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Published byAnabel Treadway Modified over 4 years ago

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By: Zack Broadhead

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When you take the derivative of a number with no variable it is just ‘0’. Y=5x+4, (dy/dx)=5+0 When you antidifferentiate an equation the result must have a “+ C” on the end because you don’t know what the constant number was.

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Integrate equation. Tack “+C” onto the end. Plug in initial values. Solve for “C”. Plug “C” into integrated equation.

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∫ (dy)=∫x dx Y=x 2 /2+C Y=3 and x=2 so: (3)=(2) 2 /2+C C=1 Y=x 2 +1

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∫(dy)=∫Cos(x) dx Y=Sin(x)+C Y=1 and x=п/2 so: 1=Sin(п/2)+C C=0 Y=Sin(x)+0

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∫(dy)=∫(x 2 +x) dx Y=(x 3 /3)+(x 2 /2)+C Y=1, and x=1 so: 1=1/3+1/2+C 1=5/6+C C=1/6 Y=(x 3 /3)+(x 2 /2)+1/6

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∫(dy)=∫(1/x) dx Y=ln(x)+C Y=2 and x=1, so: 2=ln(1)+C C=2 Y=ln(x)+2

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∫(dy)=∫(Sin(2x)) dx Y=-(Cos(2x))/2+C Y=1/2 and x=0, so: ½=-(Cos(2(0)))/2+C C=1 Y=-Cos(2x)/2+1

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A. y=x 3 -2 May have accidentally subtracted “C”. B. y=6x-2 May have derived instead of integrated. C. y=x 3 +2 Correct answer! D. y=x 3 -998 Plugged wrong points in.

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A. y=√((4/5)x 5 –(123/5)) Used wrong points. B. y=√((4/5)x 5 +(16/5)) Correct! C. y 2 /2=(2/5)x 5 +(8/5) Not simplified. D. y=8x 3 -6 Derived instead of integrated.

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Answer: Y=x 2 -x+5 ∫(dy)=∫(2x-1) dx Y=x 2 -x+C Y=11 and x=3, so: 11=(3) 2 –(3)+C C=5 Y=x 2 –x+5

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http://planetmath.org/encyclopedia/InitialValueProblem.html http://en.wikipedia.org/wiki/Initial_value_problem http://www.vias.org/calculus/14_differential_equations_01_split006.html

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By: Zack Broadhead

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