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In a differential equations class the professor stated that the general solution of a homogeneous second-order linear ODE would be in the form: $$y = c_1y_1 + c_2y_2$$

Where $y_1$ and $y_2$ were distinct solutions of the ODE: $$\frac{d^2y}{dx^2} + A\frac{dy}{dx} + By = 0$$

Where $A$ and $B$ are constant coefficients. I understand how to get the distinct solutions however what i don't get is why the general solution is in the form it is and what is the proof behind it?

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2 Answers 2

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Take a solution $y$, and for now assume that $y=c_1y_1+c_2y_2$. Consider $y(t_0)=y_0$ and $y'(t_0)=y_0'$. Thus we have $$y(t_0)=c_1y_1(t_0)+c_2y_2(t_0)=y_0$$ $$\text{and}$$ $$y'(t_0)=c_1y_1'(t_0)+c_2y_2'(t_o)=y_0'$$ We now must solve for $c_1$ and $c_2$. We have the linear system $$\left(\begin{array}{cc} y_1(t_0) & y_2(t_0) \\ y_1'(t_0) & y_2'(t_0) \end{array}\right) \left(\begin{array}{c} c_1 \\ c_2 \end{array}\right)= \left(\begin{array}{c} y_0 \\ y_0' \end{array}\right)\qquad (1)$$ Now since $y_1$ and $y_2$ are fundamental solutions, meaning their Wronskian $$W[y_1,y_2]=\left| \begin{array}{cc} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t) \end{array} \right|\neq 0$$ $\forall t$ in the interval upon which $y_1$ and $y_2$ are being taken as solutions we have the explicit solution to $(1)$ which is $$\left(\begin{array}{c} c_1 \\ c_2 \end{array}\right)=\frac{1}{W[y_1,y_2](t_0)}\left(\begin{array}{cc} y_2'(t_0) & -y_2(t_0) \\ -y_1'(t_0) & y_1(t_0) \end{array}\right)\left(\begin{array}{c} y_0 \\ y_0' \end{array}\right)$$ Hence there exists a $c_1$ and $c_2$ such that $y=c_1y_1+c_2y_2$ and by the Existence-Uniqueness Theorem this is the only possible $y$ such that $y(t_0)=y_0$ and $y'(t_0)=y_0'$, forcing $y$ to be of this form. QED

Above I gave the proof for why all solutions to the homogeneous linear ODE must to be of the form $y=c_1y_1+c_2y_2$. But if you only wish to convince yourself why $y=c_1y_1+c_2y_2$ is a solution then realize that since the operator $L[y]$, $$L[y]=y''+Ay'+By$$ is linear we have $$L[c_1y_1+c_2y_2]=L[c_1y_1]+L[c_2y_2]=c_1L[y_1]+c_2L[y_2]=0$$ because, since $y_1$ and $y_2$ are solutions to the DE, $L[y_1]=L[y_2]=0$ and since $L[c_1y_1+c_2y_2]=0$ it is also a solution to the DE.

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Another option is to explicitly construct the general solution to the differential equation $$y''+Ay'+By=0,$$ where $y=y(x)$. This can be done by transforming it to a first order linear differential equation, for which we know how to find the general solution.

If $B=0$ we can simply integrate both sides to get a first order differential equation (with general solution $y=c_1+c_2e^{-Ax}$), so we may henceforth assume that $B \neq 0$.

Now let $z=z(x)$. A change of variables $y=ze^{rx}$ (for any constant $r\neq0$) makes $y'=(z'+rz)e^{rx}$ and $y''=(z''+2rz'+r^2z)e^{rx}$. Insertion into the original equation yields $$ y''+Ay'+By = (z''+2rz'+r^2z)e^{rx}+A(z'+rz)e^{rx}+Bze^{rx} = e^{rx}(z''+(2r+A)z'+(r^2+Ar+B)z)=0 \Longleftrightarrow \\ z''+(2r+A)z'+(r^2+Ar+B)z=0. $$ Now we choose $r$ such that $r^2+Ar+B=0$. Here we must consider two cases:

  1. The equation $r^2+Ar+B=0$ has a double root $r$. In this case $z''=0 \Rightarrow z=c_1x+c_2$, so $y=ze^{rx}=(c_1x+c_2)e^{rx}$.
  2. The equation $r^2+Ar+B=0$ has distinct roots $r_1$ and $r_2$ (assume we use $r=r_1$). In this case $$ z''+(2r_1+A)z' = z''+(r_1-r_2)z'=0 \Longrightarrow z'+(r_1-r_2)z=M, $$ which has the general solution $z=\frac{M}{r_1-r2}+c_2e^{(r_2-r_1)x}=c_1+c_2e^{(r_2-r_1)x}$. This means that $y=c_1e^{r_1x}+c_2e^{r_2x}$.
We have now exhausted the possibilities, and in each case the general solution is of the form $y=c_1y_1+c_2y_2$, as desired.
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  • $\begingroup$ Are $y$ and $z$ functions of $x$? $\endgroup$ Commented Jul 6, 2021 at 10:15
  • $\begingroup$ @GiorgioPastasciutta: Yes, all functions in this case depend on $x$. $\endgroup$
    – Mårten W
    Commented Jul 6, 2021 at 10:35

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