CUET Preparation Today
Solution of the differential equation $\frac{dy}{dx}=1+x+y+x y$ are : Where C is an arbitrary constant |
$\log (1+y)=x-\frac{x^2}{2}+C$ $\log (1-y)=x+\frac{x^2}{2}+C$ $\log (1+y)=x+\frac{x^2}{2}+C$ $\log (1+y)=-x-\frac{x^2}{2}+C$ |
$\log (1+y)=x+\frac{x^2}{2}+C$ |
The correct answer is Option (3) → $\log (1+y)=x+\frac{x^2}{2}+C$ $\frac{dy}{dx}=(1+x)+y(1+x)$ $⇒\frac{1}{1+y}dy=1+xdx$ $⇒\log|1+y|=x+\frac{x^2}{2}+C$ |
