The solution of the differential equation ({e | vedclass

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(D) Given the differential equation: $({e^x} + 1)y \, dy = (y + 1){e^x} \, dx$ Separate the variables: $\frac{y}{y + 1} \, dy = \frac{{e^x}}{{e^x} + 1

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$c(y + 1)({e^x} + 1) = {e^y}$

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(D) Given the differential equation: $({e^x} + 1)y \, dy = (y + 1){e^x} \, dx$ Separate the variables: $\frac{y}{y + 1} \, dy = \frac{{e^x}}{{e^x} + 1} \, dx$ Rewrite the left side: $\left( 1 - \frac{1}{y + 1} \right) \, dy = \frac{{e^x}}{{e^x} + 1} \, dx$ Integrate both sides: $\int \left( 1 - \frac{1}{y + 1} \right) \, dy = \int \frac{{e^x}}{{e^x} + 1} \, dx$ Performing the integration: $y - \ln|y + 1| = \ln|{e^x} + 1| + C$ Rearrange the terms: $y = \ln|y + 1| + \ln|{e^x} + 1| + \ln|c|$ Using logarithmic properties: $y = \ln|c(y + 1)({e^x} + 1)|$ Exponentiating both sides: ${e^y} = c(y + 1)({e^x} + 1)$.

The solution of the differential equation $({e^x} + 1)y \, dy = (y + 1){e^x} \, dx$ is:

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