e^x(y-3(e^x+1)^2)dx+(e^x+1)^2dy

asked by guest
on Nov 19, 2024 at 8:36 am



You asked:

Evaluate the expression: \({e}^{x \cdot \left(y - 3 \cdot {\left( {e}^{x} + 1 \right)}^{2}\right)} \cdot d x + {\left( {e}^{x} + 1 \right)}^{2} \cdot d y\)

MathBot Answer:

Evaluated



\(\displaystyle {e}^{x \cdot \left(y - 3 \cdot {\left( {e}^{x} + 1 \right)}^{2}\right)} \cdot d x + {\left( {e}^{x} + 1 \right)}^{2} \cdot d y = d x e^{x \left(y - 3 \left(e^{x} + 1\right)^{2}\right)} + d y \left(e^{x} + 1\right)^{2} \)

Expanded

\[{e}^{x \cdot \left(y - 3 \cdot {\left( {e}^{x} + 1 \right)}^{2}\right)} \cdot d x + {\left( {e}^{x} + 1 \right)}^{2} \cdot d y = d x e^{- 3 x} e^{x y} e^{- 3 x e^{2 x}} e^{- 6 x e^{x}} + d y e^{2 x} + 2 d y e^{x} + d y\]

Factored

\[{e}^{x \cdot \left(y - 3 \cdot {\left( {e}^{x} + 1 \right)}^{2}\right)} \cdot d x + {\left( {e}^{x} + 1 \right)}^{2} \cdot d y = d \left(x e^{- 3 x} e^{x y} e^{- 3 x e^{2 x}} e^{- 6 x e^{x}} + y e^{2 x} + 2 y e^{x} + y\right)\]

\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).

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