### Differential equations: Solution methods for linear-second order ODEs

### From one to two solutions

*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:

Second solution of a homogeneous linear second-order ODE

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:

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

\[ \left(t-5\right)^2\cdot y''-2\cdot \left(t-5\right)\cdot y'+2\cdot y = 0\]

with unknown #y# defined for #t\gt 5#. It is given that #y_1(t)=\left(t-5\right)^2# 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) =# #5-t#

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

\[ y''-{{2}\over{t-5}}\cdot y'+{{2}\over{t^2-10\cdot t+25}}\cdot y = 0\]

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)= -{{2}\over{t-5}}#. Thus we can take #P(t)=-2\cdot \ln \left(t-5\right)#. Substituting

\[

\e^{-P(t)}=\left(t-5\right)^2 \qquad \text{ and } \qquad y_1=\left(t-5\right)^2

\]

in the function rule for #c# gives

\[\begin{array}{rcl}

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

&=&\displaystyle \int {{1}\over{t^2-10\cdot t+25}} \dd t \\

&=&\displaystyle -{{1}\over{t-5}} \\

\end{array}\]

As a consequence, #y_2(t)=y_1(t)\cdot c(t)=5-t# is a second solution.

**Pass Your Math**independent of your university. See pricing and more.

Or visit omptest.org if jou are taking an OMPT exam.