Getting into differential equations. I'm just getting into differential equations now and I've got to show that the given y(x) is a solution to the differential equation: u'+u=0, y(x)=Ce^{-x}.

lunja55 2022-09-22 Answered
Getting into differential equations
I'm just getting into differential equations now and I've got to show that the given y(x) is a solution to the differential equation:
u + u = 0   ,   y ( x ) = C e x
How do I tackle this?
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Answers (2)

Jade Mejia
Answered 2022-09-23 Author has 8 answers
Step 1
You know that the derivative of y(x)expx is a constant, so just compute it:
0 = d d x [ exp ( x ) y ( x ) ] = exp ( x ) y ( x ) + exp ( x ) y ( x )
and as the exp is never = 0, y ( x ) + y ( x ) = 0.
Step 2
It also works the other way: if you consider a solution of the equation, then
d d x [ exp ( x ) y ( x ) ] = 0
and then it has the given form.
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Corbin Bradford
Answered 2022-09-24 Author has 1 answers
Step 1
The point is that you plug y ( x ) = C e x into the differential equation at the place of u , and of course the derivative of y(x) at the place of u′ (do you know how to differentiate y(x)?).
Step 2
Then you must show that the equality in the differential equation holds, that is y ( x ) + y ( x ) = 0. So
y ( x ) + y ( x ) = C e x + C e x = ( C + C ) e x = 0 e x = 0.
So it is, in fact, a solution since it equals zero as it should.
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