anonymous 3 years ago If dy/dx=2xy and y(0)=1. What is y(1) equal to?

1. FibonacciChick666

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.

2. anonymous

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