3dy/dx=3y/x+sin(y/x)
I would imagine it would be a substitution, but I have no clue how to even begin.
I would imagine it would be a substitution, but I have no clue how to even begin.
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3 dy/dx = 3y/x + sin(y/x)
dy/dx = y/x + 1/3 sin(y/x)
u = y/x
ux = y
u + x du/dx = dy/dx
u + x du/dx = y/x + 1/3 sin(y/x)
u + x du/dx = u + 1/3 sin(u)
x du/dx = 1/3 sin(u)
csc(u) du = 1/(3x) dx
Integrate both sides:
ln|csc(u)−cot(u)| = 1/3 ln|x| + ln|C|
csc(u) − cot(u) = C∛x
1/sin(u) − cos(u)/sin(u) = C∛x
(1−cos(u))/sin(u) = C∛x
tan(u/2) = C∛x
u/2 = tan⁻¹(C∛x)
u = 2 tan⁻¹(C∛x)
y/x = 2 tan⁻¹(C∛x)
y = 2x tan⁻¹(C∛x)
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If you check WolframAlpha, they give y = 2x cot⁻¹(e^c₁/∛x)
But since cot(x) = 1/tan(x), then cot⁻¹(x) = tan⁻¹(1/x)
Therefore: cot⁻¹(e^c₁/∛x) = tan⁻¹(∛x/e^c₁) = tan⁻¹(C∛x) = my answer
where arbitrary constant 1/e^c₁ is replaced with yet another arbitrary constant C
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Regarding answer to your other question:
http://answers.yahoo.com/question/index;…
The reason I chose substitution u = xy² is because when I solved for dy/dx I got:
dy/dx = y(xy²+1)/x
I wasn't as concerned with the y and x outside parentheses, since they can always be separated. But I wanted a single variable inside parentheses, which contains xy²+1 (and which cannot be separated), so I decided to use substitution u = xy². This doesn't always work, but it was worth a shot, and it paid off.
dy/dx = y/x + 1/3 sin(y/x)
u = y/x
ux = y
u + x du/dx = dy/dx
u + x du/dx = y/x + 1/3 sin(y/x)
u + x du/dx = u + 1/3 sin(u)
x du/dx = 1/3 sin(u)
csc(u) du = 1/(3x) dx
Integrate both sides:
ln|csc(u)−cot(u)| = 1/3 ln|x| + ln|C|
csc(u) − cot(u) = C∛x
1/sin(u) − cos(u)/sin(u) = C∛x
(1−cos(u))/sin(u) = C∛x
tan(u/2) = C∛x
u/2 = tan⁻¹(C∛x)
u = 2 tan⁻¹(C∛x)
y/x = 2 tan⁻¹(C∛x)
y = 2x tan⁻¹(C∛x)
------------------------------
If you check WolframAlpha, they give y = 2x cot⁻¹(e^c₁/∛x)
But since cot(x) = 1/tan(x), then cot⁻¹(x) = tan⁻¹(1/x)
Therefore: cot⁻¹(e^c₁/∛x) = tan⁻¹(∛x/e^c₁) = tan⁻¹(C∛x) = my answer
where arbitrary constant 1/e^c₁ is replaced with yet another arbitrary constant C
------------------------------
Regarding answer to your other question:
http://answers.yahoo.com/question/index;…
The reason I chose substitution u = xy² is because when I solved for dy/dx I got:
dy/dx = y(xy²+1)/x
I wasn't as concerned with the y and x outside parentheses, since they can always be separated. But I wanted a single variable inside parentheses, which contains xy²+1 (and which cannot be separated), so I decided to use substitution u = xy². This doesn't always work, but it was worth a shot, and it paid off.
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Wow---go Mathmom!!
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