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help1219
03.04.2021 •
Mathematics
Find the general solution of DE 4y "+ y '= 0
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Ответ:
y=a+be^(-1/4t)
Step-by-step explanation:
We can solve the characteristic equation, 4r^2+r=0 to help.
Factor:
r(4r+1)=0
This implies r=0 or 4r+1=0.
The second equation can be solved by subtracting 1 on both sides and then dividing boths sides by 4 like so:
4r+1=0
4r+1-1=0-1
4r=-1
4r/4=-1/4
r=-1/4
So r=0 and r=-1/4 which means the general solution is y=ae^(0t)+be^(-1/4t)
Simplifying gives y=a+be^(-1/4t) since e^0=1.
Let's check our solution.
y=a+be^(-1/4t)
y'=0+-1/4be^(-1/4t)
y''=0+1/16be^(-1/4t)
Plug in:
4y''+y'=0
4*1/16be^(-1/4t)+-1/4be^(-1/4t)=0
1/4be^(-1/4t)+-1/4be^(-1/4t)=0
0=0 is a true equation so the solution has been verified.
Ответ:
G(62.5)
Step-by-step explanation:
30/48 × x/100=62.5/2.08
to solve x
100÷48=62.5(x)
62.5(x)×30=62.5
Answer is F