*Previously*, we saw how to find a particular solution of a linear second-order differential equation when two linearly independent solutions of the homogeneous ODE are given. Thus, the problem of solving a linear second-order ODE is reduced to finding all homogeneous solutions. This is difficult. But if we have one nonzero solution of the homogeneous equation, we can find a second one:

Suppose that #y_1# is a nonzero solution of the ODE

\[y''+p(t)\cdot y'+ q(t)\cdot y = 0\]

where #p# and #q# are continuous functions. Let

- #P(t)# be an antiderivative of #p(t)# and
- #c(t)# an antiderivative of \(\frac{\e^{-P(t)}}{y_1^2}\).

Then

\[y_2(t) =c(t)\cdot y_1(t)\] is a solution of the ODE that is linearly independent of #y_1#.

It is not necessary to memorise the formula. A second solution #y_2# can be found from the definition of the *Wronskian*:

\[W=y_1\cdot y_2'-y_2\cdot y_1'\]

According to *The Wronskian of a lineair second-order differential equation*, the Wronskian is known: #W = \e^{-P(t)}#. Also #y_1# and its derivative #y_1'# are known. Therefore, the equation is a linear first-order ODE, of which it is *known* how it can be solved.

Since #W = \e^{-P(t)}#, another expression for the function rule of #c' # is:

\[c'(t) = \frac{W}{y_1^2}\]

We begin by showing that #y_2(t) =c(t)\cdot y_1(t)# is a solution of the ODE. By use of the sum and product rule for differentiation, we first determine the first two derivatives of #y_2#:

\[\begin{array}{rcl}y_2' &=& c' \cdot y_1+c\cdot y_1'\\ y_2''&=&c''\cdot y_1 + 2c'\cdot y_1' +c\cdot y_1'' \\&=& c''\cdot y_1 + 2c'\cdot y_1' -c\cdot\left (p\cdot y_1'+ q\cdot y_1\right)\end{array}\]

Substitution of these function rules in the left hand side of the ODE gives

\[\begin{array}{rcl}y_2''+p\cdot y_2'+q\cdot y_2&=& c''\cdot y_1 +2c'\cdot y_1' -c\cdot\left (p\cdot y_1'+ q\cdot y_1\right)\\ &&+p\cdot (c' \cdot y_1+c\cdot y_1')+q\cdot c\cdot y_1 \\ &=& c''\cdot y_1 +2c'\cdot y_1' +p\cdot c' \cdot y_1\\ &=& c''\cdot y_1 +(2 y_1' +p\cdot y_1)\cdot c'\end{array}\]

It is given that #c'=\frac{\e^{-P(t)}}{y_1^2}#. In view of the product rule for differentiation this implies

\[c'' = -2y_1'\cdot\frac{\e^{-P(t)}}{y_1^3}-p\cdot \frac{\e^{-P(t)}}{y_1^2}\]

Filling these function rules into the previous result, we find

\[\begin{array}{rcl}y_2''+p\cdot y_2'+q\cdot y_2&=&c''\cdot y_1 +(2 y_1' +p\cdot y_1)\cdot c'\\ &=&-2y_1'\cdot\dfrac{\e^{-P(t)}}{y_1^2}-p\cdot \dfrac{\e^{-P(t)}}{y_1}\\&&+(2 y_1' +p\cdot y_1)\cdot\dfrac{\e^{-P(t)}}{y_1^2} \\&=&0\end{array}\]

This proves that #y_2# satisfies the differential equation. We still need to check that #y_2# is linearly independent of #y_1#. We make use of the *Linear independence test by use of the Wronskian *#W# of #y_1# and #y_2#:

\[\begin{array}{rcl}W&=&y_1\cdot y_2'-y_2\cdot y_1'\\ &=&y_1\cdot\left( c\cdot y_1'+c'\cdot y_1\right)-y_1\cdot c\cdot y_1'\\&=&y_1^2\cdot c'\\&=&\e^{-P(t)}\\&\ne&0\end{array}\]

Since #W\ne0#, it follows from the test that #y_1# and #y_2# are linearly independent.

In the derivation of the formula for #y_2#, two methodst play a role. In the first place (again), *variation of a constant,* because the constant factor #1# of the solution #y_1# is replaced by the function #c(t)#. In the second place, the idea that the *Wronskian* of the two solutions is equal to #\e^{-P(t)}# can be chosen, so that #y_2# satisfies the first-order differential equation #y_1\cdot y_2'-y_1'\cdot y_2=W#. This is a reduction of the problem of finding #y_2# to solving a first-order equation. Such an approach is called **order reduction.**

We examine what the statement means in the case of constant coefficients. To this end, we consider the homogeneous differential equation \[y''+p\cdot y'+q\cdot y=0\]

where #p# and #q# are constants. Let #\lambda# be a real number that satisfies the *characteristic equation* \( \lambda^2+p\cdot \lambda+q=0\). Then #y_1=\e^{\lambda\cdot t}# is a solution of the ODE. In this case, #p\cdot t# is an antiderivative of the constant function #p#.

If #p\ne -2\lambda#, then we have

\[\int\frac{ \e^{-p\cdot t}}{\e^{2\lambda\cdot t}}\,\dd t=\frac{1}{-p-2\lambda}\e^{(-p-2\lambda)\cdot t}+A\]

where #A# is an integration constant, so we can choose#c(t) = \e^{(-p-2\lambda)\cdot t}# (the constant factor is not important), so a second solution of the ODE is:

\[y_2=c(t)\cdot y_1 =\e^{(-p-\lambda)\cdot t}\]

If #\mu# is a second solution (in addition to #\lambda#) of the characteristic equation, then #\mu# satisfies #\lambda+\mu = -p#, so #-p-\lambda=\mu# and #y_2=\e^{\mu\cdot t}#. This is in accordance with the *General solution of a homogeneous linear second-order ODE with constant coefficients*.

This deals completely with the case of two different solutions of the characteristic equation. The case of a single solution corresponds to the situation in which #p= -2\lambda#. In this case, #c(t)# is an antiderivative of the constant function and we find #y_2=t\cdot y_1= t\cdot \e^{\lambda\cdot t}# for a second solution. Again, this is in line with *General solution of a homogeneous linear second-order ODE with constant coefficients*.

In the case where there are no real solutions, the sine solution gives the cosine, and vice versa.

Consider the ODE

\[ t^2\cdot y''-5\cdot t\cdot y'+8\cdot y = 0\]

with unknown #y# defined for #t\gt 0#. It is given that #y_1(t)=t^4# is a solution. Find a solution #y_2# such that #y_1# and #y_2# are linearly independent.

Give your answer in the form of a function rule in terms of #t#.

#y_2(t) =# #-{{t^2}\over{2}}#

In order to see this, we start with the equation in standard form:

\[ y''-{{5}\over{t}}\cdot y'+{{8}\over{t^2}}\cdot y = 0\]

We search a second solution #y_2# which is linearly independent of #y_1#. According to *Second solution of a homogeneous linear second-order ODE*, #y_2(t) = c(t) \cdot y_1(t)# is a suitable solution if

\[\begin{array}{rcl}

c(t) &=& \displaystyle \int \frac{\e^{ -P(t) }}{y_1(t)^2} \dd t\\

\end{array}\]

where #P(t)# is an antiderivative of #p(t)= -{{5}\over{t}}#. Thus we can take #P(t)=-5\cdot \ln \left(t\right)#. Substituting

\[

\e^{-P(t)}=t^5 \qquad \text{ and } \qquad y_1=t^4

\]

in the function rule for #c# gives

\[\begin{array}{rcl}

c(t) &=&\displaystyle \int \frac{t^5}{\left(t^4\right)^2} \dd t \\

&=&\displaystyle \int {{1}\over{t^3}} \dd t \\

&=&\displaystyle -{{1}\over{2\cdot t^2}} \\

\end{array}\]

As a consequence, #y_2(t)=y_1(t)\cdot c(t)=-{{t^2}\over{2}}# is a second solution.