Calculus

Lionking21

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Jun 16, 2019
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Hi, Can someone help me with These questions?

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Hello, and welcome to FMH! :)

For the first question, what are the characteristic/auxiliary roots?
 
I don't know what you mean by D = 0, but that's the correct root. What is the multiplicity of the root?
 
D is just the discriminant. So i was thinking something like. D=b^2-4ac. Where the discriminant I Got to 0. This means 1 root, whereas the formula used is -b/2a. Where i get 2/2=1. I am just in doubt about if (c1+c2t)e^t is correct?
 
Yes, you have a root of multiplicity 2 (repeated root) and so your two linearly independent solutions are:

[MATH]y_1(x)=c_1e^t[/MATH]
[MATH]y_2(x)=c_2te^t[/MATH]
And so by the principle of superposition, the general solution is:

[MATH]y(x)=y_1(x)+y_2(x)=(c_1+c_2t)e^t[/MATH]
 
For the next question, I would write:

[MATH]f(x,y,z)=\frac{x^2+y^2+z^2}{x^2+y^2}[/MATH]
Now, we know we cannot have division by zero, so where in the 3-space would division by zero occur?
 
X+Y>0? In first question and second question a sphere?

Suppose \((x,y)=(1,-1)\). This would violate \(x+y>0\), but would it cause division by zero? There is only one ordered pair \((x,y)\in\mathbb{R}\) that can cause division by zero. What would that be?
 
If both \(x\) and \(y\) are zero, that is \((x,y)=0\), then the denominator is zero, since \(0^2+0^2=0\). Where in the 3-space is this true?
 
The locus of points we therefore wish to exclude from the domain is \((0,0,z)\), which is the \(z\)-axis.

What did you get for the next problem?
 
If we take the original form of the function, we get:

[MATH]1+\frac{z^2}{x^2+y^2}=1[/MATH]
[MATH]\frac{z^2}{x^2+y^2}=0[/MATH]
This implies:

[MATH]z=0[/MATH]
This is the \(xy\)-plane, but recall we must exclude the origin of that plane, as it lies outside of the function's domain.
 
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