How to solve a differential equation with the characteristic equation plus some conditions
$begingroup$
I have the beginner equation
$frac{d^2s}{dt^2} + 2frac{ds}{dt} + 2s = 0$ with the conditions $t_0 = 0, s_0 = 1, s'_0 = 1$.
Solving the characteristic polynomial I get the solutions $lambda_1 = -1 pm i$
which I have incorporated in the solution $s(t) = c_1 e^{(-1+i)t} + c_2 e^{(-1-i)t}$
How do I incorporate the conditions into the final answer?
Edit: $s(t)$ is wrong. Because the solutions of the polynomial equation are imaginary and of special form, the correct solution involves writing $s(t)$ in terms of trigonometric functions.
ordinary-differential-equations
asked Jan 23 at 20:03
OscarOscar
351111
$endgroup$
add a comment |
$begingroup$
I have the beginner equation
$frac{d^2s}{dt^2} + 2frac{ds}{dt} + 2s = 0$ with the conditions $t_0 = 0, s_0 = 1, s'_0 = 1$.
Solving the characteristic polynomial I get the solutions $lambda_1 = -1 pm i$
which I have incorporated in the solution $s(t) = c_1 e^{(-1+i)t} + c_2 e^{(-1-i)t}$
How do I incorporate the conditions into the final answer?
Edit: $s(t)$ is wrong. Because the solutions of the polynomial equation are imaginary and of special form, the correct solution involves writing $s(t)$ in terms of trigonometric functions.
ordinary-differential-equations
asked Jan 23 at 20:03
OscarOscar
351111
$endgroup$
add a comment |
$begingroup$
I have the beginner equation
$frac{d^2s}{dt^2} + 2frac{ds}{dt} + 2s = 0$ with the conditions $t_0 = 0, s_0 = 1, s'_0 = 1$.
Solving the characteristic polynomial I get the solutions $lambda_1 = -1 pm i$
which I have incorporated in the solution $s(t) = c_1 e^{(-1+i)t} + c_2 e^{(-1-i)t}$
How do I incorporate the conditions into the final answer?
Edit: $s(t)$ is wrong. Because the solutions of the polynomial equation are imaginary and of special form, the correct solution involves writing $s(t)$ in terms of trigonometric functions.
ordinary-differential-equations
asked Jan 23 at 20:03
OscarOscar
351111
$endgroup$
I have the beginner equation
$frac{d^2s}{dt^2} + 2frac{ds}{dt} + 2s = 0$ with the conditions $t_0 = 0, s_0 = 1, s'_0 = 1$.
Solving the characteristic polynomial I get the solutions $lambda_1 = -1 pm i$
which I have incorporated in the solution $s(t) = c_1 e^{(-1+i)t} + c_2 e^{(-1-i)t}$
How do I incorporate the conditions into the final answer?
Edit: $s(t)$ is wrong. Because the solutions of the polynomial equation are imaginary and of special form, the correct solution involves writing $s(t)$ in terms of trigonometric functions.
ordinary-differential-equations
ordinary-differential-equations
asked Jan 23 at 20:03
OscarOscar
351111
asked Jan 23 at 20:03
OscarOscar
351111
edited Jan 23 at 22:51
Oscar
asked Jan 23 at 20:03
OscarOscar
351111
asked Jan 23 at 20:03
OscarOscar
351111
asked Jan 23 at 20:03
OscarOscar
351111
351111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It should be $s(t) = c_1 e^{(-1+i)t} + c_2 e^{(-1-i)t}$.
Plug in $t=0$ and $s=1$ to find $1=c_1+c_2$.
Now differentiate, and plug in $t=0$, $s'=1$. Then solve simultaneous equations.
answered Jan 23 at 20:06
AlexandrosAlexandros
9281412
$endgroup$
$begingroup$
Ah- of course. Great, I will accept this solution as soon as the minimum wait time is over.
$endgroup$
– Oscar
Jan 23 at 20:11
add a comment |
$begingroup$
Your solution is not "wrong," but if you want to express the complex exponential in terms of trigonometric functions, use Euler's formula:
$$e^{ix}=cos x+isin x.$$
answered Jan 24 at 12:56
MathIsFunMathIsFun
1715
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It should be $s(t) = c_1 e^{(-1+i)t} + c_2 e^{(-1-i)t}$.
Plug in $t=0$ and $s=1$ to find $1=c_1+c_2$.
Now differentiate, and plug in $t=0$, $s'=1$. Then solve simultaneous equations.
answered Jan 23 at 20:06
AlexandrosAlexandros
9281412
$endgroup$
$begingroup$
Ah- of course. Great, I will accept this solution as soon as the minimum wait time is over.
$endgroup$
– Oscar
Jan 23 at 20:11
add a comment |
$begingroup$
It should be $s(t) = c_1 e^{(-1+i)t} + c_2 e^{(-1-i)t}$.
Plug in $t=0$ and $s=1$ to find $1=c_1+c_2$.
Now differentiate, and plug in $t=0$, $s'=1$. Then solve simultaneous equations.
answered Jan 23 at 20:06
AlexandrosAlexandros
9281412
$endgroup$
$begingroup$
Ah- of course. Great, I will accept this solution as soon as the minimum wait time is over.
$endgroup$
– Oscar
Jan 23 at 20:11
add a comment |
$begingroup$
It should be $s(t) = c_1 e^{(-1+i)t} + c_2 e^{(-1-i)t}$.
Plug in $t=0$ and $s=1$ to find $1=c_1+c_2$.
Now differentiate, and plug in $t=0$, $s'=1$. Then solve simultaneous equations.
answered Jan 23 at 20:06
AlexandrosAlexandros
9281412
$endgroup$
It should be $s(t) = c_1 e^{(-1+i)t} + c_2 e^{(-1-i)t}$.
Plug in $t=0$ and $s=1$ to find $1=c_1+c_2$.
Now differentiate, and plug in $t=0$, $s'=1$. Then solve simultaneous equations.
answered Jan 23 at 20:06
AlexandrosAlexandros
9281412
answered Jan 23 at 20:06
AlexandrosAlexandros
9281412
answered Jan 23 at 20:06
AlexandrosAlexandros
9281412
answered Jan 23 at 20:06
AlexandrosAlexandros
9281412
9281412
$begingroup$
Ah- of course. Great, I will accept this solution as soon as the minimum wait time is over.
$endgroup$
– Oscar
Jan 23 at 20:11
add a comment |
$begingroup$
Ah- of course. Great, I will accept this solution as soon as the minimum wait time is over.
$endgroup$
– Oscar
Jan 23 at 20:11
$begingroup$
Ah- of course. Great, I will accept this solution as soon as the minimum wait time is over.
$endgroup$
– Oscar
Jan 23 at 20:11
$begingroup$
Ah- of course. Great, I will accept this solution as soon as the minimum wait time is over.
$endgroup$
– Oscar
Jan 23 at 20:11
add a comment |
$begingroup$
Your solution is not "wrong," but if you want to express the complex exponential in terms of trigonometric functions, use Euler's formula:
$$e^{ix}=cos x+isin x.$$
answered Jan 24 at 12:56
MathIsFunMathIsFun
1715
$endgroup$
add a comment |
$begingroup$
Your solution is not "wrong," but if you want to express the complex exponential in terms of trigonometric functions, use Euler's formula:
$$e^{ix}=cos x+isin x.$$
answered Jan 24 at 12:56
MathIsFunMathIsFun
1715
$endgroup$
add a comment |
$begingroup$
Your solution is not "wrong," but if you want to express the complex exponential in terms of trigonometric functions, use Euler's formula:
$$e^{ix}=cos x+isin x.$$
answered Jan 24 at 12:56
MathIsFunMathIsFun
1715
$endgroup$
Your solution is not "wrong," but if you want to express the complex exponential in terms of trigonometric functions, use Euler's formula:
$$e^{ix}=cos x+isin x.$$
answered Jan 24 at 12:56
MathIsFunMathIsFun
1715
answered Jan 24 at 12:56
MathIsFunMathIsFun
1715
answered Jan 24 at 12:56
MathIsFunMathIsFun
1715
answered Jan 24 at 12:56
MathIsFunMathIsFun
1715
1715
add a comment |
add a comment |
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