Solving Systems of Linear Differential Equations by Elimination

Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

Solving linear differential equations may seem tough, but there's a tried and tested way to do it! We'll explore solving such equations and how this relates to the technique of elimination from algebra. We will also solve a population growth system problem using what we've learned.

Yaks, Krills and Differential Equations

A differential equation is an equation with derivatives. A differential equation is linear if there are no products of dependent variables and if all the derivatives and dependent variables are raised to the first power. When there are two or more equations, we have a system. This system is coupled when different dependent variables are present in the same equation.

Now, with the basic terms out of the way, let's get to solving! Suppose we have two dependent variables, x and y, representing the population of krills and yaks. Imagine the rate of change of yaks, dx/dt, depending negatively on the number of krills. Similarly, in this hypothetical world, the rate of change of krills, dy/dt, depends negatively on the number of yaks. Thus our hypothetical coupled system of linear differential equations is:


Two unknowns and two equations, suggests the elimination method from algebra. As we'll see, writing dx/dt as Dx looks like D is multiplying x. The D, however, is an operator on x where the operation is differentiation. Actually, even multiplication, like the 4 times x in 4x, is a 4 operating on the x.

On to the krill-yak system!

Step 1: Use the D notation for the derivative.

Replace dx/dt with Dx and dy/dt with Dy.


Step 2: Organize the equations.

Putting x first, we get:


Step 3: Solve by elimination.

Multiplying equation (2) with D, we get:


Multiplying equation (1) with 4, we have:


We now eliminate 4Dx from the equation:


By subtracting one equation from the other, we have eliminated 4Dx.

Step 4: Solve the differential equation.

How do we solve this equation?

  1. First, let y = eat.
  2. Using the differentiation operator D on y, we have y = aeat
  3. This gives us D2y = a2 eat.

Substituting into -36y + D2y = 0:

-36eat + a2 eat = 0.

Dividing through by eat:

-36 + a2 = 0.

Solving for a:

a = ±6.


  • y = c1 e6t + c2 e-6t.

Step 5: Using elimination, solve for the other variable(s).

This is a repeat of Step 3, but y is eliminated.

Multiply equation (1) with D:


Multiply equation (2) with 9:


Eliminating 9Dy gives -36x + D2x = 0.

The form of this equation in x is the same as we had for y. Thus, the solutions for x and y are the same except for the subscripts on the constants:

  • x = c3 e6t + c4 e-6t.

Step 6: Using initial conditions, solve for the constants.

Initial conditions are the variable and it's first derivative values at time t = 0.

Imagine at the start, y = 0 and dy/dt = 12.

Substituting t = 0 in


we get


Since, e0 = 1,

y = c1 (1) + c2 (1).

At time t = 0, y = 0. Thus,

c1 + c2 = 0.

Now, for the initial condition on dy/dt.


At t = 0, this becomes:


At time t = 0, dy/dt = 12;

6c1 - 6c2 = 12.

Dividing through by 6:

c1 - c2 = 2.

Great! Two equation and two unknowns:

c1 + c2 = 0

c1 - c2 = 2

Solving for c1 and c2:

c1 = 1 and c2 = -1. Thus,


The solution for y, is used to find c3 and c4 in the x solution.

Differentiate the solution for x and substitute into equation (1): Dx = -9y:


Equating the e6t terms:

6c3 = -9 or c3 = -3/2.

Equating the e-6t terms:

-6c4 = 9 or c4 = -3/2.



Step 7: Check the solution.

If the expressions for x and y are valid, substituting back into the original system of equations makes these equations true.

We are going to check if the original hypothetical coupled system is satisifed by these values of x and y.






LHS (left-hand side):


simplifies to

-9e6t + 9e-6t.

RHS (right-hand side): substitute into -9y to get

-9e6t + 9e-6t.


For the second equation,

LHS: dy/dt = 6e6t + 6e-6t.

RHS: substitute into -4x to get


which simplies to

6e6t + 6e-6t.


Step 8: Plot and comment on the solution.

Plot of the solutions

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