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1 answer:
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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:
The graph has a removable discontinuity at x=-2.5 and asymptoe at x=2, and passes through (6,-3)
Step-by-step explanation:
A rational equation is a equation where
where both are polynomials and q(x) can't equal zero.
1. Discovering asymptotes. We need a asymptote at x=2 so we need a binomial factor of
in our denomiator.
So right now we have
2. Removable discontinues. This occurs when we have have the same binomial factor in both the numerator and denomiator.
We can model -2.5 as
So we have as of right now.
Now let see if this passes throught point (6,-3).
So this doesn't pass through -3 so we need another term in the numerator that will make 6,-3 apart of this graph.
If we have a variable r, in the numerator that will make this applicable, we would get
Plug in 6 for the x values.
So our rational equation will be
or
We can prove this by graphing
