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Question

(A) The given differential equation is: $(e^{x}+e^{-x}) dy = (e^{x}-e^{-x}) dx$. Separating the variables,we get: $dy = \frac{e^{x}-e^{-x}}{e^{x}+e^{-

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$y = \log(e^{x} + e^{-x}) + C$

Vedclass · Answer

(A) The given differential equation is: $(e^{x}+e^{-x}) dy = (e^{x}-e^{-x}) dx$. Separating the variables,we get: $dy = \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} dx$. Integrating both sides: $\int dy = \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} dx + C$. Let $t = e^{x}+e^{-x}$. Then $dt = (e^{x}-e^{-x}) dx$. Substituting these into the integral: $y = \int \frac{1}{t} dt + C$. Integrating,we get: $y = \log|t| + C$. Substituting back $t = e^{x}+e^{-x}$,the general solution is: $y = \log(e^{x}+e^{-x}) + C$.

Find the general solution of the differential equation: $(e^{x}+e^{-x}) dy - (e^{x}-e^{-x}) dx = 0$.

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