Examples of using General solution in English and their translations into Turkish
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Colloquial
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Ecclesiastic
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Ecclesiastic
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Computer
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Programming
Get your general solution.
Now let's take the derivative of our general solution.
You get the general solution of the homogeneous.
You can figure out a general solution.
And that it's the general solution for this non-homogeneous equation.
And now we have a truly general solution.
Maybe my general solution is just y is equal to some constant times e to the minus 2x.
That is the general solution.
So the general solution to this differential equation is y squared over 2 minus x squared over 2 is equal to c.
But it's not the general solution.
We know that the general solution to the homogeneous equation is C1e to the 4x plus C2e to the minus x, right?
This differential equations problem was literally just a problem in using the quadratic equation.And once you figure out the r's you have your general solution.
I will keep our general solution there.
But the general solution of the ordinary three-dimensional wave equation is well known, and can be given rather explicit form.
I haven't proven that is the most general solution, but I think you have the intuition, right?
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.
Furthermore, Bateman asserted that the most general solution of the Laplace equation arises in this way.
That the general solution of this non-homogeneous equation is actually the general solution of the homogeneous equation plus a particular solution. .
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?
You do that by getting the characteristic equation r squared minus 3r minus 4 is equal to 0. You get the solutions, r is equal equal to 4, r is equal to minus 1,and then you get that general solution.
And now if we want the general solution, we add to that the general solution of the homogeneous equation.
But what this gives us, if we make that simplification,we actually get a pretty straightforward, general solution to our differential equation, where the characteristic equation has complex roots.
So we can call this the most general solution-- I don't know. I will just call it y. It is our general solution C1e to the 4x plus C2e to the minus x plus our particular solution we found.
We learned that if the roots of our characteristic equation are r is equal to lambda plus orminus mu i, that the general solution for our differential equation is y is equal to e to the lambda x times c1 with some constant cosine of mu x, plus c2 times sine of mu x.
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.
Provided that the speed c{\displaystyle c} is a constant, not dependent on frequency(the dispersionless case),then the most general solution is p f( c t- x)+ g( c t+ x){\displaystyle p= f( ct-x)+ g( ct+x)} where f{\displaystyle f} and g{\displaystyle g} are any two twice-differentiable functions.
So the most general solution to this differential equation is y-- we could say y of x, just to hit it home that this is definitely a function of x-- y of x is equal to c1e to the minus 2x, plus c2e to the minus 3x.
So the general solution to this differential equation is y is equal to e to the lambda x-- well lambda is minus 2-- minus 2x times c1 cosine of mu x-- but mu is just 1-- so c1 cosine of x, plus c2 sine of mu x, when mu is 1, so sine of x.
(1991) and Gillman(1992) both show a more general solution where the car is(uniformly) randomly placed but the host is not constrained to pick uniformly randomly if the player has initially selected the car, which is how they both interpret the statement of the problem in Parade despite the author's disclaimers.