Is the system linear or nonlinear

physicsgirl199
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Homework Statement


3y(t)+2=x(t)

Homework Equations



k1y1(t) + k2y2(t) + 2(k1+k2) = k1x1(t)+k2x2(t)

The Attempt at a Solution



I know the system is non linear but I cannot explain why. It has something to do with 2(k1+k2) but I am unsure.
 
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physicsgirl199 said:

Homework Statement


3y(t)+2=x(t)

Homework Equations



k1y1(t) + k2y2(t) + 2(k1+k2) = k1x1(t)+k2x2(t)

The Attempt at a Solution



I know the system is non linear but I cannot explain why. It has something to do with 2(k1+k2) but I am unsure.
Why do you say you know it's non-linear?
 
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How are you defining the term "linear"? In some contexts, this term implies that the graph of the relationship is a straight line. In the context of linear transformations, the conditions for linearity are that ##L(x_1 + x_2) = L(x_1) + L(x_2)## and that ##L(cx) = cL(x)##.

physicsgirl199 said:
k1y1(t) + k2y2(t) + 2(k1+k2) = k1x1(t)+k2x2(t)
You put this equation in the Relevant equations section. How is it relevant to this problem?
 
it is both the properties you have described but I have combined them into one equation.

also it was marked as nonlinear in the solution manual
 
You could write the given equation as ##y(t) = \frac 1 3 x(t) - \frac 2 3##
Now check the two properties separately.
1) Is ##y(t_1 + t_2) = y(t_1) + y(t_2)##?
2) Is ##y(k \cdot t_1) = k \cdot y(t_1)##?

If both of the above are true for all values of t, the relationship is linear; otherwise, it's nonlinear.
 
physicsgirl199 said:
it is both the properties you have described but I have combined them into one equation.

also it was marked as nonlinear in the solution manual

I think it is a bit ambiguous.

If your equation ##3y(t)+2 = x(t)## describes a "dynamical system", then it is not linear because if ##(x_1(t),y_(t))## and ##(x_2(t),y_2(t))## are two solutions, the pair ##(x_1(t)+x_2(t),y_1(t)+y_2(t))## is not a solution, nor is ##(cx_1(t),cy_2(t))## for a constant ##c \neq 1##.

On the other hand the equation ##x - 3y = 2## is what would normally be called "linear equation"---meaning that the left-hand-side in ##f(x,y) = c## is a linear function of ##x,y##. Perhaps that is a bit of abuse of language, but it is nevertheless standard usage in describing equations. We do something similar when we call a differential equation such as ##dy/dx + 2 y = x^2## a (non-homogeneous) linear differential equation!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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