Confused about an example in my textbook (involving differential equations)











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I am trying to understand an example in my textbook where they solve the differential equation:



$$y''(t)-4y'(t)+13y(t)=145cos(2t)$$



Later in the example they rewrite the right side of the equation to



$$145e^{2it}$$



I know it has something to with the complex exponential function but according to the definition:



$$e^{2it}=e^{0}(cos(2t)+isin(2t))ne cos(2t)$$



So why it legal to rewrite the right side like that?










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    up vote
    1
    down vote

    favorite












    I am trying to understand an example in my textbook where they solve the differential equation:



    $$y''(t)-4y'(t)+13y(t)=145cos(2t)$$



    Later in the example they rewrite the right side of the equation to



    $$145e^{2it}$$



    I know it has something to with the complex exponential function but according to the definition:



    $$e^{2it}=e^{0}(cos(2t)+isin(2t))ne cos(2t)$$



    So why it legal to rewrite the right side like that?










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I am trying to understand an example in my textbook where they solve the differential equation:



      $$y''(t)-4y'(t)+13y(t)=145cos(2t)$$



      Later in the example they rewrite the right side of the equation to



      $$145e^{2it}$$



      I know it has something to with the complex exponential function but according to the definition:



      $$e^{2it}=e^{0}(cos(2t)+isin(2t))ne cos(2t)$$



      So why it legal to rewrite the right side like that?










      share|cite|improve this question













      I am trying to understand an example in my textbook where they solve the differential equation:



      $$y''(t)-4y'(t)+13y(t)=145cos(2t)$$



      Later in the example they rewrite the right side of the equation to



      $$145e^{2it}$$



      I know it has something to with the complex exponential function but according to the definition:



      $$e^{2it}=e^{0}(cos(2t)+isin(2t))ne cos(2t)$$



      So why it legal to rewrite the right side like that?







      differential-equations






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      asked yesterday









      Boris Grunwald

      1417




      1417






















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          Consider $y(t)$ as the real part of a complex function $F(t)=y(t)+iz(t)$ such that
          $$F''(t)-4F'(t)+13F(t)=145e^{2it}=145(cos(2t)+isin(2t)).$$
          Then by taking the real part of both sides we obtain (note that the ODE is linear and with real coefficients),
          $$y''(t)-4y'(t)+13y(t)=145cos(2t).$$
          On the other hand, by taking the imaginary part of both sides we have
          $$z''(t)-4z'(t)+13z(t)=145sin(2t).$$






          share|cite|improve this answer























          • Thanks, this makes sense.
            – Boris Grunwald
            yesterday











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          1 Answer
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          1 Answer
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          active

          oldest

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          active

          oldest

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          active

          oldest

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          up vote
          4
          down vote



          accepted










          Consider $y(t)$ as the real part of a complex function $F(t)=y(t)+iz(t)$ such that
          $$F''(t)-4F'(t)+13F(t)=145e^{2it}=145(cos(2t)+isin(2t)).$$
          Then by taking the real part of both sides we obtain (note that the ODE is linear and with real coefficients),
          $$y''(t)-4y'(t)+13y(t)=145cos(2t).$$
          On the other hand, by taking the imaginary part of both sides we have
          $$z''(t)-4z'(t)+13z(t)=145sin(2t).$$






          share|cite|improve this answer























          • Thanks, this makes sense.
            – Boris Grunwald
            yesterday















          up vote
          4
          down vote



          accepted










          Consider $y(t)$ as the real part of a complex function $F(t)=y(t)+iz(t)$ such that
          $$F''(t)-4F'(t)+13F(t)=145e^{2it}=145(cos(2t)+isin(2t)).$$
          Then by taking the real part of both sides we obtain (note that the ODE is linear and with real coefficients),
          $$y''(t)-4y'(t)+13y(t)=145cos(2t).$$
          On the other hand, by taking the imaginary part of both sides we have
          $$z''(t)-4z'(t)+13z(t)=145sin(2t).$$






          share|cite|improve this answer























          • Thanks, this makes sense.
            – Boris Grunwald
            yesterday













          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          Consider $y(t)$ as the real part of a complex function $F(t)=y(t)+iz(t)$ such that
          $$F''(t)-4F'(t)+13F(t)=145e^{2it}=145(cos(2t)+isin(2t)).$$
          Then by taking the real part of both sides we obtain (note that the ODE is linear and with real coefficients),
          $$y''(t)-4y'(t)+13y(t)=145cos(2t).$$
          On the other hand, by taking the imaginary part of both sides we have
          $$z''(t)-4z'(t)+13z(t)=145sin(2t).$$






          share|cite|improve this answer














          Consider $y(t)$ as the real part of a complex function $F(t)=y(t)+iz(t)$ such that
          $$F''(t)-4F'(t)+13F(t)=145e^{2it}=145(cos(2t)+isin(2t)).$$
          Then by taking the real part of both sides we obtain (note that the ODE is linear and with real coefficients),
          $$y''(t)-4y'(t)+13y(t)=145cos(2t).$$
          On the other hand, by taking the imaginary part of both sides we have
          $$z''(t)-4z'(t)+13z(t)=145sin(2t).$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered yesterday









          Robert Z

          90.2k1056128




          90.2k1056128












          • Thanks, this makes sense.
            – Boris Grunwald
            yesterday


















          • Thanks, this makes sense.
            – Boris Grunwald
            yesterday
















          Thanks, this makes sense.
          – Boris Grunwald
          yesterday




          Thanks, this makes sense.
          – Boris Grunwald
          yesterday


















           

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