Solve the differential equation $(D^3-3D^2+4D-2)y = (e^x +\cos x)$, where $D = \frac{\mathrm{d}}{\mathrm{d}x}$.

Question

Solve the differential equation $$(D^3-3D^2+4D-2)y = (e^x +\cos x)\,,$$ where $D = \dfrac{\mathrm{d}}{\mathrm{d}x}$.

Here, for $e^x$, we place coefficient of $x$ in place of $D$ which is $1$ but $D=1$ is solution for given equation. So how to solve this for particular integral?

Answer 1

$$(D^3-3D^2+4D-2)y = \cos x$$ For this particular integral: $$y_p =\dfrac 1 {(D^3-3D^2+4D-2)} \cos x$$ $$y_p =\dfrac 1 {3D+1} \cos x$$ $$y_p =\dfrac {3D-1} {9D^2-1} \cos x$$ $$-10y_p = ({3D-1}) \cos x$$ $$-10y_p =-3\sin x -\cos x$$ $$y_p =\dfrac {3\sin x +\cos x}{10}$$

---

$$(D^3-3D^2+4D-2)y = e^x$$ For this particular integral: $$y_p =\dfrac 1 {(D-1)(D^2-2D+2)}e^x$$ $$y_p =\dfrac 1 {(D-1)}e^x=e^x\dfrac 1 {D}1$$ $$ \implies y_p =xe^x$$

---

Finally the particular integral is: $$\boxed {y_p =\dfrac {3\sin x +\cos x}{10}+xe^x}$$

Answer 2

As an alternative method, you could find the particular integral through the method of undetermined coefficients.

$$(D^3-3D^2+4D-2)y = e^x +\cos x$$ $$y'''-3y''+4y'-2y = e^x +\cos x$$

Complementary Solution:

$$y'''-3y''+4y'-2y=0$$ The characteristic equation for this differential equation and its roots are: \begin{align}r^3-3r^2+4r-2&=(r - 1) (r^2 - 2 r + 2)\\&= (r-1)(r-(1+i))(r-(1-i))\\&=0 \end{align} which implies that $$r_1=1,~r_2=1+i,~r_3=1-i$$

The complementary solution is then $$y_c=c_1e^x+c_2e^{x}\sin x+c_3e^x\cos x$$

Particular Solution:

We apply the method of undetermined coefficients. As the right-hand side is of the form $e^x +\cos x$, a reasonable guess would be $y_p=Ae^x+B\sin x + C\cos x$. However, $e^x$ is already part of the complementary solution, so we instead make the guess

$$y_p=Axe^x+B\sin x + C\cos x$$

Differentiating thrice we find

$$y'_p=Ae^x(x+1)-C\sin x+B\cos x$$ $$y''_p=Ae^x(x+2)-B\sin x-C\cos x$$ $$y'''_p=Ae^x(x+3)+C\sin x-B\cos x$$

Therefore

$$y'''_p-3y''_p+4y'_p-2y_p =Ae^x+(B-3C)\sin x+(3B+C)\cos x=e^x+\cos x$$

$$\begin{cases} A=1\\ B-3C=0 \\ 3B+C=1 \end{cases}\implies B=\frac{3}{10},~ C=\frac{1}{10}$$

Thus, the particular solution takes the form

$$y_p=xe^x+\frac{3}{10}\sin x + \frac{1}{10}\cos x$$