Calculus Examples

$(x + 1) d y + (2 y + 1 - 2 \cos (x)) d x = 0$

Step 1

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

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Step 1.1

Rewrite.

$(2 y + 1 - 2 \cos (x)) d x + (x + 1) d y = 0$

Step 2

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 2 y + 1 - 2 \cos (x)$ .

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Step 2.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 y + 1 - 2 \cos (x)]$

Step 2.2

By the Sum Rule, the derivative of $2 y + 1 - 2 \cos (x)$ with respect to $y$ is $\frac{d}{d y} [2 y] + \frac{d}{d y} [1] + \frac{d}{d y} [- 2 \cos (x)]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 y] + \frac{d}{d y} [1] + \frac{d}{d y} [- 2 \cos (x)]$

Step 2.3

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

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Step 2.3.1

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

$\frac{\partial M}{\partial y} = 2 \frac{d}{d y} [y] + \frac{d}{d y} [1] + \frac{d}{d y} [- 2 \cos (x)]$

Step 2.3.2

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

$\frac{\partial M}{\partial y} = 2 \cdot 1 + \frac{d}{d y} [1] + \frac{d}{d y} [- 2 \cos (x)]$

Step 2.3.3

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = 2 + \frac{d}{d y} [1] + \frac{d}{d y} [- 2 \cos (x)]$

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

Since $1$ is constant with respect to $y$ , the derivative of $1$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = 2 + 0 + \frac{d}{d y} [- 2 \cos (x)]$

Step 2.4.2

Since $- 2 \cos (x)$ is constant with respect to $y$ , the derivative of $- 2 \cos (x)$ with respect to $y$ is $0$ .

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

Step 2.5

Combine terms.

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Step 2.5.1

Add $2$ and $0$ .

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

Step 2.5.2

Add $2$ and $0$ .

$\frac{\partial M}{\partial y} = 2$

Step 3

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = x + 1$ .

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Step 3.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x + 1]$

Step 3.2

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 3.3

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

$\frac{\partial N}{\partial x} = 1 + \frac{d}{d x} [1]$

Step 3.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\frac{\partial N}{\partial x} = 1 + 0$

Step 3.5

Add $1$ and $0$ .

$\frac{\partial N}{\partial x} = 1$

Step 4

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

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Step 4.1

Substitute $2$ for $\frac{\partial M}{\partial y}$ and $1$ for $\frac{\partial N}{\partial x}$ .

$2 = 1$

Step 4.2

Since the left side does not equal the right side, the equation is not an identity.

$2 = 1$ is not an identity.

Step 5

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} d x}$ .

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Step 5.1

Substitute $2$ for $\frac{\partial M}{\partial y}$ .

$\frac{2 - \frac{\partial N}{\partial x}}{N}$

Step 5.2

Substitute $1$ for $\frac{\partial N}{\partial x}$ .

$\frac{2 - 1}{N}$

Step 5.3

Substitute $x + 1$ for $N$ .

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Step 5.3.1

Substitute $x + 1$ for $N$ .

$\frac{2 - 1}{x + 1}$

Step 5.3.2

Subtract $1$ from $2$ .

$\frac{1}{x + 1}$

Step 5.4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} d x}$ .

$μ (x, y) = e^{\int \frac{1}{x + 1} d x}$

Step 6

Evaluate the integral $e^{\int \frac{1}{x + 1} d x}$ .

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Step 6.1

Let $u = x + 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.1.1

Let $u = x + 1$ . Find $\frac{d u}{d x}$ .

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Step 6.1.1.1

Differentiate $x + 1$ .

$\frac{d}{d x} [x + 1]$

Step 6.1.1.2

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 6.1.1.3

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

$1 + \frac{d}{d x} [1]$

Step 6.1.1.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$1 + 0$

Step 6.1.1.5

Add $1$ and $0$ .

$1$

Step 6.1.2

Rewrite the problem using $u$ and $d u$ .

$μ (x, y) = e^{\int \frac{1}{u} d u}$

Step 6.2

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$μ (x, y) = e^{\ln (| u |) + C}$

Step 6.3

Simplify.

$μ (x, y) = e^{\ln (| u |)} + C$

Step 6.4

Exponentiation and log are inverse functions.

$μ (x, y) = | u | + C$

Step 6.5

Replace all occurrences of $u$ with $x + 1$ .

$μ (x, y) = | x + 1 | + C$

Step 7

Multiply both sides of $(2 y + 1 - 2 \cos (x)) d x + (x + 1) d y = 0$ by the integration factor $x + 1$ .

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Step 7.1

Multiply $2 y + 1 - 2 \cos (x)$ by $x + 1$ .

$(2 y + 1 - 2 \cos (x)) (x + 1) d x + (x + 1) d y = 0$

Step 7.2

Expand $(2 y + 1 - 2 \cos (x)) (x + 1)$ by multiplying each term in the first expression by each term in the second expression.

$(2 y x + 2 y \cdot 1 + 1 x + 1 \cdot 1 - 2 \cos (x) x - 2 \cos (x) \cdot 1) d x + (x + 1) d y = 0$

Step 7.3

Simplify each term.

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Step 7.3.1

Multiply $2$ by $1$ .

$(2 y x + 2 y + 1 x + 1 \cdot 1 - 2 \cos (x) x - 2 \cos (x) \cdot 1) d x + (x + 1) d y = 0$

Step 7.3.2

Multiply $x$ by $1$ .

$(2 y x + 2 y + x + 1 \cdot 1 - 2 \cos (x) x - 2 \cos (x) \cdot 1) d x + (x + 1) d y = 0$

Step 7.3.3

Multiply $1$ by $1$ .

$(2 y x + 2 y + x + 1 - 2 \cos (x) x - 2 \cos (x) \cdot 1) d x + (x + 1) d y = 0$

Step 7.3.4

Multiply $- 2$ by $1$ .

$(2 y x + 2 y + x + 1 - 2 \cos (x) x - 2 \cos (x)) d x + (x + 1) d y = 0$

Step 7.4

Reorder factors in $2 y x + 2 y + x + 1 - 2 \cos (x) x - 2 \cos (x)$ .

$(2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)) d x + (x + 1) d y = 0$

Step 7.5

Multiply $x + 1$ by $x + 1$ .

$(2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)) d x + (x + 1) (x + 1) d y = 0$

Step 7.6

Expand $(x + 1) (x + 1)$ using the FOIL Method.

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Step 7.6.1

Apply the distributive property.

$(2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)) d x + (x (x + 1) + 1 (x + 1)) d y = 0$

Step 7.6.2

Apply the distributive property.

$(2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)) d x + (x \cdot x + x \cdot 1 + 1 (x + 1)) d y = 0$

Step 7.6.3

Apply the distributive property.

$(2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)) d x + (x \cdot x + x \cdot 1 + 1 x + 1 \cdot 1) d y = 0$

Step 7.7

Simplify and combine like terms.

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Step 7.7.1

Simplify each term.

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Step 7.7.1.1

Multiply $x$ by $x$ .

$(2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)) d x + (x^{2} + x \cdot 1 + 1 x + 1 \cdot 1) d y = 0$

Step 7.7.1.2

Multiply $x$ by $1$ .

$(2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)) d x + (x^{2} + x + 1 x + 1 \cdot 1) d y = 0$

Step 7.7.1.3

Multiply $x$ by $1$ .

$(2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)) d x + (x^{2} + x + x + 1 \cdot 1) d y = 0$

Step 7.7.1.4

Multiply $1$ by $1$ .

$(2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)) d x + (x^{2} + x + x + 1) d y = 0$

Step 7.7.2

Add $x$ and $x$ .

$(2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)) d x + (x^{2} + 2 x + 1) d y = 0$

Step 8

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

$f (x, y) = \int x^{2} + 2 x + 1 d y$

Step 9

Integrate $N (x, y) = x^{2} + 2 x + 1$ to find $f (x, y)$ .

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Step 9.1

Apply the constant rule.

$f (x, y) = (x^{2} + 2 x + 1) y + C$

Step 10

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

$f (x, y) = (x^{2} + 2 x + 1) y + g (x)$

Step 11

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

$\frac{\partial f}{\partial x} = 2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)$

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $x$ .

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

Step 12.2

By the Sum Rule, the derivative of $(x^{2} + 2 x + 1) y + g (x)$ with respect to $x$ is $\frac{d}{d x} [(x^{2} + 2 x + 1) y] + \frac{d}{d x} [g (x)]$ .

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

Step 12.3

Evaluate $\frac{d}{d x} [(x^{2} + 2 x + 1) y]$ .

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Step 12.3.1

Since $y$ is constant with respect to $x$ , the derivative of $(x^{2} + 2 x + 1) y$ with respect to $x$ is $y \frac{d}{d x} [x^{2} + 2 x + 1]$ .

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

Step 12.3.2

By the Sum Rule, the derivative of $x^{2} + 2 x + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

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

Step 12.3.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 12.3.4

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

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

Step 12.3.5

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

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

Step 12.3.6

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$y (2 x + 2 \cdot 1 + 0) + \frac{d}{d x} [g (x)] = 2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)$

Step 12.3.7

Multiply $2$ by $1$ .

$y (2 x + 2 + 0) + \frac{d}{d x} [g (x)] = 2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)$

Step 12.3.8

Add $2 x + 2$ and $0$ .

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

Step 12.4

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

$y (2 x + 2) + \frac{d g}{d x} = 2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)$

Step 12.5

Simplify.

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Step 12.5.1

Apply the distributive property.

$y (2 x) + y \cdot 2 + \frac{d g}{d x} = 2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)$

Step 12.5.2

Combine terms.

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Step 12.5.2.1

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

$2 \cdot y x + y \cdot 2 + \frac{d g}{d x} = 2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)$

Step 12.5.2.2

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

$2 y x + 2 y + \frac{d g}{d x} = 2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)$

Step 12.5.3

Reorder terms.

$\frac{d g}{d x} + 2 x y + 2 y = 2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)$

Step 13

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

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Step 13.1

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

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Step 13.1.1

Subtract $2 x y$ from both sides of the equation.

$\frac{d g}{d x} + 2 y = 2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x) - 2 x y$

Step 13.1.2

Subtract $2 y$ from both sides of the equation.

$\frac{d g}{d x} = 2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x) - 2 x y - 2 y$

Step 13.1.3

Combine the opposite terms in $2 y x + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x) - 2 x y - 2 y$ .

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Step 13.1.3.1

Reorder the factors in the terms $2 y x$ and $- 2 x y$ .

$\frac{d g}{d x} = 2 x y + 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x) - 2 x y - 2 y$

Step 13.1.3.2

Subtract $2 x y$ from $2 x y$ .

$\frac{d g}{d x} = 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x) + 0 - 2 y$

Step 13.1.3.3

Add $2 y + x + 1 - 2 x \cos (x) - 2 \cos (x)$ and $0$ .

$\frac{d g}{d x} = 2 y + x + 1 - 2 x \cos (x) - 2 \cos (x) - 2 y$

Step 13.1.3.4

Subtract $2 y$ from $2 y$ .

$\frac{d g}{d x} = x + 1 - 2 x \cos (x) - 2 \cos (x) + 0$

Step 13.1.3.5

Add $x + 1 - 2 x \cos (x) - 2 \cos (x)$ and $0$ .

$\frac{d g}{d x} = x + 1 - 2 x \cos (x) - 2 \cos (x)$

Step 14

Find the antiderivative of $x + 1 - 2 x \cos (x) - 2 \cos (x)$ to find $g (x)$ .

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Step 14.1

Integrate both sides of $\frac{d g}{d x} = x + 1 - 2 x \cos (x) - 2 \cos (x)$ .

$\int \frac{d g}{d x} d x = \int x + 1 - 2 x \cos (x) - 2 \cos (x) d x$

Step 14.2

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

$g (x) = \int x + 1 - 2 x \cos (x) - 2 \cos (x) d x$

Step 14.3

Split the single integral into multiple integrals.

$g (x) = \int x d x + \int d x + \int - 2 x \cos (x) d x + \int - 2 \cos (x) d x$

Step 14.4

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

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

Step 14.5

Apply the constant rule.

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

Step 14.6

Since $- 2$ is constant with respect to $x$ , move $- 2$ out of the integral.

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

Step 14.7

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x$ and $d v = \cos (x)$ .

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

Step 14.8

The integral of $\sin (x)$ with respect to $x$ is $- \cos (x)$ .

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

Step 14.9

Since $- 2$ is constant with respect to $x$ , move $- 2$ out of the integral.

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

Step 14.10

The integral of $\cos (x)$ with respect to $x$ is $\sin (x)$ .

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

Step 14.11

Simplify.

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

Step 14.12

Reorder terms.

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

Step 15

Substitute for $g (x)$ in $f (x, y) = (x^{2} + 2 x + 1) y + g (x)$ .

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

Step 16

Simplify each term.

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Step 16.1

Apply the distributive property.

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

Step 16.2

Multiply $y$ by $1$ .

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

Step 16.3

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

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