Basic differential equations help
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Basic differential equations help

[From: ] [author: ] [Date: 11-05-18] [Hit: ]
i thought it was 2x. i havent had a math course in a while and am a little rusty here. also does dy/dx = y?Thanks-dy/dx = y ======> meaning this already derived.........
question. my professor wrote on the board the following: dy/dx=x^2. then solved it to be y= 1/3x^3 + C. is this correct. what process was done here to obtain that answer? i thought it was 2x. i havent had a math course in a while and am a little rusty here. also does dy/dx = y'?
Thanks

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dy/dx = y ' ======> meaning this already derived........

dy/dx is derivative

∫ means integrate

dy/dx = y ' = x^2 =====> let's start integrating by applying the ∫

∫ dy/dx = ∫ y ' = ∫ x^2

y = y = (1/3) * x^3 + C

y = (1/3) * x^3 + C

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welcome :)

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dy/dx is y'

So y' = x^2 dx
Integrate both sides
If you have y' = x^n
then when you integrate, y = [x^(n+1)] / n

y= 1/3 x^3 + C

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dy/dx=y'

dy/dx=x^2.
y= (1/3)x^3 + C this is the answer

if
dy/dx=x^n
then
y=[1/(n+1)]x^(n+1) +C for n different of -1

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thats partial differential equation solving method. please refer partial differential equation books for futher details
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