As others have pointed out, you skipped the last step in both solutions, which is to plug them back into the original DE to check if they’re valid. Doing so for the first solution, we get kx+C = kx + e^k This means the solution is only valid for C=e^k. Thus the actual general linear solution is y=kx+e^k. k can still be any real number, and this solution is valid over all of R. For the second solution, using the fact that y’ = ln(-x), we get xln(-x)-x+C = xln(-x) + e^ln(-x) -> xln(-x)-x+C = xln(-x) -x So C=0, and we only get one solution y=xln(-x)-x. Obviously this is only defined for x
It seemed interesting to me (I don't remember if mentioned in the video) the fact that the linear solutions of the equation are tangent lines to the other solution x(ln(-x) - 1).
Nice,but the solution is not kx+c,but y=kx+e^k.and y=c is not asolution unless c=1.and also y= xln(-x)-x+c is not asolution unless c=0. Since you differenciated the original equation you should plug the solutions you got and see which constant c would give you the correct answer😃
Very nice ODE👍
Really liked the explanation!
Glad to hear that
As others have pointed out, you skipped the last step in both solutions, which is to plug them back into the original DE to check if they’re valid.
Doing so for the first solution, we get
kx+C = kx + e^k
This means the solution is only valid for C=e^k. Thus the actual general linear solution is y=kx+e^k. k can still be any real number, and this solution is valid over all of R.
For the second solution, using the fact that y’ = ln(-x), we get
xln(-x)-x+C = xln(-x) + e^ln(-x)
-> xln(-x)-x+C = xln(-x) -x
So C=0, and we only get one solution y=xln(-x)-x. Obviously this is only defined for x
Good points!
It seemed interesting to me (I don't remember if mentioned in the video) the fact that the linear solutions of the equation are tangent lines to the other solution x(ln(-x) - 1).
does this form of D.E. have any applications?
@ 3:32 Shouldn’t you substitute into the original equation to see which constants work.
Yess
Nice,but the solution is not kx+c,but y=kx+e^k.and y=c is not asolution unless c=1.and also y= xln(-x)-x+c is not asolution unless c=0. Since you differenciated the original equation you should plug the solutions you got and see which constant c would give you the correct answer😃
@@mrityunjaykumar4202 try to plug the solutions in the original solution,and you will understand.
This is the reason for checking solutions to make certain that they are not extraneous.
Only the linear solution is valid. The other one doesnt check when you put it back in the original equation.
And also not all linear solution works but y=kx+e^k.
@@yoav613 Yes true.
Y=1
@@bobjazz2000 That is not a new solution. It corresponds to the linear solution suggested above when k=0.
You should have known your linear solution was incorrect because there is only one constant of integration, not two.
y=1
x=t-2t/lnt...y=tlnt-t
dude are we in same class bcs i was learning clairaut today.
No way! 😁
❤❤❤
1st!