# Eliminating the Arbitrary Constants of the Equation $y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x}$

I was eliminating the arbitrary constants of the equation $y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x}$.

My work:

I see three arbitrary constants, so I need to differentiate the equation $y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x}$ thrice. Now differentiating, I get....

$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} \space \space \space -----> (1)$$ $$y' = c_1 e^x + 2c_2 e^{2x} + 3c_3 3e^{3x} \space \space \space -----> (2)$$ $$y'' = c_1 e^x + 4c_2 e^{2x} + 9c_3 3e^{3x} \space \space \space -----> (3)$$ $$y''' = c_1 e^x + 8c_2 e^{2x} + 27c_3 3e^{3x} \space \space \space -----> (4)$$

I'm eliminating the $c_1$. Looking at equation's $(1)$ and $(2)$:

$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} \space \space \space -----> (1)$$ $$y' = c_1 e^x + 2c_2 e^{2x} + 3c_3 3e^{3x} \space \space \space -----> (2)$$

Then I multiply equation $(2)$ by $-1$ and adding it to equation $(1)$, we get:

$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} \space \space \space -----> (1)$$ $$+ (-1)\left(y' = c_1 e^x + 2c_2 e^{2x} + 3c_3 3e^{3x} \right) \space \space \space -----> (2)$$ $--------------------------------------$ $$y -y' = -c_2e^{2x} - 2c_3e^{3x} -----------> (A)$$

I'm eliminating another $c_1$. Looking at equation's $(1)$ and $(3)$:

$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} \space \space \space -----> (1)$$ $$y'' = c_1 e^x + 4c_2 e^{2x} + 9c_3 3e^{3x} \space \space \space -----> (3)$$

Then I multiply equation $(3)$ by $-1$ and adding it to equation $(1)$, we get:

$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} \space \space \space -----> (1)$$ $$+(-1)\left(y'' = c_1 e^x + 4c_2 e^{2x} + 9c_3 3e^{3x} \right) \space \space \space -----> (3)$$ $--------------------------------------$ $$y -y'' = -3c_2e^{2x} - 8c_3e^{3x} ------------> (B)$$

I'm eliminating the last $c_1$. Looking at equation's $(1)$ and $(4)$:

$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} \space \space \space -----> (1)$$ $$y''' = c_1 e^x + 8c_2 e^{2x} + 27c_3 3e^{3x} \space \space \space -----> (4)$$

Then I multiply equation $(4)$ by $-1$ and adding it to equation $(1)$, we get:

$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} \space \space \space -----> (1)$$ $$+(-1)\left(y''' = c_1 e^x + 8c_2 e^{2x} + 27c_3 3e^{3x} \right) \space \space \space -----> (4)$$ $--------------------------------------$ $$y -y''' = -7c_2e^{2x} - 26c_3e^{3x} ----------> (C)$$

After all those process, I got to eliminate $c_2$. Looking at equations $(A)$ and $(B)$:

$$y -y' = -c_2e^{2x} - 2c_3e^{3x} ----------> (A)$$ $$y -y'' = -3c_2e^{2x} - 8c_3e^{3x} -----------> (B)$$

Then I multiply equation $(A)$ by $-3$ and adding it to equation $(B)$, we get:

$$(-3)\left(y -y' = -c_2e^{2x} - 2c_3e^{3x} \right)------------> (-3A)$$ $$+(y -y'' = -3c_2e^{2x} - 8c_3e^{3x}) -----------> (B)$$

$--------------------------------------$ $$-2y+3y'-y'' = -2c_3e^{3x} ------------> (\alpha)$$

I got to eliminate another $c_2$ finally. Looking at equations $(A)$ and $(C)$:

$$y -y' = -c_2e^{2x} - 2c_3e^{3x} -----------> (A)$$ $$y -y''' = -7c_2e^{2x} - 26c_3e^{3x} ------------> (C)$$

Then I multiply equation $(A)$ by $-7$ and adding it to equation $(C)$, we get:

$$(-7)\left(y -y' = -c_2e^{2x} - 2c_3e^{3x} \right)-----------> (-7A)$$ $$+(y -y''' = -7c_2e^{2x} - 26c_3e^{3x}) -----------> (C)$$ $--------------------------------------$ $$-6y+7y'-y''' = -12c_3e^{3x} ------------> (\beta)$$

Now getting their $c_3$'s in $(\alpha)$ and $(\beta)$, we get:

for $(\alpha)$ : $$c_3 = \frac{-2y+3y'-y''}{-2e^{3x}}$$

for $(\beta)$ : $$c_3 = \frac{-6y+7y'-y'''}{-12e^{3x}}$$

Then equate their $c_3$'s :

$$\frac{-2y+3y'-y''}{-2e^{3x}} = \frac{-6y+7y'-y'''}{-12e^{3x}}$$

Then....

$$\frac{-2y+3y'-y''}{-2e^{3x}} = \frac{-6y+7y'-y'''}{-12e^{3x}}$$ $$\left( \frac{-2y+3y'-y''}{-2e^{3x}} = \frac{-6y+7y'-y'''}{-12e^{3x}} \right) (12e^{3x})$$ $$(-6)(-2y+3y'-y''= -6y+7y'-y''')$$ $$12y - 18y' + 6y'' = 36y -42y' + 42y'''$$ $$42y''' -6y'' - 24y' + 24y = 0$$

Ultimately, when I eliminate arbitrary constants of the equation $y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x}$, I got the new equation $$7y''' -y'' -4y' + 4y = 0$$

I thought I got the answer correctly because when I solved it, it was smooth.....but I couldn't verify the true answer.

My question is: Is my answer ($7y''' -y'' -4y' + 4y = 0$) correct?

• You might want to check that no function $$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x}$$ solves $$7y''' -y'' -4y' + 4y = 0$$ except for $$c_1=c_2=c_3=0$$
– Did
Sep 12, 2017 at 9:34

You can work this out by Gaussian elimination

$$\begin{matrix}y&1&1&1\\y'&1&2&3\\y''&1&4&9\\y'''&1&8&27\\\end{matrix}$$

$$\begin{matrix}y&1&1&1\\y'-y&0&1&2\\y''-y&0&3&8\\y'''-y&0&7&26\\\end{matrix}$$

$$\begin{matrix}y&1&1&1\\y'-y&0&1&2\\y''-3y'+2y'&0&0&2\\y'''-3y''+2y'&0&0&12\\\end{matrix}$$

$$\begin{matrix}y&1&1&1\\y'-y&0&1&2\\y''-3y'+2y&0&0&2\\y'''-6y''+11y'-6y&0&0&0\\\end{matrix}$$

But it is much simpler to consider that the operator $D-a$ eliminates the exponential $e^{ax}$, so that

$$(D-1)(D-2)(D-3)y=(D^3-6D+11D-6)y=0.$$

Indeed, $$(D-1)y=c_1e^x-c_1e^x+2c_2e^{2x}-c_2e^{2x}+3c_3e^{3x}-c_3e^{3x}=c'_2e^{2x}+c'_3e^{3x}$$ $$(D-2)(D-1)y=2c'_2e^{2x}-2c'_2e^{2x}+3c_3e^{3x}-2c_3e^{3x}=c''_3e^{3x}$$

$$(D-3)(D-2)(D-1)y=3c''_2e^{3x}-3c''_2e^{3x}=0.$$

• For those studying these things, just to note that the last table mirrors the elimination method I used in my answer - everything here is related. Noticing the pattern enables us to prove it - and having shown it works, it radically simplifies the arithmetic we have to do, and makes the solution much easier. It also gives better insight into what is going on. Sep 12, 2017 at 9:45
• @MarkBennet: I was addressing the OP's approach, who painfully and error-pronely performed elimination "by hand", while a systematic algorithm is available.
– user65203
Sep 12, 2017 at 9:51
• @YvesDaoust What kind of differential equation did I had? and what topic in Differential Equations could I see these kinds of differential equation? Sep 12, 2017 at 10:01
• @PalautotKa This is an homogenous linear ODE with constant coefficients. They are the simplest and explained in any introductory course.
– user65203
Sep 12, 2017 at 10:37

If you want to set up the differential equation you could simply note that your eigenvalues are $\lambda = 1,2,3$. Hence the characteristic polynomials is $(\lambda-1)(\lambda-2)(\lambda-3)=\lambda^3-6\lambda^2+11\lambda-6$

Hence, the differential equation should be $y'''-6y''+11y'-6y=0$.

EDIT: When solving a homogeneous linear differential equation with constant coefficients $a_ny^{(n)}+...+a_1y^{(1)}+a_0=0$ we always use Euler's exponential ansatz $y=e^{\lambda x}$. Using this ansatz and we obtain $P(\lambda)=a_n\lambda^n+...+a_1\lambda+a_0=(\lambda-\lambda_1)\cdot \ldots \cdot (\lambda-\lambda_n)=0$ as the characteristic polymonial. If we know all the eigenvalues $\lambda_1,\lambda_2,...,\lambda_n$ (I will assume that we have no multiple eigenvalues) then we can express the general solution to our ODE as:

$$y(x)=c_1e^{\lambda_1 x}+\cdots+c_ne^{\lambda_n x}.$$

Comparision of this expression and your general solution will give you the form of the characteristic polynomial and consequently for the ODE.

If your general solution looks like this

$$y(x)=P_1(x)e^{\lambda_1 x}+\cdots+P_k(x)e^{\lambda_k x}$$

in which $P_i(x)$ are polynomials in $x$ with degree $d_i$, then you can infer from the degree that the multiplicity of the eigenvalue $\lambda_i$ is $d_i$. Hence, the characteristic polynomial is

$$P(\lambda)=(\lambda-\lambda_1)^{d_1}\cdot \ldots \cdot (\lambda-\lambda_k)^{d_k}.$$

And then derive the ODE from this after expanding the polynomial.

• Never heard of that tactic.....I never thought there are eigenvalues involved there.....Please explain further of your solution...Don't worry...I'm familiar with eigenvalues..... Sep 12, 2017 at 9:19

Your answer is not correct. The function $y(x)=e^x$ does not stisfy the differential equation $7y''' -y'' -4y' + 4y = 0$.

A short approach: write the polynomial $(x-1)(x-2)(x-3)$ in the form $a_0+a_1x+a_2x^2+a_3x^3$.

Then

$$a_3y'''+a_2y''+a_1y'+a_0y=0$$

is a differential equation you are looking for.

• I don't understand.......Why does a polynomial involved in the differential equation there? How do you convert $y = c_1e^x + c_2e^{2x}+ c_3e^{3x}$ to a polynomial form? I never heard of that tactic before.... Sep 12, 2017 at 9:22

You have a very long-winded way of doing this and another solution has a nifty short cut. But even longhand (and choosing names for constants to avoid unnecessary subscripts), consider $$y=ae^x+be^{2x}+ce^{3x}$$ then $$y'=ae^x+2be^{2x}+3ce^{3x}$$Now do your first elimination $$y'-y=be^{2x}+2ce^{3x}$$ and $$y''-y'=2be^{2x}+6ce^{3x}$$Second elimination $$(y''-y')-2(y'-y)=y''-3y'+2y=2ce^{3x}$$ and $$y'''-3y''+2y'=6ce^{3x}$$ so that $$y'''-3y''+2y'-3(y''-3y'+2y)=y'''-6y''+11y'-6y=0$$

If you look carefully at how this works you will see how the components of the answer which uses the characteristic equation $(\lambda-1)(\lambda-2)(\lambda-3)=0$ come together in the calculation. If you do the eliminations in a different order, you will change the order in which the factors arise on the left-hand side.

Note that the numbers $1,2,3$ are the numbers which appear as $n$ in the exponential terms $e^{nx}$ - the method of proceeding depends on these, with the constant multipliers being arbitrary. If you test out with sums of one and two exponentials you may get a better feel for what is going on here.

For example suppose we have $y=Ae^{\alpha x}+Be^{\beta x}$ then $$y'=A\alpha e^{\alpha x}+B\beta e^{\beta x}$$ Then $$y'-\alpha y = B(\beta-\alpha)e^{\beta x}$$ and $$y''-\alpha y'=B\beta(\beta-\alpha)e^{\beta x}$$ so that $$y''-\alpha y'-\beta (y'-\alpha y)=y''-(\alpha +\beta)y'+\alpha \beta y=0$$

If we define $D$ as the operation of differentiating with respect to $x$, we can write this as $$(D-\alpha)(D-\beta)y=D^2y-(\alpha+\beta)Dy+\alpha\beta y=y''-(\alpha +\beta)y'+\alpha \beta y=0$$ and you can see the factors highlighted in the other answers.

This does not depend on $\alpha, \beta$ being particular or nicely related numbers - the elimination works all the same.

• That short-cut is only applicable to $y = c_1e^{x} + c_2e^{2x} + c_3e^{3x} +......+ c_ne^{nx}$? Sep 12, 2017 at 9:28
• @PalautotKa No - you can have any exponentials, and every time you differentiate you can use a simple linear combination to eliminate another one of the exponential terms. Sep 12, 2017 at 9:32
• @PalautotKa I've added an example which shows how it works generally for two exponentials. Sep 12, 2017 at 9:41