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'?
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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
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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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
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
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