The equations (2) and (3) you referred to are unavailable, but it is clear that you are trying to show that two set of solutions y1 and y2, to a (second-order) differential equation are solutions, and form a fundamental set. This will be explained.
Answer:
SOLUTION OF A DIFFERENTIAL EQUATION.
Two functions y1 and y2 are set to be solutions to a differential equation if they both satisfy the said differential equation.
Suppose we have a differential equation
y'' + py' + qy = r
If y1 satisfies this differential equation, then
y1'' + py1' + qy1 = r
FUNDAMENTAL SET OF DIFFERENTIAL EQUATION.
Two functions y1 and y2 are said to form a fundamental set of solutions to a second-order differential equation if they are linearly independent. The functions are linearly independent if their Wronskian is different from zero.
If W(y1, y2) ≠ 0
Then solutions y1 and y2 form a fundamental set of the given differential equation.
The answers are 347 and 865. To solve this problem you set up two equations, x+y=1212 and x-y=518. You then get the second x by itself by adding y to both sides. You then have x+y=1212 and x=y+518. After this you substitute the x into the first equation and have y+518+y=1212. Then simplify to 2y=694 and y=347
Now that y=347 you put it in the original equation. Solve and get that x=865
F(x) can be written as:
f(x) = Asin(2x); where A is the amplitude and the period of the function is half that of a normal sin function.
f(π/4) = 4
4 = Asin(2(π/4))
4 = Asin(π/2)
A = 4
Amplitude of g(x) = 1/2 * amplitude of f(x)
A for g(x) = 2
g(x) = 2sin(x)
Answer:
sure, !whats the question?
Step-by-step explanation:
Answer:5
Step-by-step explanation:
