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

Question

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?

Answer 1

You can work this out by Gaussian elimination

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

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

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

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

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

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

Answer 2

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.

Answer 3

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.

Answer 4

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.

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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.