Ejercicios de repaso sección 4.5

Encuentre f(t)=L−1F(s)f(t)=\mathscr{L}^{-1}F(s) o F(s)=Lf(t)F(s)= \mathscr{L}f(t)

  1. L{(t−4)μ(t−4)\hspace{0.5cm} \mathscr{L}\left\{\left(t-4\right)\mu\left(t-4\right)\right.

  2. L{e2−tμ(t−2)\hspace{0.5cm} \mathscr{L}\left\{e^{2-t}\mu\left(t-2\right)\right.

  3. L{cos2tμ(t−π)\hspace{0.5cm} \mathscr{L}\left\{cos2t\mu\left(t-\pi\right)\right.

  4. L{sentμ(t−π2)\hspace{0.5cm} \mathscr{L}\left\{sent\mu\left(t-\dfrac{\pi}{2}\right)\right.

  5. L−1{e−2ss3}\hspace{0.5cm} \mathscr{L}^{-1}\left\{\dfrac{e^{-2s}}{s^3}\right\}

  6. L−1{e−πss2+1}\hspace{0.5cm} \mathscr{L}^{-1}\left\{\dfrac{e^{-\pi s}}{s^2+1}\right\}

  7. L−1{se−π2ss2+4}\hspace{0.5cm} \mathscr{L}^{-1}\left\{\dfrac{se^{-\frac{\pi}{2}s}}{s^2+4}\right\}

  8. L−1{e−ss(s+1)}\hspace{0.5cm} \mathscr{L}^{-1}\left\{\dfrac{e^{-s}}{s\left(s+1\right)}\right\}

  9. L−1{e−2ss2(s−1)}\hspace{0.5cm} \mathscr{L}^{-1}\left\{\dfrac{e^{-2s}}{s^2\left(s-1\right)}\right\}

    10. En los ejercicios 10 a 13, evaluar la función

  10. f(t)={t,0≤t<10,t≥1\hspace{0.5cm} f(t)=\left\{\begin{matrix}t,&0\le t<1\\0,&t\geq1\\\end{matrix}\right.

  11. f(t)={1,0≤t<40,4≤t<51,t≥5\hspace{0.5cm} f(t)=\left\{\begin{matrix}1,&0\le t<4\\0,&4\le t<5\\1,&t\geq5\\\end{matrix}\right.

  12. f(t)={0,0≤t<1t2,t≥1\hspace{0.5cm} f(t)=\left\{\begin{matrix}0,&0\le t<1\\t^2,&t\geq1\\\end{matrix}\right.

  13. f(t)={sent,0≤t<2π0,t≥2π\hspace{0.5cm} f(t)=\left\{\begin{matrix}sent,&0\le t<2\pi\\0,&t\geq2\pi\\\end{matrix}\right.

    14. En los ejercicios 14 a 17 resolver el PVI

  14. y′+y=f(t)\hspace{0.5cm} y\prime+y=f(t) sujeta a la condición y(0)=5y(0)=5 donde f(t)={0,0≤t<π3cost,t≥πf(t)=\left\{\begin{matrix}0,&0\le t<\pi\\3cos{t},&t\geq\pi\\\end{matrix}\right.

  15. y′+y=f(t)\hspace{0.5cm} y\prime+y=f(t) sujeta a la condición y(0)=0y(0)=0 donde f(t)={0,0≤t<15,t≥1f(t)=\left\{\begin{matrix}0,&0\le t<1\\5,&t\geq1\\\end{matrix}\right.

  16. y′′+4y=f(t)\hspace{0.5cm} y\prime\prime+4y=f(t) sujeta a la condición y(0)=0    ∧    y´(0)=−1y(0)=0 \;\; \wedge \;\; y´(0)=-1 donde f(t)={1,0≤t<10,t≥1f(t)=\left\{\begin{matrix}1,&0\le t<1\\0,&t\geq1\\\end{matrix}\right.

  17. y′′+4y=sentμ(t−2π)\hspace{0.5cm} y\prime\prime+4y=sent\mu(t-2\pi) sujeta a la condición y(0)=1    ∧    y´(0)=0y(0)=1 \;\; \wedge \;\; y´(0)=0