Answer:
x^2-3x=4
x^2-3x-4=0
x^2-4x+x-4=0
x(x-4)+1(x-4)=0
(x+1)(x-4)=0
x=-1,4
x- intercepts in the point (-1,0) and (4,0)
Step-by-step explanation:
Answer:
x⁵ +2x³ -3x², degree 5, 3 terms
Step-by-step explanation:
We assume you intend your expression to be ...
2x³ -3x² +x⁵
The superscript numbers are exponents. Each indicates the degree of the term. In standard form, terms are listed in decreasing order by degree:
x⁵ +2x³ -3x² . . . . standard form
The degree of the polynomial is the degree of the highest-degree term: 5.
The number of terms is the number of products in the sum: 3.
This one was though, but I think it might be y= -1/12x - 53/12.
Hope I helped!
Answer:
(3,4)
Step-by-step explanation:
Answer:
Please read the complete procedure below:
Step-by-step explanation:
You have the following initial value problem:
a) The algebraic equation obtain by using the Laplace transform is:
![L[y']+2L[y]=4L[t]\\\\L[y']=sY(s)-y(0)\ \ \ \ (1)\\\\L[t]=\frac{1}{s^2}\ \ \ \ \ (2)\\\\](https://tex.z-dn.net/?f=L%5By%27%5D%2B2L%5By%5D%3D4L%5Bt%5D%5C%5C%5C%5CL%5By%27%5D%3DsY%28s%29-y%280%29%5C%20%5C%20%5C%20%5C%20%281%29%5C%5C%5C%5CL%5Bt%5D%3D%5Cfrac%7B1%7D%7Bs%5E2%7D%5C%20%5C%20%5C%20%5C%20%5C%20%282%29%5C%5C%5C%5C)
next, you replace (1) and (2):
(this is the algebraic equation)
b)
![sY(s)+2Y(s)-8=\frac{4}{s^2}\\\\Y(s)[s+2]=\frac{4}{s^2}+8\\\\Y(s)=\frac{4+8s^2}{s^2(s+2)}](https://tex.z-dn.net/?f=sY%28s%29%2B2Y%28s%29-8%3D%5Cfrac%7B4%7D%7Bs%5E2%7D%5C%5C%5C%5CY%28s%29%5Bs%2B2%5D%3D%5Cfrac%7B4%7D%7Bs%5E2%7D%2B8%5C%5C%5C%5CY%28s%29%3D%5Cfrac%7B4%2B8s%5E2%7D%7Bs%5E2%28s%2B2%29%7D)
(this is the solution for Y(s))
c)
![y(t)=L^{-1}Y(s)=L^{-1}[\frac{4}{s^2(s+2)}+\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+L^{-1}[\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}](https://tex.z-dn.net/?f=y%28t%29%3DL%5E%7B-1%7DY%28s%29%3DL%5E%7B-1%7D%5B%5Cfrac%7B4%7D%7Bs%5E2%28s%2B2%29%7D%2B%5Cfrac%7B8%7D%7Bs%2B2%7D%5D%5C%5C%5C%5C%3DL%5E%7B-1%7D%5B%5Cfrac%7B4%7D%7Bs%5E2%28s%2B2%29%7D%5D%2BL%5E%7B-1%7D%5B%5Cfrac%7B8%7D%7Bs%2B2%7D%5D%5C%5C%5C%5C%3DL%5E%7B-1%7D%5B%5Cfrac%7B4%7D%7Bs%5E2%28s%2B2%29%7D%5D%2B8e%5E%7B-2t%7D)
To find the inverse Laplace transform of the first term you use partial fractions:
Thus, you have:
![y(t)=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}\\\\y(t)=L^{-1}[\frac{-1}{s}+\frac{2}{s^2}]+L^{-1}[\frac{1}{s+2}]+8e^{-2t}\\\\y(t)=-1+2t+e^{-2t}+8e^{-2t}=-1+2t+9e^{-2t}](https://tex.z-dn.net/?f=y%28t%29%3DL%5E%7B-1%7D%5B%5Cfrac%7B4%7D%7Bs%5E2%28s%2B2%29%7D%5D%2B8e%5E%7B-2t%7D%5C%5C%5C%5Cy%28t%29%3DL%5E%7B-1%7D%5B%5Cfrac%7B-1%7D%7Bs%7D%2B%5Cfrac%7B2%7D%7Bs%5E2%7D%5D%2BL%5E%7B-1%7D%5B%5Cfrac%7B1%7D%7Bs%2B2%7D%5D%2B8e%5E%7B-2t%7D%5C%5C%5C%5Cy%28t%29%3D-1%2B2t%2Be%5E%7B-2t%7D%2B8e%5E%7B-2t%7D%3D-1%2B2t%2B9e%5E%7B-2t%7D)
(this is the solution to the differential equation)
