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Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 15

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Answer: c=e^{x}+e^{-y}+\frac{x^{4}}{4}

Hint: Separate the terms of x and y and then integrate them.

Given: \frac{d y}{d x}=e^{x+y}+e^{y} x^{3}

Solution: \frac{d y}{d x}=e^{x+y}+e^{y} x^{3}

        \frac{d y}{d x}=e^{y}\left(e^{x}+x^{3}\right) \\

        \begin{aligned} &\frac{d y}{e^{y}}=\left(e^{x}+x^{3}\right) d x \\\\ &e^{-y} d y=\left(e^{x}+x^{3}\right) d x \end{aligned}

          Integrating both sides

        \begin{aligned} &\int e^{-y} d y=\int e^{x} d x+\int x^{3} d x \\\\ &-e^{-y}+c=e^{x}+\frac{x^{4}}{4} \\\\ &c=e^{x}+e^{-y}+\frac{x^{4}}{4} \end{aligned}

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