Solving $y'''+3y''+3y'+y = e^{x} -x-1$

$y'''+3y''+3y'+y = e^{x} -x-1$

Hi, i've got this method of undermined coefficients but i am unable to find the particular solution. I got the complimentary solution as $Y_c = {C_1}{e^{-x}}+ {C_2}{e^{-x}}+ {C_3}{e^{-x}}$ but keep getting the wrong answer when i try to find the particular.

• Hint: If we plug in a variation of $y=\alpha e^x$ what value of $\alpha$ would we need to obtain $e^x$ when filling it on the righthand side of your equation? – Jan Aug 9 '18 at 13:05
• @Roman What's the general solution (of homogeneous equation)? – Nosrati Aug 9 '18 at 13:18

I think you've mistyped the solution to the homogeneous equation. Try:

$$y_h=C_1e^{-x}+C_2xe^{-x}+C_3x^2e^{-x}$$

For the particular solution, try $y=a_1e^x+a_2x+a_3$. Then:

$$y' = a_1e^x + a_2$$

$$y'' = y''' = a_1e^x$$

Then:

\begin{align} y'''+3y''+3y'+y &= 8a_1e^x + a_2x + 3a_2 + a_3 \end{align}

and so put

$$a_1=\frac{1}{8} \qquad a_2=-1 \qquad a_3=2$$

to get your particular solution as

$$y=\frac{1}{8}e^x-x+2$$

Your complimentary solution is wrong. It is correct that the roots of the characteristic equation are all equal to $\lambda=-1$; in other words, there's only one root of multiplicity $3$. But when there's a multiple root, you don't set up the same solution multiple times — doing so, as you did, still only gives you one function, not three (in this example).

Let me re-explain. For a third-order complimentary solution you need three functions, which you will add together with arbitrary coefficients. Alas, writing the same thing three times does not create three things. Note that your proposed complimentary solution simplifies as $$C_1e^{-x}+C_2e^{-x}+C_3e^{-x}=Ce^{-x}, \text{ where } C=C_1+C_2+C_3,$$ so you in fact have only one term (one function) here, not three.

Check out your textbook for an explanation of what happens when the characteristic equation has repeated roots. Short version that applies to this example: if $\lambda$ is a root of the characteristic equation of multiplicity $3$, then to it correspond the following three solution functions: $$e^{\lambda x}, \; xe^{\lambda x}, \; x^2e^{\lambda x}.$$ In your example, the correct complimentary solution is $$y_c(x)=C_1e^{-x}+C_2xe^{-x}+C_3x^2e^{-x}.$$

• Ah i've seen where I've gone wrong, thanks for this! – Roman Aug 9 '18 at 13:27

The solution of the homoginous equation is given by $$y_h=C_1e^{-x}+C_2xe^{-x}+C_3x^2e^{-x}$$ and for the particula equation make the ansatz

$$y_p=Ae^x+Bx+C$$

Hint : What about a solution of the form of the sum of the exponential function and a polynomial of degree = 4 ?

Using the differential operator $D$, the equation can be written

$$(D+1)^3y=(D+1)(D+1)(D+1)y=e^x-x-1.$$

You can obtain the general solution by solving $$(D+1)y=f(x)$$ three times, which amounts to writing $$y'+y=(ye^x)'e^{-x}=f(x)$$ or

$$y=e^{-x}\int f(x) e^xdx.$$

Let us go:

$$e^{-x}\int (e^x-x-1)e^xdx=e^{-x}\left(\frac12e^{2x}-xe^x+C_1\right),$$

$$e^{-x}\int \left(\frac12e^{2x}-xe^x+C\right)dx=e^{-x}\left(\frac14e^{2x}-xe^x+e^x+C_1x+C_2\right),$$

$$e^{-x}\int \left(\frac14e^{2x}-xe^x+e^x+C_1x+C_2\right)dx=e^{-x} \left(\frac18e^{2x}-xe^x+2e^x+C_1x^2+C_2x+C_3\right) \\=\frac18e^{x}-x+2+(C_1x^2+C_2x+C_3)e^{-x}.$$