TS EAMCET · Maths · Differential Equations
The general solution of the differential equation \(\frac{d y}{d x}=1+x+y+x y\) is
- A \(\log (1+x)=y+\frac{x^2}{2}+k\)
- B \(y=x+\frac{x^2}{2}+k\)
- C \(\log (1+y)=\frac{x^3}{3}+k\)
- D \(y=k e^{x+\frac{x^2}{2}}-1\)
Answer & Solution
Correct Answer
(D) \(y=k e^{x+\frac{x^2}{2}}-1\)
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} & \frac{d y}{d x}=1+x+y+x y \\ \Rightarrow & \frac{d y}{d x}=(1+x)(1+y) \Rightarrow \frac{d y}{1+y}=(1+x) d x \end{aligned} \] On integrating both the sides, we get…
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