Coupled First-Order Simultaneous Differential Equations
Reducing a pair of linked first-order equations to a single second-order equation by elimination.
Coupled systems
Many real models involve two related quantities $x$ and $y$, each changing at a rate depending on both of them — for example, predator and prey populations, or two chemicals reacting and transferring into one another:
$$\frac{dx}{dt}=ax+by$$
$$\frac{dy}{dt}=cx+dy$$
where $a,b,c,d$ are constants. These are called coupled because each equation involves both variables.
Differentiate the first equation with respect to $t$, then use both original equations to eliminate $y$ and $\dfrac{dy}{dt}$, leaving a single second-order equation in $x$ alone. Solve this using the auxiliary-equation method, then substitute back into one of the original equations to find $y$.
Eliminating y: differentiate the first equation: $\dfrac{d^2x}{dt^2}=3\dfrac{dx}{dt}+2\dfrac{dy}{dt}$.
Substitute $\dfrac{dy}{dt}=2x-2y$: $\dfrac{d^2x}{dt^2}=3\dfrac{dx}{dt}+2(2x-2y)=3\dfrac{dx}{dt}+4x-4y$.
From the first equation, $2y=\dfrac{dx}{dt}-3x$, so $4y=2\dfrac{dx}{dt}-6x$. Substitute:
$$\frac{d^2x}{dt^2}=3\frac{dx}{dt}+4x-\left(2\frac{dx}{dt}-6x\right)=\frac{dx}{dt}+10x$$
Rearranging: $\dfrac{d^2x}{dt^2}-\dfrac{dx}{dt}-10x=0$, as required.
Solving for x: the auxiliary equation $m^2-m-10=0$ gives $m=\dfrac{1\pm\sqrt{41}}2$ by the quadratic formula. Coupled systems don't always produce tidy roots — here $x=Ae^{m_1t}+Be^{m_2t}$ with $m_1,m_2=\dfrac{1\pm\sqrt{41}}2$, an entirely valid (if untidy) general solution.
Finding y: from $2y=\dfrac{dx}{dt}-3x$, differentiating $x$ and substituting expresses $y$ in terms of $A$, $B$ and $t$ in the same way. The cleaner example below shows this final step worked through in full with nicer numbers.
Exam questions are deliberately constructed so the resulting auxiliary equation has nice roots. Example 1 above was chosen to illustrate the elimination method transparently; the checkpoint questions below use coefficients that do factorise cleanly, as you should expect in practice.
Differentiate the first equation: $\dfrac{d^2x}{dt^2}=\dfrac{dx}{dt}+3\dfrac{dy}{dt}=\dfrac{dx}{dt}+3(3x+y)=\dfrac{dx}{dt}+9x+3y$.
From the first equation, $3y=\dfrac{dx}{dt}-x$. Substitute: $\dfrac{d^2x}{dt^2}=\dfrac{dx}{dt}+9x+\left(\dfrac{dx}{dt}-x\right)=2\dfrac{dx}{dt}+8x$.
Rearranged: $\dfrac{d^2x}{dt^2}-2\dfrac{dx}{dt}-8x=0$, as required.
Auxiliary equation: $m^2-2m-8=0\Rightarrow(m-4)(m+2)=0\Rightarrow m=4,-2$.
$x=Ae^{4t}+Be^{-2t}$, so $\dfrac{dx}{dt}=4Ae^{4t}-2Be^{-2t}$.
$y=\dfrac13\left(\dfrac{dx}{dt}-x\right)=\dfrac13\left(3Ae^{4t}-3Be^{-2t}\right)=Ae^{4t}-Be^{-2t}$.
At $t=0$: $x=A+B=2$ and $y=A-B=0\Rightarrow A=B=1$.
So $x=e^{4t}+e^{-2t}$ and $y=e^{4t}-e^{-2t}$.
Exam-style questions
Question 1 10 marks
Given that $\dfrac{dx}{dt}=4x+2y$ and $\dfrac{dy}{dt}=3x-y$, show that $\dfrac{d^2x}{dt^2}-3\dfrac{dx}{dt}-10x=0$, and find general solutions for $x$ and $y$ in terms of $t$.
$\dfrac{d^2x}{dt^2}=4\dfrac{dx}{dt}+2\dfrac{dy}{dt}=4\dfrac{dx}{dt}+2(3x-y)=4\dfrac{dx}{dt}+6x-2y$. (M1, A1)
From the first equation, $2y=\dfrac{dx}{dt}-4x$. Substitute: $\dfrac{d^2x}{dt^2}=4\dfrac{dx}{dt}+6x-\left(\dfrac{dx}{dt}-4x\right)=3\dfrac{dx}{dt}+10x$. (M1, A1)
Rearranged: $\dfrac{d^2x}{dt^2}-3\dfrac{dx}{dt}-10x=0$, as required. (A1)
Auxiliary equation: $m^2-3m-10=0\Rightarrow(m-5)(m+2)=0\Rightarrow m=5,-2$; so $x=Ae^{5t}+Be^{-2t}$. (M1, A1)
$y=\dfrac12\left(\dfrac{dx}{dt}-4x\right)=\dfrac12\left(Ae^{5t}-6Be^{-2t}\right)=\dfrac12Ae^{5t}-3Be^{-2t}$. (A1, A1)
Question 2 4 marks
Explain why a coupled first-order system $\dfrac{dx}{dt}=ax+by$, $\dfrac{dy}{dt}=cx+dy$ can be reduced to a single second-order equation in $x$ alone.
Differentiating the $x$-equation introduces $\dfrac{dy}{dt}$, which the $y$-equation expresses in terms of $x$ and $y$; and $y$ itself can be expressed in terms of $x$ and $\dfrac{dx}{dt}$ using the original $x$-equation. (B1, B1)
Substituting both eliminates $y$ and $\dfrac{dy}{dt}$ entirely, leaving an equation relating $\dfrac{d^2x}{dt^2}$, $\dfrac{dx}{dt}$ and $x$ only — a standard second-order linear equation that can be solved with the auxiliary-equation method. (B1, B1)
Question 3 9 marks
Given that $\dfrac{dx}{dt}=2x-3y$, $\dfrac{dy}{dt}=x-2y$, show that $\dfrac{d^2x}{dt^2}-1\cdot x=0$ is incorrect and instead find the correct second-order equation satisfied by $x$, then solve it given $x=1,y=1$ when $t=0$.
$\dfrac{d^2x}{dt^2}=2\dfrac{dx}{dt}-3\dfrac{dy}{dt}=2\dfrac{dx}{dt}-3(x-2y)=2\dfrac{dx}{dt}-3x+6y$. (M1, A1)
From the first equation, $3y=2x-\dfrac{dx}{dt}$, so $6y=4x-2\dfrac{dx}{dt}$. Substitute: $\dfrac{d^2x}{dt^2}=2\dfrac{dx}{dt}-3x+4x-2\dfrac{dx}{dt}=x$. (M1, A1)
So the correct equation is $\dfrac{d^2x}{dt^2}-x=0$. Auxiliary equation $m^2-1=0\Rightarrow m=\pm1$, so $x=Ae^t+Be^{-t}$. (M1, A1)
$y=\dfrac13\left(2x-\dfrac{dx}{dt}\right)=\dfrac13\left(2Ae^t+2Be^{-t}-Ae^t+Be^{-t}\right)=\dfrac13Ae^t+Be^{-t}$. (A1)
At $t=0$: $x=A+B=1$, $y=\dfrac13A+B=1$. Subtracting: $\dfrac23A=0\Rightarrow A=0,B=1$. So $x=e^{-t}$, $y=e^{-t}$. (A1, A1)