Question

Question #5b598

Answer 1

`y=x-1+5e^(-x)`

Explanation:

Linear differential eqn of type

`(dy)/(dx)+Py=Q`

where `P,Q` are either constants or functions of `x`

we use an integrating factor

`IF=e^(int(Pdx))`

in this case `P=1`

`(dy)/(dx)+y=x`

`IF=e^(intdx)=e^x`

multiply the ode by `IF`

`e^x(dy)/(dx)+e^xy=xe^x`

LHS is the result of the product rule for differentiation

`d/(dx)(e^xy)=xe^x`

Integrating both sides wrt `x`

`e^xy=int(xe^x)dx`

RHS is integrated by parts

`I_(RHS)=uv-intvu'dx`

`u=x=>u'=1`
`v'=e^x=>v=e^x`

so

`e^xy=xe^x-inte^xdx`

`e^xy=xe^x-e^x+C`

to find the constant we use the boundary condition

`y(0)=4`

`e^0xx4=0xxe^0-e^0+C`

`4=-1+C=>C=5`

`e^xy=xe^x-e^x+5`

tidy up by `xxe^(-x)`

`y=x-1+5e^(-x)`

Answer 2

`y=5e^-x+x-1`

Explanation:

First solve `dy/(dx)+y=0` by separable variables:
`intdy/y=-int1dx`

`ln y=-x + ln A` where `ln A` is the arbitrary constant of integration

Hence `y=Ae^-x` (by raising `e` to the power of each side).
Now for the steady state, set spot that y=x-1 is a particular solution of the original equation on the left. (Or try out `y-=ax+b` and equate coefficients to find `a` and `b`.) So the general solution is:
`y=Ae^-x+x-1` with `A` chosen to make `y(0)=4`:
`4=Ae^0+0-1` so `A=5` and the final solution is:

`y=5e^-x+x-1`

Check:
`dy/(dx)+y=(cancel((-5)e^x)+cancel 1-0) + (cancel(5e^-x)+x-cancel 1)=x`
and
`y(0)=5e^0+0-1=5 xx 1 -1 = 4`