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Solution of $\frac{d y}{d x}=e^{y+x}+e^{y-x}$ is : |
$e^x(x+1)=y$ $e^x(x+1)+1=y$ $e^x(x-1)+1=y$ None of these |
None of these |
$\frac{d y}{d x}=e^y . e^x+e^y . e^{-x}=e^y\left(e^x+e^{-x}\right)$ $\Rightarrow e^{-y} dy=\left(e^{x}+e^{-x}\right) dx \Rightarrow e^{-y}=e^{x}-e^{-x}+c$ Hence (4) is the correct answer. |
