Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 184936 by MikeH last updated on 14/Jan/23

$$\mathrm{Use}\mathrm{Laplace}\mathrm{transform}\mathrm{to}\mathrm{solve}\mathrm{the}\mathrm{differential}$$$$\mathrm{equation}$$$$\frac{d^{2}v\left(t\right)}{{dt}^2}+6\frac{dv\left(t\right)}{dt}+8v\left(t\right)=2u\left(t\right)$$$$\mathrm{when}v\left(0\right)=1\mathrm{and}\stackrel{\bullet }{v}\left(0\right)=-2$$

Answered by hmr last updated on 15/Jan/23

$$L\left\{v^{″}\right\}+6L\left\{v^{\prime }\right\}+8L\left\{v\right\}=2L\left\{u\right\}$$$$\to$$$$(s^{2}L\left\{v\right\}-sv\left(0\right)-v^{\prime }\left(0\right))$$$$+6(sL\left\{v\right\}-v\left(0\right))$$$$+8L\left\{v\right\}=2L\left\{u\right\}$$$$\to$$$$(s^2+6s+8)L\left\{v\right\}-s+2-6=2L\left\{u\right\}$$$$\to$$$$(s^2+6s+8)L\left\{v\right\}=2L\left\{u\right\}+s+4$$$$\to$$$$L\left\{v\right\}=\frac{2L\left\{u\right\}+s+4}{s^2+6s+8}$$$$\to$$$$L\left\{v\right\}=\frac{2L\left\{u\right\}+(s+4)}{(s+4)(s+2)}$$$$\to$$$$L\left\{v\right\}=\frac{2L\left\{u\right\}}{(s+4)(s+2)}+\frac{1}{s+2}$$$$\to$$$$L\left\{v\right\}=\frac{2L\left\{u\right\}}{(s+4)(s+2)}+L\left\{e^{-2t}\right\}$$$$\to$$$$L\left\{v\right\}=\frac{4L\left\{u\right\}}{s+4}-\frac{2L\left\{u\right\}}{s+2}+L\left\{e^{-2t}\right\}$$$$\to$$$$L\left\{v\right\}=4L\left\{e^{-4t}\right\}L\left\{u\right\}-2L\left\{e^{-2t}\right\}L\left\{u\right\}+L\left\{e^{-2t}\right\}$$$$Sorry{that}^{\prime }sallIknow.$$

Commented by MikeH last updated on 15/Jan/23

$$\mathrm{thanks}\mathrm{a}\mathrm{lot}$$

Commented by hmr last updated on 15/Jan/23

$${You}^{\prime }reWelcome$$

Commented by hmr last updated on 15/Jan/23

$$Ieditedthelastline,Itwaswrong!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com