I have the following coupled differential equations:

$$ 2y''- 3y' + 2z' + 3y + z = e^{2x}$$ $$y''- 3y' + z' + 2y - z = 0 $$

I'm not sure how to solve them as if I try $y = Ae^{\lambda x} $ and $z = Be^{\lambda x}$, I only get one value for $\lambda$ ($\lambda = 1$) so am not sure how to form a complementary function.

Any suggestions?


  • $\begingroup$ One technique to be aware of is to introduce a function $u = y’$, and rewrite your system as a first order system of three equations involving the functions $y,z$, and $u$. Then solve using standard techniques for a first order linear system. $\endgroup$
    – littleO
    Jan 6 at 10:03

4 Answers 4


It is visible after some contemplation that the highest derivatives of $z$ and $y$ occur in the same linear combination in both equations, marking the system as a DAE system of index at least one.

To better handle the structure, define $u=y'+z$ for this combination and eliminate $z$ from the system. \begin{align} 2u' - 4y' + 3y + u &= e^{2x}\\ u'- 2y' + 2y - u &= 0 \end{align} Again the terms in the highest derivatives $y',u'$ occur in the same linear combination. So set $v=u-2y=y'+z-2y$ and eliminate $u$ against $v$ to get an explicit index-1 DAE system. \begin{align} 2v' + 5y + v &= e^{2x}\\ v' - v &= 0 \end{align} This now can be isolated for the highest order derivatives (or non-derivatives) $(y,v')$ to get \begin{align} 5y &= e^{2x}-3v\\ v' &= v \end{align} This system, which resulted just from easily reversible variable substitutions, now indeed has only one integration constant.

So, inserting backwards, $$ v=Ce^x,\\ y=\frac15(e^{2x}-3Ce^x),\\ u=v+2y=\frac15(2e^{2x}-Ce^x)\\ z=u-y'=\frac15(2e^{2x}-Ce^x)-\frac15(2e^{2x}-3Ce^x)=\frac25Ce^x $$


Define $\mathbb{Y}(s)=\mathcal{L}(y(t))$ and $\mathbb{Z}(s)=\mathcal{L}(z(t))$ where $\mathcal{L}$ denotes the laplace transform.

Put $a=y(0),b=y'(0),c=z(0)$.

With a little algebra your DE becomes $$(2s^2-3s+3)\mathbb{Y}(s)+(2s+1)\mathbb{Z}(s)=-as+\frac{1}{s-2}+2b+2c$$ $$(s^2-3s+2)\mathbb{Y}(s)+(s-1)\mathbb{Z}(s)=as-3a+b+c$$ This can be expressed as a linear system $$\begin{pmatrix}2s^2-3s+3&2s+1\\ s^2-3s+2&s-1\end{pmatrix}\begin{pmatrix}\mathbb{Y}(s)\\ \mathbb{Z}(s)\end{pmatrix}=\begin{pmatrix}-as+\frac{1}{s-2}+2b+2c\\ as-3a+b+c\end{pmatrix}$$ Can you finish? Keep in mind, once you solve this system via the inverse laplace transform $\mathcal{L}^{-1},$ the constants $a,b,c,d$ will be your arbitrary constants in your general solution.


We need to relate $y(x)$ with $z(x)$ or viceversa. Substracting two times the second equation to the first one, we have $$ \begin{array}{r} & 2y'' & - & 3y' & + & 2z' & + & 3y & + & z & = & e^{2x} \\ - & 2y'' & - & 6y' & + & 2z' & + & 4y & - & 2z & = & 0 \\ \hline & & & 3y' & & & - & y & + & 3z & = & e^{2x} \end{array} $$

Solving for $z$, $$ z = -y' + \frac{y}{3} + \frac{e^{2x}}{3} \Rightarrow z' = -y'' + \frac{y'}{3} + \frac{2e^{2x}}{3}.\tag{1}\label{eq:z} $$

Substituting back into your second equation $$ y'' - 3y' + \left(\frac{y'}{3} + \frac{2e^{2x}}{3} -y''\right) + 2y - \left(\frac{y}{3} + \frac{e^{2x}}{3} -y'\right) = 0 \Rightarrow 5y'-5y=e^{2x}.\tag{2}\label{eq:y} $$

Considering the homogeneous case of \eqref{eq:y}, $$ y'_h - y_h = 0 \Rightarrow y_h(x) = c_1e^x. $$ It is easy to see that a particular solution of \eqref{eq:y} will be $y_p(x) = e^{2x}/5$ (just consider $y_p(x) = c_2e^{2x}$ and solve for $c_2$). Therefore, $$ \boxed{y(x) = y_h(x) + y_p(x) = c_1e^x + \frac{e^{2x}}{5}.} $$

Using this result in \eqref{eq:z}, $$ \boxed{z(x) = -\frac{2c_1e^x}{3}.} $$

  • $\begingroup$ Your approach is definitely more slick, but there's no way that this is the general solution. The general solution will certainly contain $3$ arbitrary constants. $\endgroup$
    – Matthew H.
    Jan 6 at 4:03
  • 2
    $\begingroup$ @MatthewH. Not if the implicit DE system turns out to be a DAE system. This is the case here as in the first difference no equation for $z'$ results, it is not possible to find an equation for $z'$ that does not contain $y''$. Or in other words, an explicit system that has $(y'',z')$ on the left side and only lower order derivatives on the right. $\endgroup$ Jan 6 at 9:27
  • $\begingroup$ @LutsLehmann I did not know this! I will have to follow through with my Laplace transform method to see how the coefficients collect nicely into one term. Thank you for the enlightenment. $\endgroup$
    – Matthew H.
    Jan 6 at 15:59

Here is a pedestrian proof that Sebastián V. Romero's argument gives the general solution to the system of equations.

Heuristically, the first equation in $(1)$ of Sebastián V. Romero gives that the solution space is not more than two dimensional and the second equation in $(2)$ gives that the solution space is not more than one dimensional.

Here is a humble graph: https://www.desmos.com/calculator/1smfiqzw8s . Geometrically the solution space of the nonhomogeneous system is isomorphic (as an affine space) to the line $z=-\frac{2}{3}y+\frac{2}{15}$ in the $yz$-plane.

First observe that by linearity if $(y_1, z_1)$ and $(y_2,z_2)$ both solve the nonhomogeneous system, then $(y_1-y_2,z_1-z_2)$ solves the homogeneous system, so we focus on the homogeneous system

\begin{align*}\tag{$\ast$} 2y''- 3y' + 2z' + 3y + z = 0\\ y''- 3y' + z' + 2y - z = 0 \end{align*}

Claim: Let $(y,z)$ be a solution of $(\ast)$ (with $y$ nonvanishing). We claim that the function $\dfrac{z}{y}$ is identically $-\dfrac{2}{3}$.

Proof: Put $q=\dfrac{z}{y}$, so that $z=qy$. Plugging $z$ into the equations and some standard algebraic manipulations give $3y'+(3q-1)y=0$, whence


where $Q$ is some antiderivative of $q$. Thus we have

\begin{align*} y'&=y\left(\dfrac{1}{3}-q\right),\\ y''&=y \left(q^2-\dfrac{2}{3}q+\dfrac{1}{9}-q'\right),\\ z&= yq,\\ z' &= y\left(q'-q^2+\dfrac{1}{3}q\right). \end{align*}

Plugging these back into one of the equations and some standard algebraic manipulations give $y\left(q+\dfrac{2}{3}\right)=0$, that is,

$$0=\exp\left(\dfrac{x}{3}-Q\right)\left(Q'+\dfrac{2}{3}\right) = -\exp\left(\dfrac{x}{3}-Q\right)\left(\dfrac{1}{3}-Q'\right) + \exp\left(\dfrac{x}{3}-Q\right) = -y' +y,$$

so that $\exp\left(\dfrac{x}{3}-Q(x)\right)=y(x)=\exp(x+C)$ for some constant $C$, whence $y(x)\propto e^x$ (and so this argument turns out to stand on its own) and simultaneously $Q(x)=\dfrac{-2}{3}x-C$, so that $q(x)=Q'(x)=\dfrac{-2}{3}$.


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