Need help solving the differential equation y’-y=x

Question

Can you help me understand the steps in solving this equation? Will you always use e^() in these types of equations? 

Answer 1

 

It is linear  :

 y' + P(x)y = q(x)

  y'  + (-1)y = x

 

  P(x) = -1  and q(x) =x

 

  integrating factor :   exp(integral (-1)) = exp( -x)

 

Multiplies everything by integrating factor:

 

exp(-x)*y' - y*exp(-x) = x*exp(-x)

 

 Dx [ exp(-x)*y ] = x*exp(-x)

 

------      integrating by parts: ----------

 

U = x dv = exp(-x)

dU = dx v = -exp(-x)


-x*exp(-x) - integral -exp(-x) dx

-x(exp(-x)) + integral exp(-x) dx

-x(exp(-x)) - exp(-x)

-exp(-x) [ x + 1]


check:

-exp(-x) + (x+1)(exp(-x))

exp(-x) [ -1 + x + 1] <--- yes it is correct

 

--------------------------------------------------

 

exp(-x)*y = -exp(-x) [ x + 1] + c

 

y = -1(x+1) + c(exp(x))

 

 

check:

 

y' = -1 +c(exp(x))  

 

y' - y = -1 + c(exp(x)) - [(-1)(x+1) +c(exp(x))] =

 

            -1 + c(exp(x)) + (x+1) - c(exp(x)) =

 

             YES it works, everything cancels expect the x

 

the solution to this differential equation is :

y = -1(x+1) + c(exp(x))

 

 

 

Answer 2

y' - y = x is linear

 

Integrating factor:  e^∫(-1)dx  = e^-x

 

Multiply the differential equation by the integrating factor:

 

y'e^-x - e^-xy = xe^-x

 

From the product rule, we see that the left hand side of the equation is (e^-xy)'.

 

So, (e^-xy)' = xe^-x

 

Integrating both sides with respect to x, we get e^-xy = ∫xe^-xdx

 

   The integral on the right hand side can be evaluated by using integration by parts:

 

        u = x   du = dx

 

        dv = e^-xdx     v = ∫dv = ∫e^-xdx = -e^-x

 

So, e^-xy = -xe^-x + ∫e^-xdx

 

      e^-xy = -xe^-x - e^-x + C

 

         y = -x - 1 + Ce^x

 

Check:  Show that y' - y = x

 

           y' - y =  (-x - 1 + Ce^x)' - (-x -1 + Ce^x) 

 

                    = (-1 + Ce^x) - (-x - 1 + Ce^x) = x