Answer with explanation:
The given differential equation is
y" -y'+y=2 sin 3x------(1)
Let, y'=z
y"=z'

Substituting the value of , y, y' and y" in equation (1)
z'-z+zx=2 sin 3 x
z'+z(x-1)=2 sin 3 x-----------(1)
This is a type of linear differential equation.
Integrating factor

Multiplying both sides of equation (1) by integrating factor and integrating we get


Answer:
y = 4/3x - 4
Step-by-step explanation:
you are given b
so y = mx -4
you can look at the graph and see the slope from there. like the y-intercept and the x-intercept
(0,-4) and (3,0)
slope formula is y2-y1/x2-x1
and so you do -4-0/0-3
-4/-3.
so your slope formula would be
y = 4/3x - 4
812.30/83 = 9.78674698.....
Nearest tenth means one number under decimal place so...
812.30/83 ~ 9.8
Explanation: The 9.7... rounded up to 9.8 because the number after the tenths place (in this case the number after the 7) is greater or equal to 5.