Short Answer

Solve the exact differential equations in Exercises 63–66. $e^{y} + (x e^{y} - 7) \frac{d y}{d x} = 0$

The solution of given exact differential equation is: $x e^{y} - 7 y + C = 0$

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Exact differential equation, $e^{y} + (x e^{y} - 7) \frac{d y}{d x} = 0$

Step 2. Solving the given exact differential equation

$Given, e^{y} + (x e^{y} - 7) \frac{d y}{d x} = 0 \Rightarrow e^{y} d x + (x e^{y} - 7) d y = 0 It is a exact differential equation of the the form M d x + N d y = 0, with \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} . the solution is given by, f (x, y) = \int (treating y a s constant in M) d x + \int (terms independent of x in N) d y = 0 \Rightarrow \int e^{y} d x + \int (- 7) d y = 0 \Rightarrow x e^{y} - 7 y + C = 0$

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Calculus

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Short Answer

Solve the exact differential equations in Exercises 63–66. $e^{y} + (x e^{y} - 7) \frac{d y}{d x} = 0$

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Step 1. Given information

Step 2. Solving the given exact differential equation

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Calculus

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Solve the exact differential equations in Exercises 63–66. ey+(xey−7)dydx= 0

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Step 1. Given information

Step 2. Solving the given exact differential equation

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Solve the exact differential equations in Exercises 63–66. $e^{y} + (x e^{y} - 7) \frac{d y}{d x} = 0$