In this figure, mBDA=_ and mBCA=_. (fill in the blanks with degrees)

Mathematics

1 answer:

pickupchik [31]3 years ago

5 0

MBOA=360-250=110, so mBDA= half of mBOA=55
mBCA=180-mBOA=180-110=70

mBDA=55 and mBCA=70

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The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

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

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

So you have

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and so the general solution would be

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