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Question: Answered & Verified by Expert
The general solution of differential equation $\frac{d y}{d x}=e^{x+y}+x^2 e^{x^3+y}$ is (where $\mathrm{C}$ is a constant of integration.)
MathematicsDifferential EquationsMHT CETMHT CET 2022 (05 Aug Shift 2)
Options:
  • A $\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{y}}+\frac{1}{3} \mathrm{e}^{\mathrm{x}^3}=\mathrm{C}$
  • B $e^x-e^{-y}-\frac{1}{3} e^{x^3}=C$
  • C $e^x-e^{-y}+\frac{1}{3} e^{x^3}=C$
  • D $e^x-e^{-y}+\frac{1}{3} e^{x^3}=C$
Solution:
2659 Upvotes Verified Answer
The correct answer is: $\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{y}}+\frac{1}{3} \mathrm{e}^{\mathrm{x}^3}=\mathrm{C}$
$\begin{aligned} & \frac{d y}{d x}=e^{x+y}+x^2 e^{x^3+y} \\ & \Rightarrow \frac{d y}{d x}=e^y\left(e^x+x^2 e^{x^3}\right) \\ & \Rightarrow \int e^{-y} d y=\int\left(e^x+x^2 e^{x^3}\right) d x \\ & \Rightarrow-e^{-y}+c=e^x+\frac{1}{3} e^{x^3} \\ & \Rightarrow e^x+e^{-y}+\frac{1}{3} e^{x^3}=c\end{aligned}$

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