Non-homogeneous second order equations

Find the general solution of $y^{\mathrm{\prime }\mathrm{\prime }}-y^{\mathrm{\prime }}-2y=4x^{2}y\text{'}\text{'}\; -\; y\text{'}\; -\; 2y\; =\; 4x^2$.

Solution $r^2-r-2=(r-2)(r+1)r^2\; -\; r\; -\; 2\; =\; (r\; -\; 2)(r\; +\; 1)$, so $y_c=C_1{e}^{2x}+C_2{e}^{-x}y\_c\; =\; C\_1e^\{2x\}\; +\; C\_2e^\{-x\}$. Try $y_p=A{x}^2+Bx+Cy\_p\; =\; Ax^2\; +\; Bx\; +\; C$: $y_p^{\mathrm{\prime }}=2Ax+By\_p\text{'}\; =\; 2Ax\; +\; B$, $y_p^{\mathrm{\prime }\mathrm{\prime }}=2Ay\_p\text{'}\text{'}\; =\; 2A$. Substituting,

$$2A-(2Ax+B)-2(A{x}^2+Bx+C)=-2A{x}^2+(-2A-2B)x+(2A-B-2C)=4x^2\mathrm{.}2A\; -\; (2Ax\; +\; B)\; -\; 2(Ax^2\; +\; Bx\; +\; C)\; =\; -2Ax^2\; +\; (-2A\; -\; 2B)x\; +\; (2A\; -\; B\; -\; 2C)\; =\; 4x^2\; .$$

Compare coefficients: $-2A=4-2A\; =\; 4$, so $A=-2A\; =\; -2$; $-2A-2B=0-2A\; -\; 2B\; =\; 0$, so $B=2B\; =\; 2$; $2A-B-2C=02A\; -\; B\; -\; 2C\; =\; 0$, so $C=-3C\; =\; -3$. Hence

$$y=C_1{e}^{2x}+C_2{e}^{-x}-2x^2+2x-3.y\; =\; C\_1e^\{2x\}\; +\; C\_2e^\{-x\}\; -\; 2x^2\; +\; 2x\; -\; 3\; .$$

Find a particular integral of (a) $y^{\mathrm{\prime }\mathrm{\prime }}-3y^{\mathrm{\prime }}+2y=6e^{-x}y\text{'}\text{'}\; -\; 3y\text{'}\; +\; 2y\; =\; 6e^\{-x\}$ (b) $y^{\mathrm{\prime }\mathrm{\prime }}-3y^{\mathrm{\prime }}+2y=e^{2x}y\text{'}\text{'}\; -\; 3y\text{'}\; +\; 2y\; =\; e^\{2x\}$.

Solution The roots are 1 and 2, so $y_c=C_1{e}^x+C_2{e}^{2x}y\_c\; =\; C\_1e^\{x\}\; +\; C\_2e^\{2x\}$.

1. $-1-1$ is not a root. Try $y_p=A{e}^{-x}y\_p\; =\; Ae^\{-x\}$: $A(1+3+2)e^{-x}=6e^{-x}A(1\; +\; 3\; +\; 2)e^\{-x\}\; =\; 6e^\{-x\}$, so $A=1A\; =\; 1$ and $y_p=e^{-x}y\_p\; =\; e^\{-x\}$.
2. $e^{2x}e^\{2x\}$ is already in $y_{c}y\_c$, so $A{e}^{2x}Ae^\{2x\}$ would give 0 on the left. Since 2 is a simple root, try $y_p=Ax{e}^{2x}y\_p\; =\; Axe^\{2x\}$: $y_p^{\mathrm{\prime }}=A(1+2x)e^{2x}y\_p\text{'}\; =\; A(1\; +\; 2x)e^\{2x\}$, $y_p^{\mathrm{\prime }\mathrm{\prime }}=A(4+4x)e^{2x}y\_p\text{'}\text{'}\; =\; A(4\; +\; 4x)e^\{2x\}$, and $$A[(4+4x)-3(1+2x)+2x]e^{2x}=A{e}^{2x}=e^{2x},A\backslash left[(4\; +\; 4x)\; -\; 3(1\; +\; 2x)\; +\; 2x\backslash right]e^\{2x\}\; =\; Ae^\{2x\}\; =\; e^\{2x\},$$ so $A=1A\; =\; 1$ and $y_p=x{e}^{2x}y\_p\; =\; xe^\{2x\}$. The $xx$-terms cancel, as they must when the trial is right.

(a) Solve $y^{\mathrm{\prime }\mathrm{\prime }}+4y=3\sin xy\text{'}\text{'}\; +\; 4y\; =\; 3\backslash sin\; x$ with $y(0)=0y(0)\; =\; 0$, $y^{\mathrm{\prime }}(0)=0y\text{'}(0)\; =\; 0$. (b) Find a particular integral of $y^{\mathrm{\prime }\mathrm{\prime }}+4y=8\cos 2xy\text{'}\text{'}\; +\; 4y\; =\; 8\backslash cos\; 2x$.

Solution The roots are $\pm 2i\backslash pm\; 2i$, so $y_c=C_1\cos 2x+C_2\sin 2xy\_c\; =\; C\_1\backslash cos\; 2x\; +\; C\_2\backslash sin\; 2x$.

1. Try $y_p=A\cos x+B\sin xy\_p\; =\; A\backslash cos\; x\; +\; B\backslash sin\; x$: $y_p^{\mathrm{\prime }\mathrm{\prime }}+4y_p=3A\cos x+3B\sin x=3\sin xy\_p\text{'}\text{'}\; +\; 4y\_p\; =\; 3A\backslash cos\; x\; +\; 3B\backslash sin\; x\; =\; 3\backslash sin\; x$, so $A=0A\; =\; 0$, $B=1B\; =\; 1$. Then $y=C_1\cos 2x+C_2\sin 2x+\sin xy\; =\; C\_1\backslash cos\; 2x\; +\; C\_2\backslash sin\; 2x\; +\; \backslash sin\; x$. From $y(0)=0y(0)\; =\; 0$, $C_1=0C\_1\; =\; 0$; from $y^{\mathrm{\prime }}(0)=2C_2+1=0y\text{'}(0)\; =\; 2C\_2\; +\; 1\; =\; 0$, $C_2=-\frac{1}{2}C\_2\; =\; -\backslash frac\{1\}\{2\}$: $$y=\sin x-\frac{1}{2}\sin 2x\mathrm{.}y\; =\; \backslash sin\; x\; -\; \backslash tfrac\{1\}\{2\}\backslash sin\; 2x\; .$$
2. Here $\omega =2\backslash omega\; =\; 2$ and $\pm 2i\backslash pm\; 2i$ are roots, so try $y_p=x(A\cos 2x+B\sin 2x)y\_p\; =\; x(A\backslash cos\; 2x\; +\; B\backslash sin\; 2x)$. Substituting and simplifying gives $y_p^{\mathrm{\prime }\mathrm{\prime }}+4y_p=-4A\sin 2x+4B\cos 2xy\_p\text{'}\text{'}\; +\; 4y\_p\; =\; -4A\backslash sin\; 2x\; +\; 4B\backslash cos\; 2x$. So $A=0A\; =\; 0$ and $4B=84B\; =\; 8$: $y_p=2x\sin 2xy\_p\; =\; 2x\backslash sin\; 2x$.

Find the general solution of $y^{\mathrm{\prime }\mathrm{\prime }}-y=x+3e^{2x}y\text{'}\text{'}\; -\; y\; =\; x\; +\; 3e^\{2x\}$.

Solution $r^2-1=0r^2\; -\; 1\; =\; 0$ gives $y_c=C_1{e}^x+C_2{e}^{-x}y\_c\; =\; C\_1e^x\; +\; C\_2e^\{-x\}$. Treat the two parts separately. For $xx$: try $Ax+BAx\; +\; B$; then $0-(Ax+B)=x0\; -\; (Ax\; +\; B)\; =\; x$, so $y_{p1}=-xy\_\{p1\}\; =\; -x$. For $3e^{2x}3e^\{2x\}$: try $A{e}^{2x}Ae^\{2x\}$; then $4A-A=34A\; -\; A\; =\; 3$, so $y_{p2}=e^{2x}y\_\{p2\}\; =\; e^\{2x\}$. By superposition,

$$y=C_1{e}^x+C_2{e}^{-x}-x+e^{2x}\mathrm{.}y\; =\; C\_1e^x\; +\; C\_2e^\{-x\}\; -\; x\; +\; e^\{2x\}\; .$$