1807 Exercises

Problem 1: Find a particular solution to the DE $$x''+x=t^2+\cos(2t-1)$$ Solution: $x_p=x_{p,1}+x_{p,2}$, where $p(D)x_{p,1}=t^2$, and $p(D)x_{p,2}=\cos(2t-1)$, by superposition. Here $p(D)=D^2+1$.
try a solution of form $x_{p,1}=At^2+Bt+C$ $$x_{p,1}'=2At+B, x_{p,1}''=2A$$ $$x_{p,1}''+x_{p,1}=2A+At^2+Bt+C=t^2$$ $$A=1, B=0, C=-2$$ try a solution of form $x_{p,2}=A\cos(2t-1)+B\sin(2t-1)$ $$x_{p,2}'=-2A\sin(2t-1)+2B\cos(2t-1)$$ $$x_{p,2}''=-4A\cos(2t-1)-4B\sin(2t-1)$$ $$x_{p,2}''+x_{p,2}=-3A\cos(2t-1)-3B\sin(2t-1)=\cos(2t-1)$$ $$A=-1/3,B=0$$ so $$x_p=x_{p,1}+x_{p,2}=t^2-2-\frac{1}{3}\cos(2t-1)$$

Problem 2: Find the general solution to $$y''+y'+y=2xe^x$$ Solution: Characteristic equation: $$p(s)=s^2+s+1$$ $$r=\frac{-1\pm \sqrt{-3}}{2}$$ so the solution to the homogeneous equation is $$y_h=e^{-x/2}(c_1\cos\frac{\sqrt{3}}{2}x+c_2\sin \frac{\sqrt{3}}{2}x)$$ To find the particular solution, try $$y_p=Axe^x+Be^x$$ $$y_p'=Ae^x+Axe^x+Be^x=Axe^x+(A+B)e^e$$ $$y_p''=Ae^x+Axe^x+(A+B)e^x=Axe^x+(2A+B)e^e$$ $$y_p''+y_p'+y_p=3Axe^x+(3A+3B)e^x=2xe^x$$ $$A=2/3, B=-2/3$$ Thus the general solution is $$y=e^{-x/2}(c_1\cos\frac{\sqrt{3}}{2}x+c_2\sin \frac{\sqrt{3}}{2}x)+\frac{2}{3}e^x(x-1)$$

Problem 3: Find a particular solution to the DE $$y^{(4)}-2y''+y=xe^x$$ Solution:
$$p(s)=s^4-2s^2+1=(s^2-1)^2$$ so $p(1) = 0$ repeated root (order 2), so try $$y_p=x^2(Ax+B)e^x$$ Use the exponential shift rule $$p(D)e^xu=e^xp(D+1)u$$ so $$p(D+1)(x^2(Ax+B))=x$$ $$(Ax^3+Bx^2)'=3Ax^2+2Bx\\ (Ax^3+Bx^2)''=6Ax+2B\\ (Ax^3+Bx^2)'''=6A\\ (Ax^3+Bx^2)^{(4)}=0$$ $$p(D+1)=(D+1)^4-2(D+1)^2+1=D^4+4D^3+4D^2$$ $$\begin{aligned} &p(D+1)(x^2(Ax+B))\\ =&24A+24Ax+8B\\ =&x \end{aligned}$$ $$A=1/24, B=-3/24$$ so $$x_p=\frac{x^2e^x}{24}(x-3)$$