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  • The general solution of differential equation e^{x}+1 y d y=(y+1) e^{x} d x is (a) (y+1)=k\left(e^{x}+1

The general solution of differential equation \left(e^{x}+1\right) y d y=(y+1) e^{x} d x is
(a) (y+1)=k\left(e^{x}+1\right)
(b) y+1=e^{x}+1+k
(c) y=\log \left\{k(y+1)\left(e^{x}+1\right)\right\}
(d) y=\log \left\{\frac{e^{x}+1}{y+1}\right\}+k

Answers (1)

Ans: - The answer is the option (c) y=\log \left\{k(y+1)\left(e^{x}+1\right)\right\}

Explanation: -

\left(e^{x}+1\right) y d y=(y+1) e^{x} d x

\\ \Rightarrow \frac{y d y}{y+1}=\frac{e^{x}}{e^{x}+1} d x \\ \Rightarrow \int\left(1-\frac{1}{y+1}\right) d y=\int \frac{e^{x}}{e^{x}+1} d x \\ \Rightarrow y-\log (y+1)=\log \left(e^{x}+1\right)+\log k \\ \Rightarrow y=\log (y+1)+\log \left(1+e^{x}\right)+\log k \\ \Rightarrow y=\log \left(k(1+y)\left(1+e^{x}\right)\right)

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