The general solution of the differential equation `e^x" "dy" "+" "(y" "e^x" "+" "2x)" "dx" "=" "0` is (A) `x""e^y+x^2=C` (B) `x""e^y+y^2=C` (C) `y""e^x+x^2=C` (D) `y""e^y+x^2=C`

Explanation

`e^{x} d y =-(y e^{x}+2 x) d x`

` Rightarrow frac{d y}{d x} =-frac{(y e^{x}+2 x)}{e^{x}} `

`
frac{d y}{d x} =-(y+2 x e^{-x}) `

` Rightarrow frac{d y}{d y}+y =-2 x e^{-x} `


Here `p=1, Q= -2x e^x`

`
I F=e^{int P d x}=e^{int d x}=e^x`


Now ,`y times I F =int Q x I F+C . `

` Rightarrow y times e^{x} =int-2 x e^{-x} times e^{x} d x+C . `

` =-int 2 x d x+C . `

`Rightarrow y e^{x}=-x^{2}+c `

`Rightarrow y cdot e^{x}+x^{2}=C `

so correct answer is C