Calculus Examples

$(2 x + Y) d x + (2 Y + x) d y = 0$

Step 1

Subtract $(2 x + Y) d x$ from both sides of the equation.

$(2 Y + x) d y = - (2 x + Y) d x$

Step 2

Multiply both sides by $\frac{1}{2 Y + x}$ .

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $2 Y + x$ .

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Step 3.1.1

Cancel the common factor.

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

Step 3.1.2

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

Apply the distributive property.

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

Step 3.4

Multiply $- \frac{1}{2 Y + x} (2 x)$ .

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Step 3.4.1

Multiply $2$ by $- 1$ .

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

Step 3.4.2

Combine $- 2$ and $\frac{1}{2 Y + x}$ .

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

Step 3.4.3

Combine $\frac{- 2}{2 Y + x}$ and $x$ .

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

Step 3.5

Combine $Y$ and $\frac{1}{2 Y + x}$ .

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

Step 3.6

Combine the numerators over the common denominator.

$1 d y = \frac{- 2 x - Y}{2 Y + x} d x$

Step 3.7

Factor $- 1$ out of $- 2 x$ .

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

Step 3.8

Factor $- 1$ out of $- Y$ .

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

Step 3.9

Factor $- 1$ out of $- (2 x) - (Y)$ .

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

Step 3.10

Rewrite $- (2 x + Y)$ as $- 1 (2 x + Y)$ .

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

Step 3.11

Move the negative in front of the fraction.

$1 d y = - \frac{2 x + Y}{2 Y + x} d x$

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

$\int d y = \int - \frac{2 x + Y}{2 Y + x} d x$

Step 4.2

Apply the constant rule.

$y + C_{1} = \int - \frac{2 x + Y}{2 Y + x} d x$

Step 4.3

Integrate the right side.

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Step 4.3.1

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

$y + C_{1} = - \int \frac{2 x + Y}{2 Y + x} d x$

Step 4.3.2

Let $u = 2 Y + x$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.3.2.1

Let $u = 2 Y + x$ . Find $\frac{d u}{d x}$ .

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Step 4.3.2.1.1

Differentiate $2 Y + x$ .

$\frac{d}{d x} [2 Y + x]$

Step 4.3.2.1.2

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

$\frac{d}{d x} [2 Y] + \frac{d}{d x} [x]$

Step 4.3.2.1.3

Since $2 Y$ is constant with respect to $x$ , the derivative of $2 Y$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [x]$

Step 4.3.2.1.4

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

$0 + 1$

Step 4.3.2.1.5

Add $0$ and $1$ .

$1$

Step 4.3.2.2

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

$y + C_{1} = - \int \frac{2 (u - 2 Y) + Y}{u} d u$

Step 4.3.3

Split the fraction into multiple fractions.

$y + C_{1} = - \int \frac{2 (u - 2 Y)}{u} + \frac{Y}{u} d u$

Step 4.3.4

Split the single integral into multiple integrals.

$y + C_{1} = - (\int \frac{2 (u - 2 Y)}{u} d u + \int \frac{Y}{u} d u)$

Step 4.3.5

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

$y + C_{1} = - (2 \int \frac{u - 2 Y}{u} d u + \int \frac{Y}{u} d u)$

Step 4.3.6

Split the fraction into multiple fractions.

$y + C_{1} = - (2 \int \frac{u}{u} + \frac{- 2 Y}{u} d u + \int \frac{Y}{u} d u)$

Step 4.3.7

Split the single integral into multiple integrals.

$y + C_{1} = - (2 (\int \frac{u}{u} d u + \int \frac{- 2 Y}{u} d u) + \int \frac{Y}{u} d u)$

Step 4.3.8

Cancel the common factor of $u$ .

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Step 4.3.8.1

Cancel the common factor.

$y + C_{1} = - (2 (\int \frac{u}{u} d u + \int \frac{- 2 Y}{u} d u) + \int \frac{Y}{u} d u)$

Step 4.3.8.2

Rewrite the expression.

$y + C_{1} = - (2 (\int d u + \int \frac{- 2 Y}{u} d u) + \int \frac{Y}{u} d u)$

Step 4.3.9

Apply the constant rule.

$y + C_{1} = - (2 (u + C_{2} + \int \frac{- 2 Y}{u} d u) + \int \frac{Y}{u} d u)$

Step 4.3.10

Move the negative in front of the fraction.

$y + C_{1} = - (2 (u + C_{2} + \int - \frac{2 Y}{u} d u) + \int \frac{Y}{u} d u)$

Step 4.3.11

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

$y + C_{1} = - (2 (u + C_{2} - \int \frac{2 Y}{u} d u) + \int \frac{Y}{u} d u)$

Step 4.3.12

Since $2 Y$ is constant with respect to $u$ , move $2 Y$ out of the integral.

$y + C_{1} = - (2 (u + C_{2} - (2 Y \int \frac{1}{u} d u)) + \int \frac{Y}{u} d u)$

Step 4.3.13

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

$y + C_{1} = - (2 (u + C_{2} - (2 Y (\ln (| u |) + C_{3}))) + \int \frac{Y}{u} d u)$

Step 4.3.14

Multiply $2$ by $- 1$ .

$y + C_{1} = - (2 (u + C_{2} - 2 Y (\ln (| u |) + C_{3})) + \int \frac{Y}{u} d u)$

Step 4.3.15

Since $Y$ is constant with respect to $u$ , move $Y$ out of the integral.

$y + C_{1} = - (2 (u + C_{2} - 2 Y (\ln (| u |) + C_{3})) + Y \int \frac{1}{u} d u)$

Step 4.3.16

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

$y + C_{1} = - (2 (u + C_{2} - 2 Y (\ln (| u |) + C_{3})) + Y (\ln (| u |) + C_{4}))$

Step 4.3.17

Simplify.

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Step 4.3.17.1

Simplify.

$y + C_{1} = - (2 u - 4 Y \ln (| u |) + Y \ln (| u |)) + C_{5}$

Step 4.3.17.2

Add $- 4 Y \ln (| u |)$ and $Y \ln (| u |)$ .

$y + C_{1} = - (2 u - 3 Y \ln (| u |)) + C_{5}$

Step 4.3.18

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

$y + C_{1} = - (2 (2 Y + x) - 3 Y \ln (| 2 Y + x |)) + C_{5}$

Step 4.3.19

Simplify.

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Step 4.3.19.1

Simplify each term.

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Step 4.3.19.1.1

Apply the distributive property.

$y + C_{1} = - (2 (2 Y) + 2 x - 3 Y \ln (| 2 Y + x |)) + C_{5}$

Step 4.3.19.1.2

Multiply $2$ by $2$ .

$y + C_{1} = - (4 Y + 2 x - 3 Y \ln (| 2 Y + x |)) + C_{5}$

Step 4.3.19.2

Apply the distributive property.

$y + C_{1} = - (4 Y) - (2 x) - (- 3 Y \ln (| 2 Y + x |)) + C_{5}$

Step 4.3.19.3

Simplify.

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Step 4.3.19.3.1

Multiply $4$ by $- 1$ .

$y + C_{1} = - 4 Y - (2 x) - (- 3 Y \ln (| 2 Y + x |)) + C_{5}$

Step 4.3.19.3.2

Multiply $2$ by $- 1$ .

$y + C_{1} = - 4 Y - 2 x - (- 3 Y \ln (| 2 Y + x |)) + C_{5}$

Step 4.3.19.3.3

Multiply $- 3$ by $- 1$ .

$y + C_{1} = - 4 Y - 2 x + 3 (Y \ln (| 2 Y + x |)) + C_{5}$

$y + C_{1} = - 4 Y - 2 x + 3 Y \ln (| 2 Y + x |) + C_{5}$

Step 4.3.20

Reorder terms.

$y + C_{1} = - 4 Y + 3 Y \ln (| 2 Y + x |) - 2 x + C_{5}$

Step 4.4

Group the constant of integration on the right side as $C$ .

$y = - 4 Y + 3 Y \ln (| 2 Y + x |) - 2 x + C$