Consider the differential equation y' = y-ta) construct a slope field for this equationb) find the general solution to this differential equationc) there is exactly one solution that is given by a straight line. write the equation for this line and draw it on the slope

Accepted Solution

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-ty'(t)-y=-tFirst we find the solution of the homogenous equaiton:y'(t)-y=0Considering [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)=0r=1[tex]y(t)_{h}=c_{1}e^{t}[/tex]The particular solution is given by:y(t)=At+By'(t)=AHence,y'(t)-y=-tA-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.