{array}{cccccc} y_{1}(c)&y_{2}(c) y_{1}'(c)&y_{2}'(c) end{array} ight|=0 onumber ]

and Theorem (PageIndex{6}) implies the stated conclusion.