Let ′′() + 4′() + 3() = (). zero response

Question

Let ′′() + 4′() + 3() = ()

Answer 1

I'm guessing what this question means, but it looks like a second order differential equation:

y"(x)+4y'(x)+3y(x)=0, which can be solved.

r²+4r+3=0=(r+3)(r+1)⇒y=Ae^-3x+Be^-x where A and B are arbitrary constants.

y'=-3Ae^-3x-Be^-x, y"=9Ae^-3x+Be^-x.

4y'=-12Ae^-3x-4Be^-x, 3y=3Ae^-3x+3Be^-x, and:

y"+4y'+3=9Ae^-3x+Be^-x-12Ae^-3x-4Be^-x+3Ae^-3x+3Be^-x=

Ae^-3x(9-12+3)+Be^-x(1-4+3)=0, confirming the solution.

This is, in fact, the characteristic solution. However, this solution is based on the right-hand side of the DE being zero. "()" could be something other than zero, but there's no way of determining what. It's not clear what "zero response" means mathematically. If the right-hand side is not zero, but is some function of x then the usual procedure would be to find the particular solution by making a guess based on the "look" of the right-hand side. This guessed function would involve non-arbitrary constants which can be evaluated through the first and second derivatives of the function and plugging them into the DE, so that the result is the same as the right-hand side of the given DE. The final solution would be the sum of the characteristic and particular solutions.

EXAMPLE

If the right-hand side is sin(x), then we might guess that y=asin(x)+bcos(x) might be a particular solution, where a and b are to be found.

y'=acos(x)-bsin(x), y"=-asin(x)-bcos(x).

y"+4y'+3y=-asin(x)-bcos(x)+4acos(x)-4bsin(x)+3asin(x)+3bcos(x)=

sin(x)(-a-4b+3a)+cos(x)(-b+4a+3b)=(2a-4b)sin(x)+(2b+4a)cos(x).

This needs to be the same as sin(x), so 2a-4b=1 and 2b+4a=0, same as 4b+8a=0.

So 2a-4b+4b+8a=1+0, 10a=1, a=1/10, b=-2a=-1/5. So this particular solution is y=sin(x)/10-cos(x)/5.

The complete solution is y=Ae^-3x+Be^-x+sin(x)/10-cos(x)/5. That is, y=y_c+y_p, the sum of the characteristic and particular solutions. This is the solution of the DE:

y"(x)+4y'(x)+3y(x)=sin(x).

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