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3 years ago

The first one asks for the value of x. I need help on both. I am struggling to understand how to do this, please help

Mathematics

1 answer:

zzz [600]3 years ago

6 0

(1) since the horizontal lines are parallel, and you know the vertical intersecting line crosses at right angle, it must be that the two angles with variable x sum up to 90 degrees. So same rewritten as an equation:

x+8+6x-58 = 90

7x = 140

x = 20 (degrees)

(2) here you have two variable so you will need to find two equations.

The first one comes from the triangle and its angles summing up to 180:

63 + (3x+45) + (y-6) = 180

The second one comes from the fact that the two horizontal lines are parallel and so the angles (7x-7) and (3x+45) must be the same:

7x-7=3x+45

Let's solve the last one first:

4x = 52

x = 13

then plug this solution into the first equation and solve for y

63 + (3x+45) + (y-6) = 180

3x + y = 78

y = 78-39 = 39

so the value of y is 39 (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

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Answer:

So, solution of the differential equation is

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

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

We use the Method of Undetermined Coefficients.

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

$y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\$

It is a homogeneous solution:

$y_h(t)=c_1e^{-2i t}+c_2e^{2i t}$

Now, we finding a particular solution.

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

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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