Question #5b598

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)#

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#