Question Number 217300 by peter frank last updated on 09/Mar/25
Answered by Marzuk last updated on 09/Mar/25
![(d^2 y/dx^2 ) − 2a(dy/dx) + (a^2 + b^2 )y = 0 or,((dy′)/dx) − 2a(d/dx) [e^(ax) sin(bx) ] + a^2 y+b^2 y =0 Here : ((dy′)/dx) = (d/dx)((d/dx) [ e^(ax) sin(bx) ]) (d/dx) [ e^(ax) sin(bx) ] = ae^(ax) sin(bx)+ e^(ax) b cos(bx) ∴ (d/dx) [ ae^(ax) sin(bx) + e^(ax) b cos(bx)] = (d/dx)[ ae^(ax) sin(bx) ] + (d/dx) [ e^(ax) b cos(bx) ] = a^2 e^(ax) sin(bx)+ ae^(ax) b cos(bx) + ae^(ax) b cos(bx)−b^2 sin(bx) or, a^2 e^(ax) sin(bx)+ae^(ax) b cos(bx) + ae^(ax) b cos(bx) − b^2 sin(bx)−2a^2 e^(ax) sin(bx) − 2ae^(ax) b cos(bx)+ a^2 e^(ax) sin(bx) + b^2 e^(ax) sin(bx) = 0 or,a^2 e^(ax) sin(bx) + a^2 e^(ax) sin(bx) − 2a^2 e^(ax) sin(bx) + ae^(ax) b cos(bx) + ae^(ax) b cos(bx) − 2ae^(ax) b cos(bx) + b^2 e^(ax) sin(bx) + b^2 e^(ax) sin(bx) − 2b^2 e^(ax) sin(bx) = 0 or, 0 + 0 + 0 = 0 or, 0 = 0 ∴ (d^2 y/dx^2 ) − 2a(dy/dx) + (a^2 + b^2 )y = 0](https://www.tinkutara.com/question/Q217306.png)
$$\:\:\:\:\:\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:−\:\mathrm{2}{a}\frac{{dy}}{{dx}}\:+\:\left({a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \right){y}\:=\:\mathrm{0} \\ $$$${or},\frac{{dy}'}{{dx}}\:−\:\mathrm{2}{a}\frac{{d}}{{dx}}\:\left[{e}^{{ax}} {sin}\left({bx}\right)\:\right]\:+\:{a}^{\mathrm{2}} {y}+{b}^{\mathrm{2}} {y}\:=\mathrm{0} \\ $$$${Here}\::\:\frac{{dy}'}{{dx}}\:=\:\frac{{d}}{{dx}}\left(\frac{{d}}{{dx}}\:\left[\:{e}^{{ax}} {sin}\left({bx}\right)\:\right]\right) \\ $$$$\:\:\frac{{d}}{{dx}}\:\left[\:{e}^{{ax}} {sin}\left({bx}\right)\:\right]\:=\:{ae}^{{ax}} {sin}\left({bx}\right)+\:{e}^{{ax}} {b}\:{cos}\left({bx}\right) \\ $$$$\therefore\:\frac{{d}}{{dx}}\:\left[\:{ae}^{{ax}} {sin}\left({bx}\right)\:+\:{e}^{{ax}} {b}\:{cos}\left({bx}\right)\right] \\ $$$$=\:\frac{{d}}{{dx}}\left[\:{ae}^{{ax}} {sin}\left({bx}\right)\:\right]\:+\:\frac{{d}}{{dx}}\:\left[\:{e}^{{ax}} {b}\:{cos}\left({bx}\right)\:\right] \\ $$$$=\:{a}^{\mathrm{2}} {e}^{{ax}} {sin}\left({bx}\right)+\:{ae}^{{ax}} {b}\:{cos}\left({bx}\right)\:+\:{ae}^{{ax}} {b}\:{cos}\left({bx}\right)−{b}^{\mathrm{2}} {sin}\left({bx}\right) \\ $$$${or},\:{a}^{\mathrm{2}} {e}^{{ax}} {sin}\left({bx}\right)+{ae}^{{ax}} {b}\:{cos}\left({bx}\right)\:+\:{ae}^{{ax}} {b}\:{cos}\left({bx}\right)\:−\:{b}^{\mathrm{2}} \:{sin}\left({bx}\right)−\mathrm{2}{a}^{\mathrm{2}} {e}^{{ax}} {sin}\left({bx}\right)\:−\:\mathrm{2}{ae}^{{ax}} {b}\:{cos}\left({bx}\right)+\:{a}^{\mathrm{2}} {e}^{{ax}} {sin}\left({bx}\right)\:+\:{b}^{\mathrm{2}} {e}^{{ax}} {sin}\left({bx}\right)\:=\:\mathrm{0} \\ $$$${or},{a}^{\mathrm{2}} {e}^{{ax}} {sin}\left({bx}\right)\:+\:{a}^{\mathrm{2}} {e}^{{ax}} {sin}\left({bx}\right)\:−\:\mathrm{2}{a}^{\mathrm{2}} {e}^{{ax}} {sin}\left({bx}\right)\:+\:{ae}^{{ax}} {b}\:{cos}\left({bx}\right)\:+\:{ae}^{{ax}} {b}\:{cos}\left({bx}\right)\:−\:\mathrm{2}{ae}^{{ax}} {b}\:{cos}\left({bx}\right)\:+\:{b}^{\mathrm{2}} {e}^{{ax}} \:{sin}\left({bx}\right)\:+\:{b}^{\mathrm{2}} {e}^{{ax}} \:{sin}\left({bx}\right)\:−\:\mathrm{2}{b}^{\mathrm{2}} {e}^{{ax}} \:{sin}\left({bx}\right)\:=\:\mathrm{0} \\ $$$${or},\:\mathrm{0}\:+\:\mathrm{0}\:+\:\mathrm{0}\:=\:\mathrm{0} \\ $$$${or},\:\mathrm{0}\:=\:\mathrm{0}\: \\ $$$$\therefore\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:−\:\mathrm{2}{a}\frac{{dy}}{{dx}}\:+\:\left({a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:\right){y}\:=\:\mathrm{0} \\ $$
Answered by som(math1967) last updated on 10/Mar/25
![y=e^(ax) sinbx y_1 =ae^(ax) sinbx+be^(ax) cosbx y_1 =ay+be^(ax) cosbx [y=e^(ax) sinbx[ y_2 =ay_1 +abe^(ax) cosbx−b^2 e^(ax) sinbx y_2 =ay_1 +a(y_1 −ay) −b^2 y [ be^(ax) cosbx=y_1 −ay] y_2 −2ay_1 +(a^2 +b^2 )y=0 [y_1 =(dy/dx) y_2 =(d^2 y/dx^2 )]](https://www.tinkutara.com/question/Q217319.png)
$$\:{y}={e}^{\boldsymbol{{ax}}} \boldsymbol{{sinbx}} \\ $$$$\:\boldsymbol{{y}}_{\mathrm{1}} =\boldsymbol{{ae}}^{\boldsymbol{{ax}}} \boldsymbol{{sinbx}}+\boldsymbol{{be}}^{\boldsymbol{{ax}}} \boldsymbol{{cosbx}} \\ $$$$\:\boldsymbol{{y}}_{\mathrm{1}} =\boldsymbol{{ay}}+\boldsymbol{{be}}^{\boldsymbol{{ax}}} \boldsymbol{{cosbx}}\:\:\left[\boldsymbol{{y}}=\boldsymbol{{e}}^{\boldsymbol{{ax}}} \boldsymbol{{sinbx}}\left[\right.\right. \\ $$$$\:\boldsymbol{{y}}_{\mathrm{2}} =\boldsymbol{{ay}}_{\mathrm{1}} +\boldsymbol{{abe}}^{\boldsymbol{{ax}}} \boldsymbol{{cosbx}}−\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{e}}^{\boldsymbol{{ax}}} \boldsymbol{{sinbx}} \\ $$$$\:\boldsymbol{{y}}_{\mathrm{2}} =\boldsymbol{{ay}}_{\mathrm{1}} +\boldsymbol{{a}}\left(\boldsymbol{{y}}_{\mathrm{1}} −\boldsymbol{{ay}}\right)\:−\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{y}} \\ $$$$\left[\:\boldsymbol{{be}}^{\boldsymbol{{ax}}} \boldsymbol{{cosbx}}=\boldsymbol{{y}}_{\mathrm{1}} −\boldsymbol{{ay}}\right] \\ $$$$\:\boldsymbol{{y}}_{\mathrm{2}} −\mathrm{2}\boldsymbol{{ay}}_{\mathrm{1}} +\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} \right)\boldsymbol{{y}}=\mathrm{0} \\ $$$$\:\left[\boldsymbol{{y}}_{\mathrm{1}} =\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}\:\:\:\:\:\boldsymbol{{y}}_{\mathrm{2}} =\frac{\boldsymbol{{d}}^{\mathrm{2}} \boldsymbol{{y}}}{\boldsymbol{{dx}}^{\mathrm{2}} }\right] \\ $$