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Vladimir79 [104]
3 years ago
6

Can someone please check my answers I WIILL GOVE BRAINLY TO FIRST PERSON THAT ANSWERS!

Mathematics
1 answer:
Rainbow [258]3 years ago
5 0
The first is C. r = 4(4/9)
the second one is B. h = 130
the second one is A. k = 7

i’m so sorry, they’re all wrong
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D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x
MA_775_DIABLO [31]
The solution depends on the value of k. To make things simple, assume k>0. The homogeneous part of the equation is

\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0

and has characteristic equation

r^2-16k=0\implies r=\pm4\sqrt k

which admits the characteristic solution y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}.

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be y_p=ae^{4x}+be^x. Then

\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x

So you have

16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x
(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x

This means

16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}
b(1-16k)=30\implies b=\dfrac{30}{1-16k}

and so the general solution would be

y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x
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