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Quiz 7 solution
Find the general solution of $y''-y=t^2+1$.

Solution: First we solve the homogeneous problem $y''-y=0$. The characteristic equation is $r^2-1=0$ which has roots $r=\pm 1$. Therefore the homogeneous solution is $$y_h(t)=c_1y_1(t)+c_2y_2(t)=c_1e^t+c_2e^{-t}.$$ Now the particular solution $y_p(t)$ is given by the variation of parameters formula $y_p(t)=u_1(t)e^t+u_2(t)e^{-t}$, where $u_1(t)=-\displaystyle\int \dfrac{e^{-t}(t^2+1)}{W\{y_1,y_2\}(t)} \mathrm{d}t$ and $u_2(t)=\displaystyle\int \dfrac{e^t(t^2+1)}{W\{y_1,y_2\}(t)} \mathrm{d}t$. First compute the Wronskian $$W\{y_1,y_2\}(t)=\mathrm{det} \begin{bmatrix} e^t & e^{-t} \\ e^t & -e^{-t} \end{bmatrix}=-1-1=-2.$$ Then using integration by parts, $$u_1(t)=-\displaystyle\int \dfrac{e^{-t}(t^2+1)}{-2} \mathrm{d}t=-\dfrac{e^{-t}}{2}\Big(t^2+2t+3\Big),$$ and $$u_2(t)=\displaystyle\int \dfrac{e^t(t^2+1)}{-2} \mathrm{d}t = -\dfrac{e^t}{2}\Big(t^2-2t+3\Big).$$ Thus the particular solution is $$\begin{array}{ll} y_p(t)&=u_1(t)y_1(t)+u_2(t)y_2(t) \\ &=-\dfrac{e^{-t}}{2}\Big(t^2+2t+3\Big)e^t -\dfrac{e^t}{2}\Big(t^2-2t+3\Big)e^{-t}\\ &=-\dfrac{1}{2}\Big( t^2+2t+3+t^2-2t+3\Big) \\ &=-\dfrac{1}{2}\Big( 2t^2+6 \Big) = -t^2-3 \end{array}$$ Therefore, the general solution is $$y(t)=y_h(t)+y_p(t)=c_1 e^t + c_2 e^{-t} -t^2-3.$$