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For the 2nd order linear PDE below, please give method(s) to solve it, working, a solution, and what conditions the solution can exist?

$$\sin(t)\frac{\partial^2y}{\partial t^2}+\cos(t)\frac{\partial y}{\partial t}+\cos(t)\frac{\partial ^2y}{\partial x^2}=0$$

I tried a solution like $y=y(x,t)=Ae^{\alpha t}sin(Ct)+Be^{\beta t}sin(Dt)+Fe^{\zeta x}sin(Ex)+Ge^{\gamma x}sin(Hx)$ via method of undetermined coeffs. However I found this made coeffs hard to determine.

Can it be solved via transformations?

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    $\begingroup$ Since the equation takes the derivative of $y$ with respect to both $t$ and $x$, isn't this a partial differential equation, a PDE rather than an ODE? $\endgroup$ Jul 13 '16 at 22:39
  • $\begingroup$ Possibly, but if so I don't understand why? $\endgroup$ Jul 13 '16 at 22:54
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    $\begingroup$ It's a PDE, you can eventually solve it by separation of variabiles. Put the last term on the RHS and express $y=X(x)T(t)$ as the product of two functions. Re arrange to find an expression of only $t$ on the LHS and of $x$ on the RHS. As the two members have to be equal but vary with two different Independent variabiles you will assume that both LHS and RHS are constant. Split the problem in two ODE to be solved to find $X$ and $Y$. $\endgroup$
    – N74
    Jul 13 '16 at 23:31
  • $\begingroup$ Updated question & title - now in PDE syntax. Why was it not a PDE when in ODE notation? $\endgroup$ Jul 14 '16 at 0:44
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$$\sin(t)\frac{\partial^2y}{\partial t^2}+\cos(t)\frac{\partial y}{\partial t}+\cos(t)\frac{\partial ^2y}{\partial x^2}=0$$

You tried a solution like $y=y(x,t)=Ae^{\alpha t}sin(Ct)+Be^{\beta t}sin(Dt)+Fe^{\zeta x}sin(Ex)+Ge^{\gamma x}sin(Hx)$ via method of undetermined coefficients.

This can succeed only if each of the terms is solution of the PDE. Putting each one into the PDE shows that no one of these terms satisfies the PDE. So, the method of undetermined coefficients cannot be successful on this basis.

The PDE has an infinity of solutions. Some are easy to obtain. For example :

Case of solutions including only the variable $x \quad\to\quad \frac{d^2y}{dx^2}=0$ : $$y(x,t)=c_1t+c_2 \quad\text{ is solution of the PDE}$$

Case of solutions including only the variable $t \quad\to\quad \sin(t)\frac{d^2y}{dt^2}+\cos(t)\frac{dy}{dt}=0$ : $$y(x,t)=c_1\ln\left(\tan\left(\frac{t}{2} \right) \right)+c_2 \quad\text{ is solution of the PDE}$$

Case of solutions on the form $y(x,t)=f(x)g(t) \quad\to\quad \begin{cases} f''=C\:f \quad\text{any constant } C \\ \tan(t)g''+g'=-C\:g \end{cases}$

(As far as I know, there is no closed form for $g(t)$ : Numerical solving required in this case)

General case :

The PDE has an infinity of solutions. Without giving the boundary conditions, the question is too wide. You have to add the boundary conditions in the wording of the question if you expect a more specific answer.

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  • $\begingroup$ 1) why is there no closed form for $g(t)$? 2) the question is currently in PDE syntax, and originally in ODE syntax. Do you know why the original ODE equation should be a PDE? $\endgroup$ Jul 14 '16 at 10:01
  • $\begingroup$ 1) why is there no closed form for g(t) ? Because there is no available standard special function convenient to express the solutions of this ODE. Many special functions where defined in order to solve some kind of ODEs (About special functions : fr.scribd.com/doc/14623310/… ) But of course, this is not the case for many ODEs among the infinity of forms of ODEs. $\endgroup$
    – JJacquelin
    Jul 14 '16 at 10:26
  • $\begingroup$ 2) Do you know why the original ODE equation should be a PDE? My answer : What original ODE are you mentioning ? I don't see any in your wording. I cannot understand what is your comment about "originally ODE syntax". $\endgroup$
    – JJacquelin
    Jul 14 '16 at 10:31
  • $\begingroup$ @jjaquelin Please see earlier revision of question, where question was in ODE form. $\endgroup$ Jul 14 '16 at 10:35
  • $\begingroup$ How can I found the earlier wording of the question ? $\endgroup$
    – JJacquelin
    Jul 14 '16 at 10:38

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