Differential Equations and Checking Solutions Using A Columnar Form
by
Jerry Tobin
Basic Forms for Solutions of Homogeneous Differential Equations:
y  Ae kt or y  Ae kx
sinusoidal: y  A sin(pt)  Bcos(pt) or y  A sin(px)  Bcos(px)
Use y = Aekt for equations with derivatives with respect to t and y = Aekx for simple homogeneous
equations with derivatives with respect to x such as y // + 3y / +2y = 0
From, we get y / = kAekx and y // = k2 Aekx
Substituting into y // + 3y / +2y = 0 gives (k2 + 3k + 2) Aekx = 0 which in turn gives
(k2 + 3k + 2) = 0 which is referred to as the characteristic equation
Now, we have (k+1)(k+2)= 0 for k = -1, -2, and the solution is y = Ae-x + Be-2x
To check the solution, use the columnar form as below with the original equation shown vertically
instead of horizontally. First, take the needed derivatives.
y / = -1Ae-x - 2Be-2x and y // = 1Ae-x + 4Be-2x , so we can write
y // = +1Ae-x + 4Be-2x
3y / = - 3Ae-x - 6Be-2x
2 y = 2Ae-x + 2Be-2x
0 = 0
+ 0 -- > The solution checks!
For a homogeneous equation with only even derivatives and plus signs, the sinusoidal form applies.
Consider y // +4y = 0 and the basic form y = Aekx .
This gives y / = kAekx and y // = k2 Aekx , so we now have k2 Aekx + 4Aekx = 0 for which the
characteristic equation is k2 +4 =0 or k2 = -4 with the solution form k = +2i. This give y = A1ei2x and
y = A2e-i2x. By using Euler’s formula, this can be converted to the form y = A sin(2x) + Bcos(2x).
To check using columnar form, consider that y / = 2Acos(2x) - 2Bsin(2x) and
y // = -4Asin(2x) – 4Bcos(2x). So, using the columnar form,
y // = -4Asin(2x) – 4Bcos(2x)
+4y = +4A sin(2x) + 4Bcos(2x)
0 =
0
+ 0
-- > The solution checks!
For a non-homogeneous as above like y // + 3y / +2y = 5e-x , if the expression(s) on the right side of
the equation contain the same form as the form(s) in the solution to the homogeneous equation, then
you can increase the power of x for the solution form as many powers as is required to match the
form to the right of the equal sign. (Hint: Solve the homogeneous equation first.)
For this problem, since the solution to the homogeneous equation is known to be y = Ae-x + Be-2x ,
and the new equation contains a multiple of e-x , we realize that e-x is also x0e-x , and that the next
power of x is x1. So the new solution form to try is yp = Cxe-x .
We know that the terms for the homogeneous solution all yield zero, so we only need to test the new
form to verify the answer.
So, given that yp = Cxe-x , So, using the product rule, y/p = Ce-x - Cxe-x, and y//p = -2Ce-x + Cxe-x.
Now, we can write
y//p = -2Ce-x +1 Cxe-x
+3y/p = +3Ce-x - 3Cxe-x
+2y =
+2Cxe-x
5e-x = Ce-x
-- >
C = 5 for this example.
So, the solution to y // + 3y / +2y = 5e-x is y = Ae-x + Be-2x + 5xe-x .
For a form such as y // + 4 y = 5 sin(2x), you need to use the product of x1 and both forms [sin(ax)
and cos(ax)].
Consider that we know the solution for y // +4y = 0 to be y = Asin(2x) + Bcos(2x). Since the form to
the right of the equality sign contains the same form as one of these, we also need sinusoidal form,
so we use yp = Cxsin(2x) + Dxcos(2x)
This gives yp / = Csin(2x) + 2Cxcos(2x) + Dcos(2x) – 2Dxsin(2x) and
yp // = 4Ccos(2x) – 4Csin(2x) – 4Dsin(2x) – 4Dxcos(2x)
So, using the columnar form,
yp // = +4Ccos(2x) – 4Cxsin(2x) – 4Dsin(2x) – 4Dxcos(2x)
+4yp =
+4Cxsin(2x) +
+ 4Dxcos(2x)
5 sin(2x) = +4Ccos(2x)
+ 0
– 4Dsin(2x) +
0
For this to be true, we need 5 = – 4D or D =
-5
5
and C = 0. So, for yp = 0xsin(2x) - xcos(2x), the
4
4
solution checks!
This gives the solution for y // + 4 y = 5 sin(2x) to be y = A sin(2x) + Bcos(2x) - (5/4)xcos(2x)
Note that the last term contains cos(2x) instead of sin(2x), so we had to use both forms, Cxsin(2x)
and Dxcos(2x) to ensure that the entire solution was identified.
For a non-homogeneous as above like y // + 3y / +2y = 5ex, if the expression(s) on the right side of the
equation do not contain the same form as the form(s) in the solution to the homogeneous equation,
then you can assume that the solution form for the homogeneous equation will not be included in the
particular solution. (Hint: Be sure to solve the homogeneous equation first.)
For this problem, since the solution to the homogeneous equation is known to be y = Ae-x + Be-2x ,
and the new equation contains a multiple of ex , we seek a solution that contains ex for the particular
solution. So, the new solution form to try is yp = Cex .
So, given that yp = Cex , y/p = Cex, and y//p = Cex.
Now, we can write
y//p =
3y/p =
2 yp =
5 ex =
1 Cex
3 Cex
2 Cex
6 Cex
For this to be true, we need 5 = 6C, so C =
5
5
and yp = ex
6
6
This gives a general solution of y = yh + yp = Ae-x + Be-2x +
with no repeating roots, y // + 3y / +2y = 5ex .
5 x
e for the non-homogeneous equation
6