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never [62]
3 years ago
15

Can you solve the equation 2/3(4x-5)=8 by using the Division Property of Equality? Explain.

Mathematics
1 answer:
Nata [24]3 years ago
3 0
To solve for X you need to simplify both sides on the equation 
<span><span><span>23</span><span>(<span><span>4x</span>−5</span>)</span></span>=8

</span><span><span><span><span>(<span>23</span>)</span><span>(<span>4x</span>)</span></span>+<span><span>(<span>23</span>)</span><span>(<span>−5</span>)</span></span></span>=8 *</span>Distribute*

<span><span><span><span>8/3</span>x</span>+<span><span>−10/</span>3</span></span>=<span>8

after that you add 10/3 on both sides 

</span></span><span><span><span><span><span>8/3</span>x</span>+<span><span>−10/</span>3</span></span>+10/3</span>=<span>8+10/3</span></span><span><span><span>8/3</span>x</span>=<span>34/<span>3

finally multiply 3/8 on both sides 
</span></span></span><span><span>(<span>3/8</span>) x </span><span>(<span><span>8/3</span>x</span>)</span></span>=<span><span>(<span>3/8</span>) x </span><span>(<span>34/3</span><span>)

and your answer should be 
X= 17/4
</span></span></span>
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Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).
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Take the homogeneous part and find the roots to the characteristic equation:

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

This means the characteristic solution is y_c=C_1\cos2x+C_2\sin2x.

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form y_p=ax\cos2x+bx\sin2x. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With y_1=\cos2x and y_2=\sin2x, you're looking for a particular solution of the form y_p=u_1y_1+u_2y_2. The functions u_i satisfy

u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx

where W(y_1,y_2) is the Wronskian determinant of the two characteristic solutions.

W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2

So you have

u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx
u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x

u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx
u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x

So you end up with a solution

u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x

but since \cos2x is already accounted for in the characteristic solution, the particular solution is then

y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

so that the general solution is

y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x
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Answer:

Please see attached picture for full solution.

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