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Quiz-1 / Q1: TUT0602
« on: September 27, 2019, 02:31:55 PM »
Solve the given differential equation:
$$\frac{dy}{dx} = \frac{x-e^{-x}}{y+e^y}$$
This is a separable differential equation. Rearranging, we have
$$(y+e^y)dy = (x-e^{-x})dx\ \Rightarrow\ \int(y+e^y)dy = \int(x-e^{-x})dx\ \Rightarrow\ y^2 + 2e^y = x^2 + 2e^{-x} + C$$
is the general implicit solution.
$$\frac{dy}{dx} = \frac{x-e^{-x}}{y+e^y}$$
This is a separable differential equation. Rearranging, we have
$$(y+e^y)dy = (x-e^{-x})dx\ \Rightarrow\ \int(y+e^y)dy = \int(x-e^{-x})dx\ \Rightarrow\ y^2 + 2e^y = x^2 + 2e^{-x} + C$$
is the general implicit solution.