Differential Equation and Applications (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question
Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Differential Equation and Applications (p-1)1 Mark
Question
Solve the following differential equations:$\frac{d y}{d x}=1+ x + y + xy$
✓
Answer
$ \begin{aligned} & \frac{d y}{d x}=1+ x + y + xy \\ & \therefore \frac{d y}{d x}=(1+ x )+ y (1+ x )=(1+ x )(1+ y ) \\ & \therefore \frac{1}{1+y} dy =(1+ x ) dx \end{aligned} $ Integrating, we get $ \begin{aligned} & \int \frac{1}{1+y} d y=\int(1+x) d x \\ & \therefore|\log | 1+y \mid=x+\frac{x^2}{2}+c \end{aligned} $ This is the general solution.
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