Answer:
The anwers are:
a) The slope field is attached.
b) The general solution is [tex]y(t)=c_{1}e^t+t+1[/tex]
c) The solution that it is exactly a straight line is y(t)=t+1 (when c1=0)
Step-by-step explanation:
y'(t)=y-t
y'(t)-y=-t
First we find the solution of the homogenous equaiton:
y'(t)-y=0
Considering [tex]y(t)=e^{rt}[/tex] where r is a constant
[tex]y'(t)=re^{rt}[/tex]
[tex]re^{rt}-e^{rt}=0[/tex]
[tex](r-1)e^{rt}=0[/tex]
[tex]e^{rt}[/tex] is never zero, so:
(r-1)=0
r=1
[tex]y(t)_{h}=c_{1}e^{t}[/tex]
The particular solution is given by:
y(t)=At+B
y'(t)=A
Hence,
y'(t)-y=-t
A-At-B=-t
[tex]\left \{ {{A-C=0} \atop {-A=-1}} \right.[/tex]
A=C=1
[tex]y(t)_{p}=t+1[/tex]
The general solution is the sum of y(t)h and y(t)p:
[tex]y(t)=c_{1}e^t+t+1[/tex]
When c1=0, y(t)=t+1 which is a straight line of slope 1 and intercept 1.
