how do you check if your general solution in DE is correct? I tried solving equations using different methods and plugged in same x and y values. But the arbitrary constant ended up differently.
See once you solve a DE you get an arbitrary constant by putting the values of x and y given.. When the values change the value of arbitrary constant also changes.
Is that what you asked? or do you have a different query?
On general principles, if u have found a general solutionto a de, then u should be able to substitute it in the original de to check
ie not a particular solution
hmm. but i have rechecked both my solution and nothing seemed wrong. But when I substituted my value of x and y, they both give different constant. Does it mean that my general solution is wrong?
That's what I am saying, if u substitute the general solution into the de, you can that way guarantee that it right.
The specifics of a particular solution involve arbitrary constants whose value depends on initial conditions, that is just a calculation.
Hmm. How do I substitue my general solution in my DE?
What sort of de is?
like dy/dx = f(x,y)...
The only difference between this kind of substitution and the usual sort is that u are substituting a function....
firstorder >.<
Do u want an example..?
Yeah!
OK, y = 1/2 x^2 + 1/2 and dy/dx = x Simplest one I can think of.....
SO if u diff the y u get dy/dx = x, right?
hmmm..
Of course, your equation probably has more terms....
How am I going to substitute my general solution to my differential equation? Sorry I still don't get it. >.<
Should I take the derivative of my gen sol and DE wrt to x/y and see if they are equal?
Ok, try y = 2e^x -(x^2 +2x +2) and dy/dx = y +x^2
So diff the y to get dy/dx for one part. Then sub for y in the de and see if both sides are the same.
If they are, your solution satisfies the de.
oh so im going to isolate one of my variable and substitue it to DE?
I'm not sure about the terminology but in the example I just gave u the de is dy/dx = y +x^2 and the solution is y = 2e^x -(x^2 +2x +2) and I want to verify that the solution satisfies the de. So if I differentiate y (my solution) I can get the dy/dx part of my de. And if I substitute 2e^x -(x^2 +2x +2) for y in the de, then see what I have...
U end up with 2e^x - 2x -2 = 2e^x -2x -2 True.
oh! thank you very much! i got it nnow!
:-)
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