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Confused about an example in my textbook (involving differential equations)
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
asked Nov 21 '18 at 7:38
Boris Grunwald
1487
add a comment |
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
asked Nov 21 '18 at 7:38
Boris Grunwald
1487
add a comment |
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
asked Nov 21 '18 at 7:38
Boris Grunwald
1487
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
differential-equations
asked Nov 21 '18 at 7:38
Boris Grunwald
1487
asked Nov 21 '18 at 7:38
Boris Grunwald
1487
asked Nov 21 '18 at 7:38
Boris Grunwald
1487
asked Nov 21 '18 at 7:38
Boris Grunwald
1487
asked Nov 21 '18 at 7:38
Boris Grunwald
1487
1487
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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
$$z''(t)-4z'(t)+13z(t)=145sin(2t).$$
answered Nov 21 '18 at 7:43
Robert Z
93.5k1061132
Thanks, this makes sense.
– Boris Grunwald
Nov 21 '18 at 7:53
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
$$z''(t)-4z'(t)+13z(t)=145sin(2t).$$
answered Nov 21 '18 at 7:43
Robert Z
93.5k1061132
Thanks, this makes sense.
– Boris Grunwald
Nov 21 '18 at 7:53
add a comment |
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
$$z''(t)-4z'(t)+13z(t)=145sin(2t).$$
answered Nov 21 '18 at 7:43
Robert Z
93.5k1061132
Thanks, this makes sense.
– Boris Grunwald
Nov 21 '18 at 7:53
add a comment |
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
$$z''(t)-4z'(t)+13z(t)=145sin(2t).$$
answered Nov 21 '18 at 7:43
Robert Z
93.5k1061132
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
$$z''(t)-4z'(t)+13z(t)=145sin(2t).$$
answered Nov 21 '18 at 7:43
Robert Z
93.5k1061132
edited Nov 21 '18 at 7:50
answered Nov 21 '18 at 7:43
Robert Z
93.5k1061132
answered Nov 21 '18 at 7:43
Robert Z
93.5k1061132
answered Nov 21 '18 at 7:43
Robert Z
93.5k1061132
93.5k1061132
Thanks, this makes sense.
– Boris Grunwald
Nov 21 '18 at 7:53
add a comment |
Thanks, this makes sense.
– Boris Grunwald
Nov 21 '18 at 7:53
Thanks, this makes sense.
– Boris Grunwald
Nov 21 '18 at 7:53
Thanks, this makes sense.
– Boris Grunwald
Nov 21 '18 at 7:53
add a comment |
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