Ejercicios de repaso Capítulo 1

En los ejercicios 1 a 20, Compruebe si la función indicada es una solución o no de la E.D. dada.

  1. y=ex2\hspace{0.5cm}y=e^\frac{x}{2}

de 2y′+y=0\hspace{0.3cm}2y^{'}+y=0

  1. y=e3x+10e2x\hspace{0.5cm}y=e^{3x}+10e^{2x}

de dydx−2y=e3x\hspace{0.3cm}\dfrac{dy}{dx}-2y=e^{3x}

  1. y=xln⁡∣x∣\hspace{0.5cm}y=x\ln \left| x\right|

de y′−1xy=1\hspace{0.3cm}y^{'}-\dfrac{1}{x}y=1

  1. y=5tan5x\hspace{0.5cm}y=5tan5x

de y′=25+y2\hspace{0.3cm}y^{'}=25+y^2

  1. y=senx3x\hspace{0.5cm}y=\dfrac{senx}{3x}

de xy′+y=cosx\hspace{0.3cm}xy^{'}+y=cosx

  1. y=−33x+2\hspace{0.5cm}y=\dfrac{-3}{3x+2}

de y′=3y2\hspace{0.3cm}y^{'}=3y^2

  1. y−1cosx=0\hspace{0.5cm}y-\dfrac{1}{cosx}=0

de y′−ytanx=0\hspace{0.3cm}y^{'}-ytanx=0

  1. y=xcosx\hspace{0.5cm}y=\dfrac{x}{cosx}

de xy′−y=xtanxsecx\hspace{0.2cm}xy^{'}-y=xtanxsecx

  1. y=1+c1−x2\hspace{0.5cm}y=1+c\sqrt {1-x^{2}}

de (1−x2)y′+xy=x\hspace{0.2cm}(1-x^2)y^{'}+xy=x

  1. y=ex−e−x\hspace{0.5cm}y=e^{x}-e^{-x}

de y′=y+2e−x\hspace{0.2cm}y^{'}=y+2e^{-x}

  1. y=excosx\hspace{0.5cm}y=e^{x}cosx

de y′′−2y′+2y=0\hspace{0.2cm}y^{''}-2y^{'}+2y=0

  1. y=xex\hspace{0.5cm}y=xe^{x}

de 2y′′−3y′+y=0\hspace{0.2cm}2y^{''}-3y^{'}+y=0

  1. y=x−ln⁡∣x∣\hspace{0.5cm}y=x-\ln \left| x\right|

de x2y′′+xy′−y=ln⁡∣x∣\hspace{0.01cm}x^2y^{''}+xy^{'}-y=\ln \left| x\right|

  1. y=e3xcos2x\hspace{0.5cm}y=e^{3x}cos{2x}

de y′′−6y′+13y=0\hspace{0.2cm}y^{''}-6y^{'}+13y=0

  1. y=e−xcos12x\hspace{0.5cm}y=e^{-x}cos\frac{1}{2}x

de 4y′′+8y′+5y=0\hspace{0.2cm}4y^{''}+8y^{'}+5y=0

  1. y=2x12−x12ln⁡∣x∣\hspace{0.5cm}y=2x^\frac{1}{2}-x^\frac{1}{2}\ln \left| x\right|

de 4x2y′′+y=0\hspace{0.2cm}4x^2y^{''}+y=0

  1. y=ex(3cos2x+sen2x)\hspace{0.5cm}y=e^{x}(3cos{2x}+sen{2x})

de y′′−2y′+5y=0\hspace{0.2cm}y^{''}-2y^{'}+5y=0

  1. y=−(cosx)ln⁡∣secx+tanx∣\hspace{0.5cm}y=-(cosx)\ln \left| secx+tanx\right|

de y′′+y=tanx\hspace{0.2cm}y^{''}+y=tanx

  1. y=cosxln⁡∣cosx∣+xsenx\hspace{0.5cm}y=cosx\ln \left| cosx\right|+xsenx

de y′′+y=secx\hspace{0.2cm}y^{''}+y=secx

  1. y=xcos(ln⁡∣x∣)\hspace{0.5cm}y=xcos(\ln \left| x\right|)

de x2y′′−xy′+2y=0\hspace{0.2cm}x^2y^{''}-xy^{'}+2y=0

En los ejercicios 21 a 27, Demuestre que la función indicada es una solución implicita de la E.D. dada.

  1. c=5x22+4xy−2y4\hspace{0.5cm}c=\dfrac{5x^2}{2}+4xy-2y^4

de (4x−8y3)dydx=−5x−4y\hspace{0.2cm}(4x-8y^3)\dfrac{dy}{dx}=-5x-4y

  1. c=x2y2−3x+4y\hspace{0.5cm}c=x^2y^2-3x+4y

de (2y2x−3)=−(2yx2+4)y′\hspace{0.2cm}(2y^2x-3)=-(2yx^2+4)y^{'}

  1. c=x44+xy3\hspace{0.5cm}c=\dfrac{x^4}{4}+xy^3

de (x3+y3)dx+3xy2dy=0\hspace{0.2cm}(x^3+y^3)dx+3xy^2dy=0

  1. ln⁡∣y∣+y2=−cosx+c\hspace{0.5cm}\ln \left| y\right|+y^2=-cosx+c

de ysenxdx=(1+2y2)dy\hspace{0.2cm}ysenxdx=(1+2y^2)dy

  1. y2−x3+8=0\hspace{0.5cm}y^2-x^3+8=0

de dydx−3x22y=0\hspace{0.2cm}\dfrac{dy}{dx}-\dfrac{3x^2}{2y}=0

  1. x2y3−3x+2y=c\hspace{0.5cm}x^2y^3-3x+2y=c

de (3x2y2+2)dy=−(2xy3−3)dx\hspace{0.05cm}(3x^2y^2+2)dy=-(2xy^3-3)dx

  1. y−1y=tan−1x+c\hspace{0.5cm}y-\dfrac{1}{y}=tan^{-1}x+c

de (1+x2+y2+x2y2)dy=y2dx\hspace{0.2cm}(1+x^2+y^2+x^2y^2)dy=y^2dx

En los ejercicios 28 a 38, Se le da una ED de primer orden, su solución y una condición inicial, determinar el valor de la constante.

  1. y′=y−y2\hspace{0.2cm}y^{'}=y-y^2

y=11+ce−xy=\dfrac{1}{1+ce^{-x}}

y(0)=−13y(0)=-\dfrac{1}{3}

  1. y′=y−y2\hspace{0.2cm}y^{'}=y-y^2

y=11+ce−xy=\dfrac{1}{1+ce^{-x}}

y(−1)=2y(-1)=2

  1. y′+2xy2=0\hspace{0.2cm}y^{'}+2xy^2=0

y=1x2+cy=\dfrac{1}{x^{2}+c}

y(12)=−4y\left( \dfrac {1}{2}\right) =-4

  1. y′+2xy2=0\hspace{0.2cm}y^{'}+2xy^2=0

y=1x2+cy=\dfrac{1}{x^{2}+c}

y(0)=1y(0)=1

  1. y′+2xy2=0\hspace{0.2cm}y^{'}+2xy^2=0

y=1x2+cy=\dfrac{1}{x^{2}+c}

y(−2)=12y(-2)=\dfrac{1}{2}

  1. yy′+6x=0\hspace{0.2cm}yy^{'}+6x=0

y2=−6x2+cy^2=-6x^2+c

y(0)=4y(0)=4

  1. y′=1+y2\hspace{0.2cm}y^{'}=1+y^2

y=tanx(x+c)y=tanx(x+c)

y(π4)=1y\left( \dfrac {\pi}{4}\right) =1

  1. yy′=e2x+1\hspace{0.2cm}yy^{'}=e^{2x}+1

y2=e2x+2x+cy^2=e^{2x}+2x+c

y(0)=12y(0)=\dfrac{1}{2}

  1. y2y′−4x=0\hspace{0.2cm}y^2y^{'}-4x=0

y3=6x2+cy^3=6x^2+c

y(12)=0y\left( \dfrac {1}{2}\right) =0

  1. y′+y=0\hspace{0.2cm}y^{'}+y=0

y=ce−xy=ce^{-x}

y(0)=2y(0)=2

  1. y′+3x2y=0\hspace{0.2cm}y^{'}+3x^2y=0

y=ce−x3y=ce^{-x^3}

y(0)=7y(0)=7

En los ejercicios 39 a 47, Se le da una ED de segundo orden, su solución y las condiciones iniciales, determinar el valor de las constantes.

  1. x′′+x=0\hspace{0.2cm}x^{''}+x=0

    x=c1cost+c2sentx=c_1cost+c_2sent

      x(0)=−1x(0)=-1

        x′(0)=8x^{'}(0)=8

          1. x′′+x=0\hspace{0.2cm}x^{''}+x=0

            x=c1cost+c2sentx=c_1cost+c_2sent

              x(π2)=0x\left( \dfrac {\pi}{2}\right) =0

                x′(π2)=1x^{'}\left( \dfrac {\pi}{2}\right) =1

                  1. x′′+x=0\hspace{0.2cm}x^{''}+x=0

                    x=c1cost+c2sentx=c_1cost+c_2sent

                      x(π6)=12x\left( \dfrac {\pi}{6}\right) =\dfrac{1}{2}

                        x′(π6)=0x^{'}\left( \dfrac {\pi}{6}\right) =0

                          1. y′′−y=0\hspace{0.2cm}y^{''}-y=0

                            y=c1ex+c2e−xy=c_1e^x+c_2e^{-x}

                              y(0)=1y(0)=1

                                y′(0)=2y^{'}(0)=2

                                  1. y′′−y=0\hspace{0.2cm}y^{''}-y=0

                                    y=c1ex+c2e−xy=c_1e^x+c_2e^{-x}

                                      y(−1)=5y(-1)=5

                                        y′(−1)=−5y^{'}(-1)=-5

                                          1. 2y′′+y′−y=0\hspace{0.2cm}2y^{''}+y^{'}-y=0

                                            y=c1ex2+c2e−xy=c_1e^{\frac{x}{2}}+c_2e^{-x}

                                              y(0)=0y(0)=0

                                                y′(0)=1y^{'}(0) =1

                                                  1. y′′+y=cosx+4\hspace{0.2cm}y^{''}+y=cosx+4

                                                    y=c1xsenx+c2y=c_1xsenx+c_2

                                                      y(0)=4y(0)=4

                                                        y′(π2)=1y^{'}\left( \dfrac {\pi}{2}\right)=1

                                                          1. y′′−3y′−4y=0\hspace{0.2cm}y^{''}-3y^{'}-4y=0

                                                            y=c1e4x+c2e−xy=c_1e^{4x}+c_2e^{-x}

                                                              y(0)=1y(0)=1

                                                                y′(0)=2y^{'}(0)=2

                                                                  1. x2y′′−xy′+y=0\hspace{0.2cm}x^2y^{''}-xy^{'}+y=0

                                                                    y=c1x+c2xln⁡∣x∣y=c_1x+c_2x\ln \left| x\right|

                                                                      y(1)=3y(1)=3

                                                                        y′(1)=−1y^{'}(1)=-1