Mathematics, 24.07.2019 02:30 kdtrillz
Find an equation of the curve that passes through the point (0,1) and whose slope at (x, y) is xy.
Answers: 1
is 
, then we have ![\displaystyle\frac{dy}{dx} = xy\ \Rightarrow\ \frac{dy}{y} = x\, dx\ [y \ne 0] \Rightarrow \int \frac{dy}{y} = \int x \, dx \ \Rightarrow \\ \\ \textstyle \ln|y| = \frac{1}{2}x^2 + C.\ y(0) = 1\ \Rightarrow\ \ln 1 = 0 + C\ \Rightarrow\ C = 0. \\ \\ \text{Thus } |y| = e^{x^2 / 2}\ \Rightarrow\ y = \pm e^{x^2 / 2},\ \text{so } y = e^{x^2/2} \text{ since } y(0) = 1 \ \textgreater \ 0. \\ \\ \text{Note that $y=0$ is not a solution because it does not satisfy} \\ \text{the initial condition $y(0) = 1$}.](/tpl/images/0125/6321/58d1d.png)
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Find an equation of the curve that passes through the point (0,1) and whose slope at (x, y) is xy....
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