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)
We are given with the expression or equation s<span>in2x cosx + cos 2x sin x = √3/2. The expression can be patterned from the trigonometric identity sin (a + b) = sin a cos B + cos A sin B. In this case, the expression is equal to sin 3x = sqrt 3 /2. using arc sign, x is equal to pi/9 </span>
Answer:
is there anything else??
Step-by-step explanation:
Hey there!
So, first, you need to rearrange the second equation so y is in front.
x+y=10
Subtract x.
y=10-x
Plug that in for y.
2x+3(10-x)=45
2x+30-3x= 45
Subtract 30 on both sides.
-x= 15
Divide by -1 on both sides.
x=-15.
I hope this helps!
~kaikers
