# How to solve coupled second order differential equations

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?

Thanks.

• 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. Jan 6 at 10:03

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}, $$\boxed{z(x) = -\frac{2c_1e^x}{3}.}$$

• 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. Jan 6 at 4:03
• @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. Jan 6 at 9:27
• @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. 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

$$y=\exp\left(\dfrac{x}{3}-Q(x)\right),$$

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 $$q(x)=Q'(x)=\dfrac{-2}{3}$$.