Menu Close

Use-Laplace-transform-to-solve-the-initial-value-problem-ty-4t-2-y-13t-4-y-0-where-y-0-0-and-y-0-0-

Tinku Tara 0 Comments
Pin




Question Number 116457 by bobhans last updated on 04/Oct/20
Use Laplace transform to solve the initial value problem ty′′+(4t−2)y′+(13t−4)y=0 where y(0)=0 and y′(0)=0
$$\mathrm{Use}\:\mathrm{Laplace}\:\mathrm{transform}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{the}\: \\ $$$$\mathrm{initial}\:\mathrm{value}\:\mathrm{problem}\:\mathrm{ty}''+\left(\mathrm{4t}−\mathrm{2}\right)\mathrm{y}'+\left(\mathrm{13t}−\mathrm{4}\right)\mathrm{y}=\mathrm{0} \\ $$$$\mathrm{where}\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{0}\:\mathrm{and}\:\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$
Answered by Olaf last updated on 04/Oct/20
L(y′′) = p^2 L(p)−py(0)−y′(0) L(y′′) = p^2 L(p) L(y′) = pL(p)−y(0) L(y′) = pL(p) L(t^n f(t)) = (−1)^n ((d^n L(p))/dp^n ) L(tf(t)) = −L′(p) −(p^2 L)′−4(pL)′−2pL−13L′−4L = 0 p^2 L′+2pL+4pL′+4L+2(p+2)L+13L′ = 0 (p^2 +4p+13)L′+4(p+1)L = 0 ((L′)/L) = −4((p+1)/(p^2 +4p+13)) ⇒ integation and inverse Laplace
$$\mathrm{L}\left({y}''\right)\:=\:\mathrm{p}^{\mathrm{2}} \mathrm{L}\left(\mathrm{p}\right)−\mathrm{p}{y}\left(\mathrm{0}\right)−{y}'\left(\mathrm{0}\right) \\ $$$$\mathrm{L}\left({y}''\right)\:=\:\mathrm{p}^{\mathrm{2}} \mathrm{L}\left(\mathrm{p}\right) \\ $$$$ \\ $$$$\mathrm{L}\left({y}'\right)\:=\:\mathrm{pL}\left(\mathrm{p}\right)−{y}\left(\mathrm{0}\right) \\ $$$$\mathrm{L}\left({y}'\right)\:=\:\mathrm{pL}\left(\mathrm{p}\right) \\ $$$$ \\ $$$$\mathrm{L}\left({t}^{{n}} {f}\left({t}\right)\right)\:=\:\left(−\mathrm{1}\right)^{{n}} \frac{{d}^{{n}} \mathrm{L}\left(\mathrm{p}\right)}{{d}\mathrm{p}^{\mathrm{n}} } \\ $$$$\mathrm{L}\left({tf}\left({t}\right)\right)\:=\:−\mathrm{L}'\left(\mathrm{p}\right) \\ $$$$ \\ $$$$−\left(\mathrm{p}^{\mathrm{2}} \mathrm{L}\right)'−\mathrm{4}\left(\mathrm{pL}\right)'−\mathrm{2pL}−\mathrm{13L}'−\mathrm{4L}\:=\:\mathrm{0} \\ $$$$\mathrm{p}^{\mathrm{2}} \mathrm{L}'+\mathrm{2pL}+\mathrm{4pL}'+\mathrm{4L}+\mathrm{2}\left(\mathrm{p}+\mathrm{2}\right)\mathrm{L}+\mathrm{13L}'\:=\:\mathrm{0} \\ $$$$\left(\mathrm{p}^{\mathrm{2}} +\mathrm{4p}+\mathrm{13}\right)\mathrm{L}'+\mathrm{4}\left(\mathrm{p}+\mathrm{1}\right)\mathrm{L}\:=\:\mathrm{0} \\ $$$$\frac{\mathrm{L}'}{\mathrm{L}}\:=\:−\mathrm{4}\frac{\mathrm{p}+\mathrm{1}}{\mathrm{p}^{\mathrm{2}} +\mathrm{4p}+\mathrm{13}} \\ $$$$\Rightarrow\:\mathrm{integation}\:\mathrm{and}\:\mathrm{inverse}\:\mathrm{Laplace} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *