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zvonat [6]
3 years ago
7

Find the general solution of (2x² + y)dx + (x²y - x)dy = 0​

Mathematics
1 answer:
dmitriy555 [2]3 years ago
8 0

Multiply the original DE by xy:

xy2(1+x2y4+1−−−−−−−√)dx+2x2ydy=0(1)

Let v=xy2, so that dv=y2dx+2xydy. Then (1) becomes

x(y2dx+2xydy)+xy2x2y4+1−−−−−−−√dxxdv+vv2+1−−−−−√dx=0=0

This final equation is easily recognized as separable:

dxxln|x|+CKxvKx2y2−1K2x4y4−2Kx2y2y2=−dvvv2+1−−−−−√=ln∣∣∣v2+1−−−−−√+1v∣∣∣=v2+1−−−−−√+1=x2y4+1−−−−−−−√=x2y4=2KK2x2−1integrate both sides

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11050 divided by 26 what the remainder and turn into a fraction
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Find the circumference of a circle with a radius of 35 inches in terms of pie add to the nearest 10th of an inch
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Circumference of circle = 2 π r

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<h2>1)</h2>

(x - 4) {}^{2}  - 28 = 8 \\ (x - 4) {}^{2}  = 8 + 28 \\ (x - 4) {}^{2}  = 36

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\sqrt{(x - 4) {}^{2} }  =  ± \sqrt{36}  \\x - 4 = ±6 \\ x _{1}- 4 = 6 \:  \:  \:  \:  \:  \:  \: \:  \:  \:  x _{2}- 4 =  - 6 \\ x_{1} = 6 + 4 \:  \:  \:  \:  \:  \:  \:  \:  \:  \: x_{2} =  - 6 + 4 \\ x_{1} = 10 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \: \:  \: x_{2} =  - 2

<h2>2)</h2>

Well he started off correct to the point of completing the square.

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