9.5 Workshop Solutions 5

W5.1. a) The CF is the general solution to $y^{\prime\prime}+4y=0$, which is $A\cos 2x+B\sin 2x$. This is not an exceptional case so we look for a PI of the form $y(x)=b_{2}x^{2}+b_{1}x+b_{0}$. Then $y^{\prime}=2b_{2}x+b_{1}$ so $y^{\prime\prime}=2b_{2}$, hence $y^{\prime\prime}+4y=4b_{2}x^{2}+4b_{1}x+(4b_{0}+2b_{2})$, so for this to equal $4x^{2}+3$ we need $b_{2}=1$, $b_{1}=0$ and $b_{0}=\frac{1}{4}$. Thus the general solution to the inhomogeneous equation is $y(x)=x^{2}+\frac{1}{4}+A\cos 2x+B\sin 2x$.

b) The homogeneous equation here is the same as in (a), but this is an exceptional case because $\sin 2x$ is a solution to the homogeneous equation. Thus we look for a PI of the form $y(x)=Hx\cos 2x+Kx\sin 2x$. Then $y^{\prime}(x)=H\cos 2x+K\sin 2x-2Hx\sin 2x+2Kx\cos 2x$, and $y^{\prime\prime}=-4H\sin 2x+4K\sin 2x-4Hx\cos 2x-4Kx\sin 2x$. Thus $y^{\prime\prime}+4y=-4H\sin 2x+4K\cos 2x$, so for this to equal $\sin 2x$ we require $H=-\frac{1}{4}$ and $K=0$. So the PI is $-\frac{1}{4}x\cos 2x$, so the general solution is $-\frac{1}{4}x\cos 2x+A\cos 2x+B\sin 2x$.

W5.2. a) The auxiliary equation here is $s^{2}-s-2=(s-2)(s+1)=0$, so the CF is $Ae^{2x}+Be^{-x}$. Since $e^{x}$ is not a solution of the homogeneous equation, this is not an exceptional case and so the PI has the form $Ce^{x}$ for some $C$. Substituting into the differential equation: if $y=Ce^{x}$ then $y^{\prime\prime}=y^{\prime}=Ce^{x}$ so $y^{\prime\prime}-y^{\prime}-2y=-2Ce^{x}$, which we want to equal $e^{x}$, hence $C=-\frac{1}{2}$. Therefore the general solution to the (inhomogeneous) equation is $y(x)=$ PI + CF $=Ae^{2x}+Be^{-x}-\frac{1}{2}e^{x}$. To find the solution to the initial value problem, we first evaluate: $y(0)=A+B-\frac{1}{2}=2$, to obtain $A+B=\frac{5}{2}$. Secondly, we have $y^{\prime}(x)=2Ae^{2x}-Be^{-2x}-\frac{1}{2}e^{x}$ so that $y^{\prime}(0)=2A-B-\frac{1}{2}=1$, whence $2A-B=\frac{3}{2}$ and so, after summing the two equations together, we obtain $3A=4$ and therefore $A=\frac{4}{3}$. Subtracting this from the equation $A+B=\frac{5}{2}$, we obtain $B=\frac{7}{6}$. So the particular solution is $y(x)=\frac{4}{3}e^{2x}+\frac{7}{6}e^{-x}-\frac{1}{2}e^{x}$.

b) The auxiliary equation is the same as in part (a), so the CF is $Ae^{2x}+Be^{-x}$ as above. However, in this case the right-hand side of the differential equation is $e^{-x}$, which is a solution to the homogeneous equation and this is therefore an exceptional case. We therefore look for a PI of the form $Cxe^{-x}$. If $y(x)=Cxe^{-x}$ then $y^{\prime}(x)=Ce^{-x}-Cxe^{-x}$ and $y^{\prime\prime}(x)=Cxe^{-x}-2Ce^{-x}$. Therefore $y^{\prime\prime}-y^{\prime}-2y=C(x-2-1+x-2x)e^{-x}=-3Ce^{-x}$, hence $C=\frac{-1}{3}$. So the general solution to the inhomogeneous equation is $y(x)=Ae^{2x}+Be^{-x}-\frac{1}{3}xe^{-x}$.

To find the particular solution, we have $y(0)=A+B$ so we require $A+B=2$. Furthermore, $y^{\prime}(x)=2Ae^{2x}-Be^{-x}+\frac{1}{3}xe^{-x}-\frac{1}{3}e^{-x}$, hence $y^{\prime}(0)=2A-B-\frac{1}{3}$. Therefore we require $2A-B-\frac{1}{3}=1$ and so, after substituting $B=2-A$, we obtain $3A-2=\frac{4}{3}$, hence $A=\frac{10}{9}$, $B=\frac{8}{9}$. So the particular solution is $y(x)=\frac{10}{9}e^{2x}+\frac{8}{9}e^{-x}-\frac{1}{3}xe^{-x}$.

c) Here the auxiliary equation is $s^{2}-2s+1=(s-1)^{2}$, so the CF is $(Ax+B)e^{x}$. This is an exceptional case, since $e^{x}$ is a solution to the homogeneous equation. But in fact $xe^{x}$ is also a solution, so in this case we need to look for a PI of the form $Cx^{2}e^{x}$. If $y(x)=Cx^{2}e^{x}$ then $y^{\prime}(x)=Cx^{2}e^{x}+2Cxe^{x}$ and $y^{\prime\prime}(x)=Cx^{2}e^{x}+4Cxe^{x}+2Ce^{x}$. Hence $y^{\prime\prime}-2y^{\prime}+y=C(x^{2}+4x+2-2x^{2}-4x+x^{2})e^{x}=2Ce^{x}$, hence $C=\frac{1}{2}$. Thus the general solution to the differential equation is $y(x)=(\frac{1}{2}x^{2}+Ax+B)e^{x}$.

To find the particular solution, we first of all have $y(0)=B=2$. Next, $y^{\prime}(x)=(\frac{1}{2}x^{2}+Ax+B)e^{x}+(x+A)e^{x}$, so $y^{\prime}(0)=A+B$, which we require to equal $1$. Thus $A=1-B=-1$. So the particular solution is $y(x)=(\frac{1}{2}x^{2}-x+2)e^{x}$.

W5.3. Here the auxiliary equation is $s^{2}-6s+9=(s-3)^{2}$, so the general solution to the equation is $(Ax+B)e^{3x}$. (See Thm. 4.45(ii) in MATH101.) Since $e^{3x}$ is a solution to the homogeneous equation, then this is an exceptional case. But in fact, $xe^{3x}$ is also a solution to the homogeneous equation, so we can think of this as a ‘doubly exceptional’ case, meaning that we should look for a PI of the form $Cx^{2}e^{3x}$. (See 6.31 or 6.38(c) in the notes.)

If $y=Cx^{2}e^{3x}$, then $y^{\prime}=C(2xe^{3x}+3x^{2}e^{3x})$ and $y^{\prime\prime}=C(2e^{3x}+12xe^{3x}+9x^{2}e^{3x})$, so

$$y^{\prime\prime}-6y^{\prime}+9y=Ce^{3x}(9x^{2}+12x+2-18x^{2}-12x+18x^{2})=2Ce^{3x}.$$

So we require $2C=4$, hence $C=2$ and so the general solution is $y=(2x^{2}+Ax+B)e^{3x}$.

W5.4. a) By linearity, ${\mathcal{L}}(f(x))=3{\mathcal{L}}(x^{2})+2{\mathcal{L}}(\cos x)$. Inspecting the table on 6.44, we therefore have ${\mathcal{L}}(f(x))=\frac{6}{s^{3}}+\frac{2s}{s^{2}+1}$.

b) By linearity we have ${\mathcal{L}}(g(x))={\mathcal{L}}(e^{2x}\cos 3x)+{\mathcal{L}}(5)$. By the table on 6.44 we have ${\mathcal{L}}(e^{2x}\cos 3x)=\frac{s-2}{(s-2)^{2}+9}=\frac{s-2}{s^{2}-4s+13}$. Thus the answer is: ${\mathcal{L}}(g(x))=\frac{s-2}{s^{2}-4s+13}+\frac{5}{s}$.

W5.5. i) We have $f(x)=\frac{e^{2x}}{2}+\frac{1}{2}$, so ${\mathcal{L}}(f(x))=\frac{1}{2}{\mathcal{L}}(e^{2x})+{\mathcal{L}}(\frac{1}{2})=\frac{1}{2(s-2)}+\frac{1}{2s}$.

ii) Taking the Laplace transform of both sides, we have:

$${\mathcal{L}}(y^{\prime})+2{\mathcal{L}}(y)=-y(0)+s{\mathcal{L}}(y)+2{\mathcal{L}}(y)=(s+2){\mathcal{L}}(y)=\frac{1}{2(s-2)}+\frac{1}{2s}.$$

Therefore ${\mathcal{L}}(y)=\frac{1}{2(s^{2}-4)}+\frac{1}{2s(s+2)}$. We recall that $\frac{2}{s^{2}-4}$ is the Laplace transform of $\sinh 2x$, so

$${\mathcal{L}}^{-1}\left(\frac{1}{2(s^{2}-4)}\right)=\frac{1}{4}\sinh 2x.$$

Similarly, $\frac{1}{s^{2}-1}$ is the Laplace transform of $\sinh x$, so by the shift formula, $\frac{1}{(s+1)^{2}-1}=\frac{1}{s(s+2)}$ is the Laplace transform of $e^{-x}\sinh x$. Thus

$${\mathcal{L}}^{-1}\left(\frac{1}{2s(s+2)}\right)=\frac{1}{2}e^{-x}\sinh x.$$

So the solution is $y=\frac{1}{4}\sinh 2x+\frac{1}{2}e^{-x}\sinh x=\frac{1}{8}e^{2x}+\frac{1}{4}-\frac{3}{8}e^{-2x}$.