2nd order ODE with Sin function.

• rizvi71
In summary, the conversation is about solving a 2nd order differential equation with the given form d^2 X/dt^2 +dX/dt +sinX=0. The speakers discuss whether or not the equation is solvable and provide a solution for the case of "sint" being the correct function. They also mention the possibility of "sinx" being the correct function and ask for a solution in that case.
rizvi71
Hey guys!
I'm trying to solve a 2nd order differential equation. I am quite familiar with the method of solving these equations like treat them like characteristic equation ODE.
but there's a question which I really want to solve.
question is d^2 X/dt^2 +dX/dt +sinX=0.
How should I solve this??

rizvi71 said:
Hey guys!
I'm trying to solve a 2nd order differential equation. I am quite familiar with the method of solving these equations like treat them like characteristic equation ODE.
but there's a question which I really want to solve.
question is d^2 X/dt^2 +dX/dt +sinX=0.
How should I solve this??

That is not a linear ode and mathematica can't solve it... Is it solvable?

Hmmmm... my teacher gave me an assignment to solve this. So are you implying that its unsolvable??

rizvi71 said:
Hmmmm... my teacher gave me an assignment to solve this. So are you implying that its unsolvable??

Maybe he meant to write sint instead of sinx. If that's the case, then the solution is easy to find. Otherwise, unless you use some kind of "trick" I don't see how you can solve it

could you please give me some kind of a hint??
If it is "sint" how can i solve it??

rizvi71 said:
could you please give me some kind of a hint??
If it is "sint" how can i solve it??

Sure I can(if it is sint)

\begin{array}\\x''+x'+sint=0\\
x(t)=x_0(t)+x_p(t)\\
x_0''+x_0'=0\\
r^2+r=0\Leftrightarrow r(r+1)=0\Leftrightarrow r_1=0, r_2=-1\\
x_0(t)=c_1e^{0t}+c_2e^{-t}=c_1+c_2e^{-t}\end{array}

Let $$x_p=Asint+Bcost$$
Then
\begin{array}\\x''_p+x'_p+sint=0\Leftrightarrow (Asint+Bcost)''+(Asint+Bcost)'+sint=0\\-Asint-Bcost+Acost-Bsint+sint=0\Leftrightarrow sint(-A-B+1)+cost(-B+A)=0
\\\begin{Bmatrix}
-A& -B& =-1\\
-B& +A& =0
\end{Bmatrix}\Leftrightarrow \begin{Bmatrix}
B& +B& =1\\
& A& =B
\end{Bmatrix}\Rightarrow 2B=1\Leftrightarrow B=\frac{1}{2}=A\\
x_p=\frac{1}{2}(sint+cost)
\end{array}

Therefore, $$x(t)=c_1+c_2e^{-t}+\frac{1}{2}(sint+cost)$$

thanx a lot buddy !
Really appreciate it !

rizvi71 said:
thanx a lot buddy !
Really appreciate it !

No problem. Just ask your teacher and tell me whether it is sint or sinx.

Yeah sure will.

1. What is a second order ODE with a Sin function?

A second order ODE with a Sin function is a type of differential equation that involves a second derivative of a function and a Sin function. It can be written in the form of y'' + p(x)y' + q(x)y = f(x)sin(g(x)), where p(x), q(x), and f(x) are functions of x and g(x) is a Sin function.

2. What is the solution to a second order ODE with a Sin function?

The solution to a second order ODE with a Sin function depends on the specific equation and initial conditions. In general, the solution can be found by using various techniques such as variation of parameters, undetermined coefficients, or the Laplace transform.

3. What are the applications of second order ODEs with Sin functions?

Second order ODEs with Sin functions have many applications in physics, engineering, and other fields. They can be used to model systems that involve oscillations or vibrations, such as a pendulum, a spring-mass system, or an electrical circuit with an AC power source.

4. How do I determine if a second order ODE with a Sin function is linear or nonlinear?

A second order ODE with a Sin function is linear if the coefficients of y'', y', and y are constants. It is nonlinear if the coefficients are functions of x. In other words, if the equation can be written in the form y'' + p(x)y' + q(x)y = f(x)sin(g(x)), where p(x), q(x), and f(x) are functions of x, then it is a nonlinear ODE.

5. Can a second order ODE with a Sin function have non-constant coefficients?

Yes, a second order ODE with a Sin function can have non-constant coefficients. This means that the coefficients of y'', y', and y can be functions of x. In this case, the equation is a nonlinear ODE and its solution may require more advanced techniques to find.

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