Getting into differential equationsI'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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Getting into differential equationsI&#039;m just getting into differential equations now and I&#039;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?

Jade Mejia · Accepted Answer

Step 1You 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 2It also works the other way: if you consider a solution of the equation, then      d          d      x            [    exp    ⁡    (    x    )    y    (    x    )    ]    =  0and then it has the given form.

Corbin Bradford · Accepted Answer

Step 1The 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 2Then 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.

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}.

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