Question

Consider the differential equation y' = y-t

a) construct a slope field for this equation

b) find the general solution to this differential equation

c) there is exactly one solution that is given by a straight line. write the equation for this line and draw it on the slope

Answer 1

Answer:

The anwers are:

a) The slope field is attached.

b) The general solution is $y(t)=c_{1}e^t+t+1$

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 $y(t)=e^{rt}$ where r is a constant

$y'(t)=re^{rt}$

$re^{rt}-e^{rt}=0$

$(r-1)e^{rt}=0$

$e^{rt}$ is never zero, so:

(r-1)=0

r=1

$y(t)_{h}=c_{1}e^{t}$

The particular solution is given by:

y(t)=At+B

y'(t)=A

Hence,

y'(t)-y=-t

A-At-B=-t

$\left \{ {{A-C=0} \atop {-A=-1}} \right.$

A=C=1

$y(t)_{p}=t+1$

The general solution is the sum of y(t)h and y(t)p:

$y(t)=c_{1}e^t+t+1$

When c1=0, y(t)=t+1 which is a straight line of slope 1 and intercept 1.