Calculus – Second-order differential equations – Homogeneous linear differential equations

A homogeneous second-order linear differential equation has the form

\[P(x)y''(x)+Q(x)y'(x)+R(x)y(x)=0,\]

where \(P\), \(Q\) and \(R\) are continuous functions with \(P(x)\not\equiv0\). Then we have:

Theorem: If \(y_1(x)\) and \(y_2(x)\) are both solutions, then \(y(x)=c_1y_1(x)+c_2y_2(x)\) is also a solution for every \(c_1,c_2\in\mathbb{R}\).

Proof: Since \(y_1(x)\) and \(y_2(x)\) are both solutions, we have

\[P(x)y_1''(x)+Q(x)y_1'(x)+R(x)y_1(x)=0\quad\text{and}\quad P(x)y_2''(x)+Q(x)y_2'(x)+R(x)y_2(x)=0.\]

Let \(y(x)=c_1y_1(x)+c_2y_2(x)\), then we have

\begin{align*} P(x)y''(x)+Q(x)y'(x)+R(x)y(x)&=P(x)\left[c_1y_1''(x)+c_2y_2''(x)\right]+Q(x)\left[c_1y_1'(x)+c_2y_2'(x)\right] +R(x)\left[c_1y_1(x)+c_2y_2(x)\right]\\[5mm] &=c_1\left[P(x)y_1''(x)+Q(x)y_1'(x)+R(x)y_1(x)\right]+c_2\left[P(x)y_2''(x)+Q(x)y_2'(x)+R(x)y_2(x)\right]\\[5mm] &=c_1\cdot0+c_2\cdot0=0. \end{align*}

Theorem: If \(y_1(x)\) and \(y_2(x)\) are linearly independent solutions on an interval, and \(P(x)\) is never \(0\) there, then the general solution is given by \(y(x)=c_1y_1(x)+c_2y_2(x)\), where \(c_1\) and \(c_2\) are arbitrary constants.

In the case of a homogeneous second-order linear differential equation with constant coefficients

\[ay''+by'+cy=0,\quad a,b,c\in\mathbb{R},\quad a\neq0,\]

we try \(y=e^{rx}\) as a solution, then \(y'=re^{rx}\) and \(y''=t^2e^{rx}\), and substitution leads to

\[ar^2e^{rx}+bre^{rx}+ce^{rx}=0\quad\Longleftrightarrow\quad e^{rx}\left(ar^2+br+c\right)=0.\]

Since \(e^{rx}\neq0\) this leads to the characteristic equation (or auxiliary equation)

\[ar^2+br+c=0,\quad a,b,c\in\mathbb{R},\quad a\neq0.\]

Now we consider three possibilities for the discriminant \(D=b^2-4ac\): \(D>0\), \(D=0\) and \(D<0\).

If \(D>0\) the characteristic equation has two different real solutions, say \(r_1\) and \(r_2\). Then the general solution is \(y(x)=c_1e^{r_1x}+c_2e^{r_2x}\).

If \(D=0\) the characteristic equation has two equal real solutions \(r_1=r_2=r\). So we only have one linear independent solution \(y=r^{rx}\) in that case. However, then \(y=xe^{rx}\) is a solution too. In order to see this, we note that both \(ar^2+br+c=0\) and \(2ar+b=0\). Then, if a substitute \(y=xe^{rx}\), \(y'=(rx+1)e^{rx}\) and \(y''=(r^2x+2r)e^{rx}\), we find that

\[ay''+by'+cy=a(r^2x+2r)e^{rx}+b(rx+1)e^{rx}+cxe^{rx}=(ar^2+br+c)xe^{rx}+(2ar+b)e^{rx}=0+0=0.\]

Since \(y_1=e^{rx}\) and \(y_2=xe^{rx}\) are linearly independent, the general solution is \(y(x)=c_1e^{rx}+c_2xe^{rx}\) in this case.

If \(D<0\) the characteristic equation has a complex conjugate pair as solutions, say \(r=\alpha\pm i\beta\) with \(\beta\neq0\). This implies that the general (complex) solution is a (complex) linear combination of

\[e^{(\alpha+i\beta)x}=e^{\alpha x}\cdot e^{i\beta x}=e^{\alpha x}\left(\cos(\beta x)+i\sin(\beta x)\right)\quad\text{and}\quad e^{(\alpha-i\beta)x}=e^{\alpha x}\cdot e^{-i\beta x}=e^{\alpha x}\left(\cos(\beta x)-i\sin(\beta x)\right).\]

Note that it is possible to take certain (complex) linear combinations such that \(e^{\alpha x}\cos(\beta x)\) and \(e^{\alpha x}\sin(\beta x)\) are solutions. These solutions are real and linearly independent. This implies that the general solution is \(y(x)=c_1e^{\alpha x}\cos(\beta x)+c_1e^{\alpha x}\sin(\beta x)\) in this case.

Résumé: if \(r_1\) and \(r_2\) are the two solutions of the characteristic equation, then

1. if \(r_1,r_2\in\mathbb{R}\) with \(r_1\neq r_2\), then: \(y(x)=c_1e^{r_1x}+c_2e^{r_2x}\) with \(c_1,c_2\in\mathbb{R}\),


2. if \(r_1,r_2\in\mathbb{R}\) with \(r_1=r_2=r\), then: \(y(x)=c_1e^{rx}+c_2xe^{rx}\) with \(c_1,c_2\in\mathbb{R}\),


3. if \(r_1,r_2\notin\mathbb{R}\), so \(r_{1,2}=\alpha\pm i\beta\) with \(\beta\neq0\), then: \(y(x)=c_1e^{\alpha x}\cos(\beta x)+c_2e^{\alpha x}\sin(\beta x)\) with \(c_1,c_2\in\mathbb{R}\).

Examples:

1. \(y''-2y'-8y=0\) has characteristic equation \(r^2-2r-8=0\;\Longleftrightarrow\;(r-4)(r+2)=0\),
   hence: \(y(x)=c_1e^{-2x}+c_2e^{4x}\) with \(c_1,c_2\in\mathbb{R}\),


2. \(y''+6y'+9y=0\) has characteristic equation \(r^2+6r+9=0\;\Longleftrightarrow\;(r+3)^2=0\),
   hence: \(y(x)=c_1e^{-3x}+c_2xe^{-3x}\) with \(c_1,c_2\in\mathbb{R}\),


3. \(y''-4y'+13y=0\) has characteristic equation \(r^2-4r+13=0\;\Longleftrightarrow\;(r-2)^2+9=0\),
   hence: \(y(x)=c_1e^{2x}\cos(3x)+c_2e^{2x}\sin(3x)\) with \(c_1,c_2\in\mathbb{R}\).

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Last modified on March 8, 2021