Answer:
a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)
Step-by-step explanation:
Let's solve by separating variables:
a) x’=t–sin(t), x(0)=1
Apply integral both sides:
where k is a constant due to integration. With x(0)=1, substitute:
Finally:
b) x’+2x=4; x(0)=5
Completing the integral:
Solving the operator:
Using algebra, it becomes explicit:
With x(0)=5, substitute:
Finally:
c) x’’+4x=0; x(0)=0; x’(0)=1
Let 
be the solution for the equation, then:
Substituting these equations in <em>c)</em>
This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>
![x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)](https://tex.z-dn.net/?f=x%3De%5E%7B%5Calpha%20t%7D%5BAsin%5Cbeta%20t%2BBcos%5Cbeta%20t%5D%5C%5C%5C%5Cx%3De%5E%7B0%7D%5BAsin%28%282%29t%29%2BBcos%28%282%29t%29%5D%5C%5C%5C%5Cx%3DAsin%28%282%29t%29%2BBcos%28%282%29t%29)
Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:
Finally:
Answer:
Step-by-step explanation:
Since log is defined by all positive real numbers
therefore domain is all positive real number that is ( 0,∞)
Range is given by real numbers
inverse of the given function is (10^x)/7
Whose domain is all real numbers and range is all positive real number
And since we know that domain of function and range of its inverse
& range of a function and domain of its inverse is same
which we are getting in the problem
so answer is justified
Answer:
sqrt(-3)
Step-by-step explanation:
solve using sqrt( b^2 - 4ac)
substitute values from original equation to find discriminant
The first and the third equation can be solved by using the zero product property. Let us examine each equation.
4x²+16x=0
4x(x+4)=0
4x=0 and x+4=0
where x1=0 and x2=-4
Next answer is whown below:
(x+4)(x-12)=0
x1=-4 and x2=12
Therefore, the first and the third equations are the answers.
