Calculus Examples

$e^{2 x} d y + (2 e^{2 x} y - x) d x = 0$

Step 1

Rewrite the differential equation to fit the Exact differential equation technique.

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Step 1.1

Rewrite.

$(2 e^{2 x} y - x) d x + e^{2 x} d y = 0$

Step 2

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 2 e^{2 x} y - x$ .

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Step 2.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 e^{2 x} y - x]$

Step 2.2

By the Sum Rule, the derivative of $2 e^{2 x} y - x$ with respect to $y$ is $\frac{d}{d y} [2 e^{2 x} y] + \frac{d}{d y} [- x]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 e^{2 x} y] + \frac{d}{d y} [- x]$

Step 2.3

Evaluate $\frac{d}{d y} [2 e^{2 x} y]$ .

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Step 2.3.1

Since $2 e^{2 x}$ is constant with respect to $y$ , the derivative of $2 e^{2 x} y$ with respect to $y$ is $2 e^{2 x} \frac{d}{d y} [y]$ .

$\frac{\partial M}{\partial y} = 2 e^{2 x} \frac{d}{d y} [y] + \frac{d}{d y} [- x]$

Step 2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 1$ .

$\frac{\partial M}{\partial y} = 2 e^{2 x} \cdot 1 + \frac{d}{d y} [- x]$

Step 2.3.3

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = 2 e^{2 x} + \frac{d}{d y} [- x]$

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

Since $- x$ is constant with respect to $y$ , the derivative of $- x$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = 2 e^{2 x} + 0$

Step 2.4.2

Add $2 e^{2 x}$ and $0$ .

$\frac{\partial M}{\partial y} = 2 e^{2 x}$

Step 3

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = e^{2 x}$ .

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Step 3.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [e^{2 x}]$

Step 3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 2 x$ .

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Step 3.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d u} [e^{u}] \frac{d}{d x} [2 x]$

Step 3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$\frac{\partial N}{\partial x} = e^{u} \frac{d}{d x} [2 x]$

Step 3.2.3

Replace all occurrences of $u$ with $2 x$ .

$\frac{\partial N}{\partial x} = e^{2 x} \frac{d}{d x} [2 x]$

Step 3.3

Differentiate.

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Step 3.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$\frac{\partial N}{\partial x} = e^{2 x} (2 \frac{d}{d x} [x])$

Step 3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{\partial N}{\partial x} = e^{2 x} (2 \cdot 1)$

Step 3.3.3

Simplify the expression.

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Step 3.3.3.1

Multiply $2$ by $1$ .

$\frac{\partial N}{\partial x} = e^{2 x} \cdot 2$

Step 3.3.3.2

Move $2$ to the left of $e^{2 x}$ .

$\frac{\partial N}{\partial x} = 2 e^{2 x}$

Step 4

Check that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .

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Step 4.1

Substitute $2 e^{2 x}$ for $\frac{\partial M}{\partial y}$ and $2 e^{2 x}$ for $\frac{\partial N}{\partial x}$ .

$2 e^{2 x} = 2 e^{2 x}$

Step 4.2

Since the two sides have been shown to be equivalent, the equation is an identity.

$2 e^{2 x} = 2 e^{2 x}$ is an identity.

Step 5

Set $f (x, y)$ equal to the integral of $N (x, y)$ .

$f (x, y) = \int e^{2 x} d y$

Step 6

Integrate $N (x, y) = e^{2 x}$ to find $f (x, y)$ .

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Step 6.1

Apply the constant rule.

$f (x, y) = e^{2 x} y + C$

Step 7

Since the integral of $g (x)$ will contain an integration constant, we can replace $C$ with $g (x)$ .

$f (x, y) = e^{2 x} y + g (x)$

Step 8

Set $\frac{\partial f}{\partial x} = M (x, y)$ .

$\frac{\partial f}{\partial x} = 2 e^{2 x} y - x$

Step 9

Find $\frac{\partial f}{\partial x}$ .

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Step 9.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [e^{2 x} y + g (x)] = 2 e^{2 x} y - x$

Step 9.2

By the Sum Rule, the derivative of $e^{2 x} y + g (x)$ with respect to $x$ is $\frac{d}{d x} [e^{2 x} y] + \frac{d}{d x} [g (x)]$ .

$\frac{d}{d x} [e^{2 x} y] + \frac{d}{d x} [g (x)] = 2 e^{2 x} y - x$

Step 9.3

Evaluate $\frac{d}{d x} [e^{2 x} y]$ .

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Step 9.3.1

Since $y$ is constant with respect to $x$ , the derivative of $e^{2 x} y$ with respect to $x$ is $y \frac{d}{d x} [e^{2 x}]$ .

$y \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [g (x)] = 2 e^{2 x} y - x$

Step 9.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 2 x$ .

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Step 9.3.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

$y (\frac{d}{d u} [e^{u}] \frac{d}{d x} [2 x]) + \frac{d}{d x} [g (x)] = 2 e^{2 x} y - x$

Step 9.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$y (e^{u} \frac{d}{d x} [2 x]) + \frac{d}{d x} [g (x)] = 2 e^{2 x} y - x$

Step 9.3.2.3

Replace all occurrences of $u$ with $2 x$ .

$y (e^{2 x} \frac{d}{d x} [2 x]) + \frac{d}{d x} [g (x)] = 2 e^{2 x} y - x$

Step 9.3.3

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$y (e^{2 x} (2 \frac{d}{d x} [x])) + \frac{d}{d x} [g (x)] = 2 e^{2 x} y - x$

Step 9.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$y (e^{2 x} (2 \cdot 1)) + \frac{d}{d x} [g (x)] = 2 e^{2 x} y - x$

Step 9.3.5

Multiply $2$ by $1$ .

$y (e^{2 x} \cdot 2) + \frac{d}{d x} [g (x)] = 2 e^{2 x} y - x$

Step 9.3.6

Move $2$ to the left of $e^{2 x}$ .

$y (2 \cdot e^{2 x}) + \frac{d}{d x} [g (x)] = 2 e^{2 x} y - x$

Step 9.3.7

Move $2$ to the left of $y$ .

$2 y e^{2 x} + \frac{d}{d x} [g (x)] = 2 e^{2 x} y - x$

Step 9.4

Differentiate using the function rule which states that the derivative of $g (x)$ is $\frac{d g}{d x}$ .

$2 y e^{2 x} + \frac{d g}{d x} = 2 e^{2 x} y - x$

Step 9.5

Simplify.

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Step 9.5.1

Reorder terms.

$\frac{d g}{d x} + 2 e^{2 x} y = 2 e^{2 x} y - x$

Step 9.5.2

Reorder factors in $\frac{d g}{d x} + 2 e^{2 x} y$ .

$\frac{d g}{d x} + 2 y e^{2 x} = 2 e^{2 x} y - x$

Step 10

Solve for $\frac{d g}{d x}$ .

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Step 10.1

Solve for $\frac{d g}{d x}$ .

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Step 10.1.1

Reorder factors in $2 e^{2 x} y - x$ .

$\frac{d g}{d x} + 2 y e^{2 x} = 2 y e^{2 x} - x$

Step 10.1.2

Move all terms not containing $\frac{d g}{d x}$ to the right side of the equation.

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Step 10.1.2.1

Subtract $2 y e^{2 x}$ from both sides of the equation.

$\frac{d g}{d x} = 2 y e^{2 x} - x - 2 y e^{2 x}$

Step 10.1.2.2

Combine the opposite terms in $2 y e^{2 x} - x - 2 y e^{2 x}$ .

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Step 10.1.2.2.1

Subtract $2 y e^{2 x}$ from $2 y e^{2 x}$ .

$\frac{d g}{d x} = - x + 0$

Step 10.1.2.2.2

Add $- x$ and $0$ .

$\frac{d g}{d x} = - x$

Step 11

Find the antiderivative of $- x$ to find $g (x)$ .

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Step 11.1

Integrate both sides of $\frac{d g}{d x} = - x$ .

$\int \frac{d g}{d x} d x = \int - x d x$

Step 11.2

Evaluate $\int \frac{d g}{d x} d x$ .

$g (x) = \int - x d x$

Step 11.3

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$g (x) = - \int x d x$

Step 11.4

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$g (x) = - (\frac{1}{2} x^{2} + C)$

Step 11.5

Rewrite $- (\frac{1}{2} x^{2} + C)$ as $- \frac{1}{2} x^{2} + C$ .

$g (x) = - \frac{1}{2} x^{2} + C$

Step 12

Substitute for $g (x)$ in $f (x, y) = e^{2 x} y + g (x)$ .

$f (x, y) = e^{2 x} y - \frac{1}{2} x^{2} + C$

Step 13

Simplify $f (x, y) = e^{2 x} y - \frac{1}{2} x^{2} + C$ .

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Step 13.1

Combine $x^{2}$ and $\frac{1}{2}$ .

$f (x, y) = e^{2 x} y - \frac{x^{2}}{2} + C$

Step 13.2

Reorder factors in $f (x, y) = e^{2 x} y - \frac{x^{2}}{2} + C$ .

$f (x, y) = y e^{2 x} - \frac{x^{2}}{2} + C$