From one to two solutions

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:

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

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
$t^2\cdot y''-t\cdot y'-3\cdot y = 0$
with unknown $y$ defined for $t\gt 0$ . It is given that $y_1(t)={{1}\over{t}}$ 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 $t$ .

$y_2(t) =$ ${{t^3}\over{4}}$

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

$y''-{{1}\over{t}}\cdot y'-{{3}\over{t^2}}\cdot y = 0$

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 $p(t)= -{{1}\over{t}}$ . Thus we can take $P(t)=-\ln \left(t\right)$ . Substituting

$\e^{-P(t)}=t \qquad \text{ and } \qquad y_1={{1}\over{t}}$

in the function rule for $c$ gives

$\begin{array}{rcl} c(t) &=&\displaystyle \int \frac{t}{\left({{1}\over{t}}\right)^2} \dd t \\ &=&\displaystyle \int t^3 \dd t \\ &=&\displaystyle {{t^4}\over{4}} \\ \end{array}$

As a consequence, $y_2(t)=y_1(t)\cdot c(t)={{t^3}\over{4}}$ is a second solution.

New example