7.1 Second-order linear equations  (Page 6/15)

Note that all these equations are already given in standard form (step 1).

1. The characteristic equation is ${\lambda }^2+3\lambda -4=0$ (step 2). This factors into $\left(\lambda +4\right)\left(\lambda -1\right)=0,$ so the roots of the characteristic equation are ${\lambda }_1=\mathrm{-4}$ and ${\lambda }_2=1$ (step 3). Then the general solution to the differential equation is
   
   $y(x)=c_1{e}^{\mathrm{-4}x}+c_2{e}^x\text{(step 4)}\text{.}$
2. The characteristic equation is ${\lambda }^2+6\lambda +13=0$ (step 2). Applying the quadratic formula, we see this equation has complex conjugate roots $\mathrm{-3}\pm 2i$ (step 3). Then the general solution to the differential equation is
   
   $y(t)=e^{\mathrm{-3}t}\left(c_1\text{cos}2t+c_2\text{sin}2t\right)\text{(step 4)}\text{.}$
3. The characteristic equation is ${\lambda }^2+2\lambda +1=0$ (step 2). This factors into ${\left(\lambda +1\right)}^2=0,$ so the characteristic equation has a repeated real root $\lambda =\mathrm{-1}$ (step 3). Then the general solution to the differential equation is
   
   $y(t)=c_1{e}^{\text{−}t}+c_{2}t{e}^{\text{−}t}\text{(step 4).}$
4. The characteristic equation is ${\lambda }^2-5\lambda$ (step 2). This factors into $\lambda \left(\lambda -5\right)=0,$ so the roots of the characteristic equation are ${\lambda }_1=0$ and ${\lambda }_2=5$ (step 3). Note that $e^{0x}=e^0=1,$ so our first solution is just a constant. Then the general solution to the differential equation is
   
   $y(x)=c_1+c_2{e}^{5x}\text{(step 4)}\text{.}$
5. The characteristic equation is ${\lambda }^2-16=0$ (step 2). This factors into $(\lambda +4)(\lambda -4)=0,$ so the roots of the characteristic equation are ${\lambda }_1=4$ and ${\lambda }_2=\mathrm{-4}$ (step 3). Then the general solution to the differential equation is
   
   $y(x)=c_1{e}^{4x}+c_2{e}^{\mathrm{-4}x}\text{(step 4)}\text{.}$
6. The characteristic equation is ${\lambda }^2+16=0$ (step 2). This has complex conjugate roots $\pm 4i$ (step 3). Note that $e^{0x}=e^0=1,$ so the exponential term in our solution is just a constant. Then the general solution to the differential equation is
   
   $y(t)=c_1\text{cos}4t+c_2\text{sin}4t\text{(step 4)}\text{.}$