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The solution of $\frac{d y}{d x}+1=e^{x+y}$ is : |
$(x+y) e^{x+y}=0$ $(x+c) e^{x+y}=0$ $(x-y) e^{x+y}=1$ $(x-c) e^{x+y}+1=0$ |
$(x-c) e^{x+y}+1=0$ |
$\frac{d y}{d x}+1=e^{x+y}$ let $z=x+y$ $\frac{dz}{dx}=1+\frac{dy}{dx}$ $\frac{dz}{dx}=e^z$ so $\int e^{-z}dx=\int dx$ $-e^{-z}=x+c$ so $-1=(x+c)e^z$ so $(x+c)e^{x+y}+1=0$ $=(x-c) e^{x+y}+1=0$ |
