CUET Preparation Today
The general solution of the equation $\frac{d y}{d x}=1+x y$ is : |
$y=c e^{-\frac{x^2}{2}}$ $y=c e^{\frac{x^2}{2}}$ $y=(x+c) e^{-\frac{x^2}{2}}$ None of these |
None of these |
$\frac{d y}{d x}-x y=1 \Rightarrow P=-x, Q = 1$ I.F. $= e^{-\int x d x}=e^{-\frac{x^2}{2}}$ ∴ Solution is $y . e^{\frac{-x^2}{2}}=\int e^{-\frac{x^2}{2}} . 1 . d x+c$ Hence (4) is the correct answer. |
