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cander3
If dy/dx=2xy and y(0)=1. What is y(1) equal to?
So, this looks more like partial derivatives. That is How I shall approach it. We have: \[\frac{\delta~y}{\delta~x}=2xy\] To this we have taken the derivative WRT x. So the y terms are considered constant. If we integrate WRT x we should arrive at our initial function. So: \[ \int 2xy~ \delta x= x^2y+f(y)\] Now in order to determine f(y) we must look at our givens. So if y=0 then f(x,y)=1 so we know that f(y) must equal 1 at y(0) (because x^2y=0). So we can infer that f(y) is actually just 1. This yields the eq. \[f(x,y)=x^{2} y+1\] From here it is a plug and chug.
looks like a seperable ordinary differential equation to me. first solve for the equation. \[\frac{ dy }{ dx }=2xy\] if I write dy/dx as y', and divide both sides by y then we get. \[\frac{ y' }{ y }=2x\]Integrate both sides with respect to x gives: \[\ln \left| y \right|=x^{2}+C\] Therefore \[y=Ce ^{x ^{2}}\] is a solution. if y(0) = 1 then substitute this to solve for C \[1=Ce ^{0^{2}}\] so C=1 and \[y=e ^{x ^{2}}\] is the particular solution to the equation. If you substitute 1 for x into the equation, you get y(1) = e