Examples of using The general solution in English and their translations into Thai
{-}
-
Colloquial
-
Ecclesiastic
-
Ecclesiastic
-
Computer
That is the general solution.
Figure out the derivative of the general solution.
You get the general solution of the homogeneous.
When we have two real roots, then this is the general solution.
So this is the general solution to this differential equation.
So that at least gives you an intuitive feel of why this isn't the general solution.
But it's not the general solution.
The general solution to the nonhomogeneous equation is going to be the sum of the two.
And we're asked to find the general solution to this differential equation.
We said, well if those are the roots and we substitute it back into this formula for the general solution, we get all of this.
And this is the general solution of this differential equation.
Your two roots are lambda plus or minus mu i, then the general solution is going to be this.
And that it's the general solution for this non-homogeneous equation.
But you see that once you know the pattern, or once you know that this is going to be the general solution, they're pretty easy to solve.
So anyway, this was the general solution to this differential equation.
That's what I mean when I say that h is a solution-- and actually, let's just say that h is the general solution for this homogeneous equation.
And now we if we want the general solution we just have to throw this right back into that.
And so, now we just substitute our initial conditions into our general solution and the derivative of the general solution, and hopefully we can get meaningful answers.
And you know how to solve the general solution of the homogeneous equation if this were 0.
And now, like I just showed you before I cleared the screen, our general solution of this non-homogeneous equation is going to be our particular solution plus the general solution to the homogeneous equation.
So the general solution to this differential equation is y squared over 2 minus x squared over 2 is equal to c.
When you have a repeated root of your characteristic equation, the general solution is going to be-- you're going to use that e to the, that whatever root is, twice.
So the general solution of this differential equation is y is equal to c1 times e-- let's use our first r-- e to the 3/2 x, plus c2 times e to the 1/2 x.
Now the problem with this solution, and why it's not the general solution, is if you use one of these initial conditions, you can solve for a c, right?
That the general solution of this non-homogeneous equation is actually the general solution of the homogeneous equation plus a particular solution. .
And then add them to the general solution for the homogeneous equation, if this was a 0 on the right-hand side.
Because the general solution on the homogeneous one that was the most general solution, and now we're adding a particular solution that gets you the g of x on the right-hand side.
And now if we want the general solution, we add to that the general solution of the homogeneous equation.
So we get y-- the general solution is y is equal to e to the lambda x, times-- let's add up the two cosine mu x terms.
We know that the general solution to the homogeneous equation is C1e to the 4x plus C2e to the minus x, right?