Question

Question #000c1

Answer 1

`y=1-e^-x`
happens to be the solution.

Hope this may be the expected answer

Explanation:

The differential equation is given as

`f(x)=y'+2y-2=0`

`f(0)=0`

`dy/dx+2y-2=0`

`dy/dx+2(y-1)=0`

`dy/dx=-2(y-1)`

Separating the variables

`dy/(y-1)=-dx`

Integrating on both the sides

`intdy/(y-1)=-intdx`

`ln(y-1)=-x+c`

Simplifying

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

`y-1=e^-x*e^c`

if `e^c=C`

`y-1=Ce^-x`

`y=1+Ce^-x`

`f(0)=0`

`0=1+Ce^-0`

`0=1+C`

`C=-1`

Thus,

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

`y=1-e^-x`
happens to be the solution.

Hope this may be the expected answer

Answer 2

`y=2e^(-2x)`

Explanation:

`y'+2y-2=0`

After taking Laplace transform both sides,

`sY(s)-y(0)+2Y(s)-2=0`

`(s+2)*Y(s)-0-2=0`

`(s+2)*Y(s)=2`

`Y(s)=2/(s+2)`

After taking inverse Laplace transform, I found

`y=2e^(-2x)`