Calculus Examples

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

Step 1

Let $v = x + y$ . Substitute $v$ for all occurrences of $x + y$ .

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

Step 2

Find $\frac{d v}{d x}$ by differentiating $x + y$ .

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Step 2.1

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

$\frac{d v}{d x} = \frac{d}{d x} [x] + \frac{d}{d x} [y]$

Step 2.2

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

$\frac{d v}{d x} = 1 + \frac{d}{d x} [y]$

Step 2.3

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$\frac{d v}{d x} = 1 + y'$

Step 3

Substitute the derivative back in to the differential equation.

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

Step 4

Separate the variables.

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Step 4.1

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

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Step 4.1.1

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

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Step 4.1.1.1

Add $v$ to both sides of the equation.

$v \frac{d v}{d x} = 2 v + 1 + v$

Step 4.1.1.2

Add $2 v$ and $v$ .

$v \frac{d v}{d x} = 3 v + 1$

Step 4.1.2

Divide each term in $v \frac{d v}{d x} = 3 v + 1$ by $v$ and simplify.

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Step 4.1.2.1

Divide each term in $v \frac{d v}{d x} = 3 v + 1$ by $v$ .

$\frac{v \frac{d v}{d x}}{v} = \frac{3 v}{v} + \frac{1}{v}$

Step 4.1.2.2

Simplify the left side.

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Step 4.1.2.2.1

Cancel the common factor of $v$ .

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Step 4.1.2.2.1.1

Cancel the common factor.

$\frac{v \frac{d v}{d x}}{v} = \frac{3 v}{v} + \frac{1}{v}$

Step 4.1.2.2.1.2

Divide $\frac{d v}{d x}$ by $1$ .

$\frac{d v}{d x} = \frac{3 v}{v} + \frac{1}{v}$

Step 4.1.2.3

Simplify the right side.

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Step 4.1.2.3.1

Cancel the common factor of $v$ .

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Step 4.1.2.3.1.1

Cancel the common factor.

$\frac{d v}{d x} = \frac{3 v}{v} + \frac{1}{v}$

Step 4.1.2.3.1.2

Divide $3$ by $1$ .

$\frac{d v}{d x} = 3 + \frac{1}{v}$

Step 4.2

Multiply both sides by $\frac{1}{3 + \frac{1}{v}}$ .

$\frac{1}{3 + \frac{1}{v}} \frac{d v}{d x} = \frac{1}{3 + \frac{1}{v}} (3 + \frac{1}{v})$

Step 4.3

Cancel the common factor of $3 + \frac{1}{v}$ .

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Step 4.3.1

Cancel the common factor.

$\frac{1}{3 + \frac{1}{v}} \frac{d v}{d x} = \frac{1}{3 + \frac{1}{v}} (3 + \frac{1}{v})$

Step 4.3.2

Rewrite the expression.

$\frac{1}{3 + \frac{1}{v}} \frac{d v}{d x} = 1$

Step 4.4

Rewrite the equation.

$\frac{1}{3 + \frac{1}{v}} d v = 1 d x$

Step 5

Integrate both sides.

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Step 5.1

Set up an integral on each side.

$\int \frac{1}{3 + \frac{1}{v}} d v = \int d x$

Step 5.2

Integrate the left side.

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Step 5.2.1

Simplify.

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Step 5.2.1.1

Simplify the denominator.

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Step 5.2.1.1.1

To write $3$ as a fraction with a common denominator, multiply by $\frac{v}{v}$ .

$\int \frac{1}{\frac{3 v}{v} + \frac{1}{v}} d v = \int d x$

Step 5.2.1.1.2

Combine the numerators over the common denominator.

$\int \frac{1}{\frac{3 v + 1}{v}} d v = \int d x$

Step 5.2.1.2

Multiply the numerator by the reciprocal of the denominator.

$\int 1 \frac{v}{3 v + 1} d v = \int d x$

Step 5.2.1.3

Multiply $\frac{v}{3 v + 1}$ by $1$ .

$\int \frac{v}{3 v + 1} d v = \int d x$

Step 5.2.2

Divide $v$ by $3 v + 1$ .

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Step 5.2.2.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$3 v$

$1$

$v$

$0$

Step 5.2.2.2

Divide the highest order term in the dividend $v$ by the highest order term in divisor $3 v$ .

					$\frac{1}{3}$
	$3 v$	+	$1$		$v$	+	$0$

Step 5.2.2.3

Multiply the new quotient term by the divisor.

				$\frac{1}{3}$
$3 v$	+	$1$		$v$	+	$0$
			+	$v$	+	$\frac{1}{3}$

Step 5.2.2.4

The expression needs to be subtracted from the dividend, so change all the signs in $v + \frac{1}{3}$

				$\frac{1}{3}$
$3 v$	+	$1$		$v$	+	$0$
			-	$v$	-	$\frac{1}{3}$

Step 5.2.2.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$\frac{1}{3}$
$3 v$	+	$1$		$v$	+	$0$
			-	$v$	-	$\frac{1}{3}$
					-	$\frac{1}{3}$

Step 5.2.2.6

The final answer is the quotient plus the remainder over the divisor.

$\int \frac{1}{3} - \frac{1}{3 (3 v + 1)} d v = \int d x$

Step 5.2.3

Split the single integral into multiple integrals.

$\int \frac{1}{3} d v + \int - \frac{1}{3 (3 v + 1)} d v = \int d x$

Step 5.2.4

Apply the constant rule.

$\frac{1}{3} v + C_{1} + \int - \frac{1}{3 (3 v + 1)} d v = \int d x$

Step 5.2.5

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

$\frac{1}{3} v + C_{1} - \int \frac{1}{3 (3 v + 1)} d v = \int d x$

Step 5.2.6

Since $\frac{1}{3}$ is constant with respect to $v$ , move $\frac{1}{3}$ out of the integral.

$\frac{1}{3} v + C_{1} - (\frac{1}{3} \int \frac{1}{3 v + 1} d v) = \int d x$

Step 5.2.7

Let $u = 3 v + 1$ . Then $d u = 3 d v$ , so $\frac{1}{3} d u = d v$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.2.7.1

Let $u = 3 v + 1$ . Find $\frac{d u}{d v}$ .

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Step 5.2.7.1.1

Differentiate $3 v + 1$ .

$\frac{d}{d v} [3 v + 1]$

Step 5.2.7.1.2

By the Sum Rule, the derivative of $3 v + 1$ with respect to $v$ is $\frac{d}{d v} [3 v] + \frac{d}{d v} [1]$ .

$\frac{d}{d v} [3 v] + \frac{d}{d v} [1]$

Step 5.2.7.1.3

Evaluate $\frac{d}{d v} [3 v]$ .

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Step 5.2.7.1.3.1

Since $3$ is constant with respect to $v$ , the derivative of $3 v$ with respect to $v$ is $3 \frac{d}{d v} [v]$ .

$3 \frac{d}{d v} [v] + \frac{d}{d v} [1]$

Step 5.2.7.1.3.2

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

$3 \cdot 1 + \frac{d}{d v} [1]$

Step 5.2.7.1.3.3

Multiply $3$ by $1$ .

$3 + \frac{d}{d v} [1]$

Step 5.2.7.1.4

Differentiate using the Constant Rule.

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Step 5.2.7.1.4.1

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

$3 + 0$

Step 5.2.7.1.4.2

Add $3$ and $0$ .

$3$

Step 5.2.7.2

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

$\frac{1}{3} v + C_{1} - \frac{1}{3} \int \frac{1}{u} \cdot \frac{1}{3} d u = \int d x$

Step 5.2.8

Simplify.

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Step 5.2.8.1

Multiply $\frac{1}{u}$ by $\frac{1}{3}$ .

$\frac{1}{3} v + C_{1} - \frac{1}{3} \int \frac{1}{u \cdot 3} d u = \int d x$

Step 5.2.8.2

Move $3$ to the left of $u$ .

$\frac{1}{3} v + C_{1} - \frac{1}{3} \int \frac{1}{3 u} d u = \int d x$

Step 5.2.9

Since $\frac{1}{3}$ is constant with respect to $u$ , move $\frac{1}{3}$ out of the integral.

$\frac{1}{3} v + C_{1} - \frac{1}{3} (\frac{1}{3} \int \frac{1}{u} d u) = \int d x$

Step 5.2.10

Simplify.

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Step 5.2.10.1

Multiply $\frac{1}{3}$ by $\frac{1}{3}$ .

$\frac{1}{3} v + C_{1} - \frac{1}{3 \cdot 3} \int \frac{1}{u} d u = \int d x$

Step 5.2.10.2

Multiply $3$ by $3$ .

$\frac{1}{3} v + C_{1} - \frac{1}{9} \int \frac{1}{u} d u = \int d x$

Step 5.2.11

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

$\frac{1}{3} v + C_{1} - \frac{1}{9} (\ln (| u |) + C_{2}) = \int d x$

Step 5.2.12

Simplify.

$\frac{1}{3} v - \frac{1}{9} \ln (| u |) + C_{3} = \int d x$

Step 5.3

Apply the constant rule.

$\frac{1}{3} v - \frac{1}{9} \ln (| u |) + C_{3} = x + C_{4}$

Step 5.4

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

$\frac{1}{3} v - \frac{1}{9} \ln (| u |) = x + C$

Step 6

Solve for $v$ .

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Step 6.1

Simplify each term.

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Step 6.1.1

Combine $\frac{1}{3}$ and $v$ .

$\frac{v}{3} - \frac{1}{9} \ln (| u |) = x + C$

Step 6.1.2

Multiply $- \frac{1}{9} \ln (| u |)$ .

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Step 6.1.2.1

Reorder $\ln (| u |)$ and $\frac{1}{9}$ .

$\frac{v}{3} - (\frac{1}{9} \ln (| u |)) = x + C$

Step 6.1.2.2

Simplify $\frac{1}{9} \ln (| u |)$ by moving $\frac{1}{9}$ inside the logarithm.

$\frac{v}{3} - \ln ({| u |}^{\frac{1}{9}}) = x + C$

Step 6.2

Add $\ln ({| u |}^{\frac{1}{9}})$ to both sides of the equation.

$\frac{v}{3} = x + C + \ln ({| u |}^{\frac{1}{9}})$

Step 6.3

Multiply both sides of the equation by $3$ .

$3 \frac{v}{3} = 3 (x + C + \ln ({| u |}^{\frac{1}{9}}))$

Step 6.4

Simplify both sides of the equation.

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Step 6.4.1

Simplify the left side.

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Step 6.4.1.1

Cancel the common factor of $3$ .

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Step 6.4.1.1.1

Cancel the common factor.

$3 \frac{v}{3} = 3 (x + C + \ln ({| u |}^{\frac{1}{9}}))$

Step 6.4.1.1.2

Rewrite the expression.

$v = 3 (x + C + \ln ({| u |}^{\frac{1}{9}}))$

Step 6.4.2

Simplify the right side.

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Step 6.4.2.1

Apply the distributive property.

$v = 3 x + 3 C + 3 \ln ({| u |}^{\frac{1}{9}})$

Step 6.5

Simplify $3 x + 3 C + 3 \ln ({| u |}^{\frac{1}{9}})$ .

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Step 6.5.1

Simplify each term.

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Step 6.5.1.1

Simplify $3 \ln ({| u |}^{\frac{1}{9}})$ by moving $3$ inside the logarithm.

$v = 3 x + 3 C + \ln ({({| u |}^{\frac{1}{9}})}^{3})$

Step 6.5.1.2

Multiply the exponents in ${({| u |}^{\frac{1}{9}})}^{3}$ .

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Step 6.5.1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$v = 3 x + 3 C + \ln ({| u |}^{\frac{1}{9} \cdot 3})$

Step 6.5.1.2.2

Cancel the common factor of $3$ .

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Step 6.5.1.2.2.1

Factor $3$ out of $9$ .

$v = 3 x + 3 C + \ln ({| u |}^{\frac{1}{3 (3)} \cdot 3})$

Step 6.5.1.2.2.2

Cancel the common factor.

$v = 3 x + 3 C + \ln ({| u |}^{\frac{1}{3 \cdot 3} \cdot 3})$

Step 6.5.1.2.2.3

Rewrite the expression.

$v = 3 x + 3 C + \ln ({| u |}^{\frac{1}{3}})$

Step 6.5.2

Reorder $3 x$ and $3 C$ .

$v = 3 C + 3 x + \ln ({| u |}^{\frac{1}{3}})$

Step 7

Simplify the constant of integration.

$v = C + 3 x + \ln ({| u |}^{\frac{1}{3}})$

Step 8

Replace all occurrences of $v$ with $x + y$ .

$x + y = C + 3 x + \ln ({| u |}^{\frac{1}{3}})$

Step 9

Move all terms not containing $y$ to the right side of the equation.

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Step 9.1

Subtract $x$ from both sides of the equation.

$y = C + 3 x + \ln ({| u |}^{\frac{1}{3}}) - x$

Step 9.2

Subtract $x$ from $3 x$ .

$y = C + 2 x + \ln ({| u |}^{\frac{1}{3}})$