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3 years ago

Need help with this question, thank you in advance!

Mathematics

1 answer:

blondinia [14]3 years ago

3 0

Answer:

arc RP = 84 degrees

Step-by-step explanation:

If the angle NPQ is 102 degrees, we can find the angle NPM:

NPM + NPQ = 180

NPM + 102 = 180

NPM = 78 degrees.

Now, with the triangle NPM, we can find the angle MNP:

60 + 78 + MNP = 180

MNP = 180 - 60 - 78 = 42 degrees

The angle MNP is inscribed in the circle, and is related to the arc RP:

arc RP = 2 * MNP

arc RP = 2 * 42 = 84 degrees

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Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Meth

taurus [48]

Answer:

So, solution of the differential equation is

$y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

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We first solve the homogeneous differential equation y′′+4y=0.

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$y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\$

we get

$y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

So, solution of the differential equation is

$y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

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