[laplace transforms for solving differential equations] this is the heart of the post-midterm 2 material. here, i give you a number of examples. in a couple of parts, i intentionally make you deal with some "special" cases (e. g. a coefficient that come out to be zero) so that you know how to handle those. (a) for the differential equation: y' + 5y = e-4u(t), y(0) = 2 i. find the laplace transform y(s) of y(t). ii. mark the pole locations of y(s) in the complex frequency (s) plane. iii. take the inverse laplace transform of y(s) to obtain y(t) for t > 0.

idk

step-by-step explanation:

i don't know

add an image, and i will answer

step-by-step explanation:

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?