To do this question you need to separate the variables. So we split the differential and get the variable together with the appropriate differential.
AP Calc AB help!! - College Discussion::
14 posts - Last post: May 9, 2007Can someone help me with these: -The average value of cos x on teh ln 1 = 0 dy/dx = x y = 1/2x - 3/4 slope of equation is 1/2 make the
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dy / [(y - 1)(y + 2)] = x² dx
Now integrate both sides, the right hand side is simple giving x³ /3, the left hand side needs to be done by partial fractions. So I need A and B such that:
A / (y - 1) + B / (y+ 2) = 1 / [(y - 1)(y +2)]
Getting a common denominator on the left hand side, the numerator will be:
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Help me please y = 2x / ( x-(sqrt)x) y = sin (… caranya soal mtk ini (1/4) pangkat men 2/3-16 pangkat5/4+(1/8)pangkat men 1/3? tlng y,please!thanks.
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Need help to see what ive done wrong with differentiation ::
y = x - x^3 y' = 1 - 3x^2 y'(0) = 1 - 3(0)^2 = 1 - 3(0) = 1. So you have your slope, y-y1 = dy/dx (x-x1) x1=0, y1=0 dy/dx=1 y-0 = 1(x-0)
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A(y + 2) + B(y - 1) and this must be equal to the numerator of 1 on the right hand side since the denominators are equal.
Equating both numerators:
Ay + 2A + By - B = 1
(A + B) y + (2A - B) = 1
So A + B = 0 and 2A - B = 1; the first gives B = -A, so substitute this into the second and we have 2A - (-A) = 1; 3A = 1 or A = 1/3 and so B = -1/3.
The problem is now:
Derivatives::
File Format: PDF/Adobe Acrobat - View as HTML3. Fig. 2.1. The derivative is 2 then. 0. It does not exist at .. for negative x and y. =. Mx. +. B for. x 3 0. The graphs meet if . The two slopes are
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1/3 ʃ dy / (y -1) - 1/3 ʃ dy / (y + 2) = x³ /3 + C
Integrating, we have:
1/3 ln y - 1 - 1/3 ln y + 2 = x³ /3 + K
1/3 [ ln y - 1 - ln y + 2] = x³ /3 + K
Yahoo! Answers - What's the derivative of x^(1/x)?::
( dy/dx ) / y = ( 1 / x ) d( ln x )/dx + d( 1/x )/dx ln x ( dy/dx ) / y = ( 1/x )( 1/x ) + ( -1/x² ) ln x dy/dx = x^(1/x) ( 1 - ln x ) / x². 2 years ago
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Full text of "An elementary treatise on differential equations and ::
<2) xyz + xy^-y^O. (3) y n 2 = y n _ r (4) y n + y n _ 2 = 8 cos 3x. (5) (a 2 log x-x 2 )y 2 -xy l + y = 0. (6) (x 2 + 2x-l)y 2 -(3x 2 + 8x-l)y 1 + (2x 2 +
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1/3 ln (y - 1) / (y + 2) = x³ /3 + K
Now you are after a particular solution when y(0) = 3, so substitute x = 0 and y = 3 into the answer to determine the constant K.
1/3 ln (3 - 1) / (3 + 2) = 0 + K
1/3 ln (2/5) = K
Putting it all together:
1/3 ln (y - 1) / (y + 2) = x³ /3 + 1/3 ln (2/5)
or
ln (y - 1) / (y + 2) = x³ + ln (2/5)
(y - 1) / (y + 2) = e^(x³ + ln (2/5))
(y - 1) / (y + 2) = 2(e^x³) / 5
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