Ans : 2(-2)-3(3)-(-2) = -11
Looks like the function on the right hand side is

We can write it in terms of the Heaviside function,

as

Now for the ODE: take the Laplace transform of both sides:


Solve for <em>Y</em>(<em>s</em>), then take the inverse transform to solve for <em>y</em>(<em>t</em>):






Answer:
y = m x + b equation for a straight line
m m' = -1 for perpendicular lines
Thus m' = -1/2 is required
Check:
-1/2 * -6 + 2 = = 3 + 2 = 5 = y for the last equation give
Answer:
14m +4+3m-3-2m
14m+3m-2m+4-3
15m+1
Step-by-step explanation:
this is equivalent