Calculus Examples

$\frac{d y}{d x} = \frac{1}{3} (1 + 3 x) y^{4} - \frac{1}{3} y$

Step 1

Rewrite the equation as $M (x) \frac{d y}{d x} + P (x) y = Q (x)$ .

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Step 1.1

Apply the distributive property.

$\frac{d y}{d x} = (\frac{1}{3} \cdot 1 + \frac{1}{3} (3 x)) y^{4} - \frac{1}{3} y$

Step 1.2

Add $\frac{1}{3} y$ to both sides of the equation.

$\frac{d y}{d x} + \frac{1}{3} y = (\frac{1}{3} \cdot 1 + \frac{1}{3} (3 x)) y^{4}$

Step 1.3

Reorder terms.

$\frac{d y}{d x} + \frac{1}{3} y = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) y^{4}$

Step 2

To solve the differential equation, let $v = y^{1 - n}$ where $n$ is the exponent of $y^{4}$ .

$v = y^{- 3}$

Step 3

Solve the equation for $y$ .

$y = v^{- \frac{1}{3}}$

Step 4

Take the derivative of $y$ with respect to $x$ .

$y' = v^{- \frac{1}{3}}$

Step 5

Take the derivative of $v^{- \frac{1}{3}}$ with respect to $x$ .

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Step 5.1

Take the derivative of $v^{- \frac{1}{3}}$ .

$y' = \frac{d}{d x} [v^{- \frac{1}{3}}]$

Step 5.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$y' = \frac{d}{d x} [\frac{1}{v^{\frac{1}{3}}}]$

Step 5.3

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = 1$ and $g (x) = v^{\frac{1}{3}}$ .

$y' = \frac{v^{\frac{1}{3}} \frac{d}{d x} [1] - 1 \cdot 1 \frac{d}{d x} [v^{\frac{1}{3}}]}{{(v^{\frac{1}{3}})}^{2}}$

Step 5.4

Differentiate using the Constant Rule.

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Step 5.4.1

Multiply $- 1$ by $1$ .

$y' = \frac{v^{\frac{1}{3}} \frac{d}{d x} [1] - \frac{d}{d x} [v^{\frac{1}{3}}]}{{(v^{\frac{1}{3}})}^{2}}$

Step 5.4.2

Multiply the exponents in ${(v^{\frac{1}{3}})}^{2}$ .

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Step 5.4.2.1

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

$y' = \frac{v^{\frac{1}{3}} \frac{d}{d x} [1] - \frac{d}{d x} [v^{\frac{1}{3}}]}{v^{\frac{1}{3} \cdot 2}}$

Step 5.4.2.2

Combine $\frac{1}{3}$ and $2$ .

$y' = \frac{v^{\frac{1}{3}} \frac{d}{d x} [1] - \frac{d}{d x} [v^{\frac{1}{3}}]}{v^{\frac{2}{3}}}$

Step 5.4.3

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

$y' = \frac{v^{\frac{1}{3}} \cdot 0 - \frac{d}{d x} [v^{\frac{1}{3}}]}{v^{\frac{2}{3}}}$

Step 5.4.4

Simplify the expression.

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Step 5.4.4.1

Multiply $v^{\frac{1}{3}}$ by $0$ .

$y' = \frac{0 - \frac{d}{d x} [v^{\frac{1}{3}}]}{v^{\frac{2}{3}}}$

Step 5.4.4.2

Subtract $\frac{d}{d x} [v^{\frac{1}{3}}]$ from $0$ .

$y' = \frac{- \frac{d}{d x} [v^{\frac{1}{3}}]}{v^{\frac{2}{3}}}$

Step 5.4.4.3

Move the negative in front of the fraction.

$y' = - \frac{\frac{d}{d x} [v^{\frac{1}{3}}]}{v^{\frac{2}{3}}}$

Step 5.5

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{3}}$ and $g (x) = v$ .

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Step 5.5.1

To apply the Chain Rule, set $u$ as $v$ .

$y' = - \frac{\frac{d}{d u} [u^{\frac{1}{3}}] \frac{d}{d x} [v]}{v^{\frac{2}{3}}}$

Step 5.5.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{1}{3}$ .

$y' = - \frac{\frac{1}{3} u^{\frac{1}{3} - 1} \frac{d}{d x} [v]}{v^{\frac{2}{3}}}$

Step 5.5.3

Replace all occurrences of $u$ with $v$ .

$y' = - \frac{\frac{1}{3} v^{\frac{1}{3} - 1} \frac{d}{d x} [v]}{v^{\frac{2}{3}}}$

Step 5.6

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y' = - \frac{\frac{1}{3} v^{\frac{1}{3} - 1 \cdot \frac{3}{3}} \frac{d}{d x} [v]}{v^{\frac{2}{3}}}$

Step 5.7

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

$y' = - \frac{\frac{1}{3} v^{\frac{1}{3} + \frac{- 1 \cdot 3}{3}} \frac{d}{d x} [v]}{v^{\frac{2}{3}}}$

Step 5.8

Combine the numerators over the common denominator.

$y' = - \frac{\frac{1}{3} v^{\frac{1 - 1 \cdot 3}{3}} \frac{d}{d x} [v]}{v^{\frac{2}{3}}}$

Step 5.9

Simplify the numerator.

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Step 5.9.1

Multiply $- 1$ by $3$ .

$y' = - \frac{\frac{1}{3} v^{\frac{1 - 3}{3}} \frac{d}{d x} [v]}{v^{\frac{2}{3}}}$

Step 5.9.2

Subtract $3$ from $1$ .

$y' = - \frac{\frac{1}{3} v^{\frac{- 2}{3}} \frac{d}{d x} [v]}{v^{\frac{2}{3}}}$

Step 5.10

Move the negative in front of the fraction.

$y' = - \frac{\frac{1}{3} v^{- \frac{2}{3}} \frac{d}{d x} [v]}{v^{\frac{2}{3}}}$

Step 5.11

Combine $\frac{1}{3}$ and $v^{- \frac{2}{3}}$ .

$y' = - \frac{\frac{v^{- \frac{2}{3}}}{3} \frac{d}{d x} [v]}{v^{\frac{2}{3}}}$

Step 5.12

Move $v^{- \frac{2}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$y' = - \frac{\frac{1}{3 v^{\frac{2}{3}}} \frac{d}{d x} [v]}{v^{\frac{2}{3}}}$

Step 5.13

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

$y' = - \frac{\frac{1}{3 v^{\frac{2}{3}}} v'}{v^{\frac{2}{3}}}$

Step 5.14

Combine $\frac{1}{3 v^{\frac{2}{3}}}$ and $v'$ .

$y' = - \frac{\frac{v'}{3 v^{\frac{2}{3}}}}{v^{\frac{2}{3}}}$

Step 5.15

Rewrite $\frac{\frac{v'}{3 v^{\frac{2}{3}}}}{v^{\frac{2}{3}}}$ as a product.

$y' = - (\frac{v'}{3 v^{\frac{2}{3}}} \cdot \frac{1}{v^{\frac{2}{3}}})$

Step 5.16

Multiply $\frac{v'}{3 v^{\frac{2}{3}}}$ by $\frac{1}{v^{\frac{2}{3}}}$ .

$y' = - \frac{v'}{3 v^{\frac{2}{3}} v^{\frac{2}{3}}}$

Step 5.17

Multiply $v^{\frac{2}{3}}$ by $v^{\frac{2}{3}}$ by adding the exponents.

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Step 5.17.1

Move $v^{\frac{2}{3}}$ .

$y' = - \frac{v'}{3 (v^{\frac{2}{3}} v^{\frac{2}{3}})}$

Step 5.17.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y' = - \frac{v'}{3 v^{\frac{2}{3} + \frac{2}{3}}}$

Step 5.17.3

Combine the numerators over the common denominator.

$y' = - \frac{v'}{3 v^{\frac{2 + 2}{3}}}$

Step 5.17.4

Add $2$ and $2$ .

$y' = - \frac{v'}{3 v^{\frac{4}{3}}}$

Step 6

Substitute $- \frac{v'}{3 v^{\frac{4}{3}}}$ for $\frac{d y}{d x}$ and $v^{- \frac{1}{3}}$ for $y$ in the original equation $\frac{d y}{d x} + \frac{1}{3} y = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) y^{4}$ .

$- \frac{v'}{3 v^{\frac{4}{3}}} + \frac{1}{3} v^{- \frac{1}{3}} = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4}$

Step 7

Solve the substituted differential equation.

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Step 7.1

Multiply each term in $- \frac{\frac{d v}{d x}}{3 v^{\frac{4}{3}}} + \frac{1}{3} v^{- \frac{1}{3}} = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4}$ by $- (3 v^{\frac{4}{3}})$ to eliminate the fractions.

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Step 7.1.1

Multiply each term in $- \frac{\frac{d v}{d x}}{3 v^{\frac{4}{3}}} + \frac{1}{3} v^{- \frac{1}{3}} = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4}$ by $- (3 v^{\frac{4}{3}})$ .

$- \frac{\frac{d v}{d x}}{3 v^{\frac{4}{3}}} (- (3 v^{\frac{4}{3}})) + \frac{1}{3} v^{- \frac{1}{3}} (- (3 v^{\frac{4}{3}})) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2

Simplify the left side.

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Step 7.1.2.1

Simplify each term.

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Step 7.1.2.1.1

Cancel the common factor of $3 v^{\frac{4}{3}}$ .

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Step 7.1.2.1.1.1

Move the leading negative in $- \frac{\frac{d v}{d x}}{3 v^{\frac{4}{3}}}$ into the numerator.

$\frac{- \frac{d v}{d x}}{3 v^{\frac{4}{3}}} (- (3 v^{\frac{4}{3}})) + \frac{1}{3} v^{- \frac{1}{3}} (- (3 v^{\frac{4}{3}})) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.1.2

Factor $3 v^{\frac{4}{3}}$ out of $- (3 v^{\frac{4}{3}})$ .

$\frac{- \frac{d v}{d x}}{3 v^{\frac{4}{3}}} (3 v^{\frac{4}{3}} (- 1)) + \frac{1}{3} v^{- \frac{1}{3}} (- (3 v^{\frac{4}{3}})) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.1.3

Cancel the common factor.

$\frac{- \frac{d v}{d x}}{3 v^{\frac{4}{3}}} (3 v^{\frac{4}{3}} \cdot - 1) + \frac{1}{3} v^{- \frac{1}{3}} (- (3 v^{\frac{4}{3}})) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.1.4

Rewrite the expression.

$- \frac{d v}{d x} \cdot - 1 + \frac{1}{3} v^{- \frac{1}{3}} (- (3 v^{\frac{4}{3}})) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.2

Multiply $- 1$ by $- 1$ .

$1 \frac{d v}{d x} + \frac{1}{3} v^{- \frac{1}{3}} (- (3 v^{\frac{4}{3}})) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.3

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

$\frac{d v}{d x} + \frac{1}{3} v^{- \frac{1}{3}} (- (3 v^{\frac{4}{3}})) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.4

Multiply $v^{- \frac{1}{3}}$ by $v^{\frac{4}{3}}$ by adding the exponents.

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Step 7.1.2.1.4.1

Move $v^{\frac{4}{3}}$ .

$\frac{d v}{d x} + \frac{1}{3} (v^{\frac{4}{3}} v^{- \frac{1}{3}}) (- 1 \cdot (3)) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.4.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{d v}{d x} + \frac{1}{3} v^{\frac{4}{3} - \frac{1}{3}} (- 1 \cdot (3)) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.4.3

Combine the numerators over the common denominator.

$\frac{d v}{d x} + \frac{1}{3} v^{\frac{4 - 1}{3}} (- 1 \cdot (3)) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.4.4

Subtract $1$ from $4$ .

$\frac{d v}{d x} + \frac{1}{3} v^{\frac{3}{3}} (- 1 \cdot (3)) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.4.5

Divide $3$ by $3$ .

$\frac{d v}{d x} + \frac{1}{3} v^{1} (- 1 \cdot (3)) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.5

Simplify $\frac{1}{3} v^{1} (- 1 \cdot (3))$ .

$\frac{d v}{d x} + \frac{1}{3} v (- 1 \cdot (3)) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.6

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

$\frac{d v}{d x} + \frac{v}{3} (- 1 \cdot (3)) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.7

Cancel the common factor of $3$ .

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Step 7.1.2.1.7.1

Factor $3$ out of $- 1 \cdot (3)$ .

$\frac{d v}{d x} + \frac{v}{3} (3 \cdot - 1) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.7.2

Cancel the common factor.

$\frac{d v}{d x} + \frac{v}{3} (3 \cdot - 1) = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.7.3

Rewrite the expression.

$\frac{d v}{d x} + v \cdot - 1 = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.8

Move $- 1$ to the left of $v$ .

$\frac{d v}{d x} - 1 \cdot v = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.2.1.9

Rewrite $- 1 v$ as $- v$ .

$\frac{d v}{d x} - v = (3 (\frac{1}{3}) x + 1 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} (- (3 v^{\frac{4}{3}}))$

Step 7.1.3

Simplify the right side.

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Step 7.1.3.1

Simplify the expression.

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Step 7.1.3.1.1

Rewrite using the commutative property of multiplication.

$\frac{d v}{d x} - v = - 1 \cdot 3 (3 (\frac{1}{3}) x + \frac{1}{3}) {(v^{- \frac{1}{3}})}^{4} v^{\frac{4}{3}}$

Step 7.1.3.1.2

Multiply $- 1$ by $3$ .

$\frac{d v}{d x} - v = - 3 (3 (\frac{1}{3}) x + \frac{1}{3}) {(v^{- \frac{1}{3}})}^{4} v^{\frac{4}{3}}$

Step 7.1.3.2

Simplify each term.

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Step 7.1.3.2.1

Cancel the common factor of $3$ .

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Step 7.1.3.2.1.1

Cancel the common factor.

$\frac{d v}{d x} - v = - 3 (3 (\frac{1}{3}) x + \frac{1}{3}) {(v^{- \frac{1}{3}})}^{4} v^{\frac{4}{3}}$

Step 7.1.3.2.1.2

Rewrite the expression.

$\frac{d v}{d x} - v = - 3 (1 x + \frac{1}{3}) {(v^{- \frac{1}{3}})}^{4} v^{\frac{4}{3}}$

Step 7.1.3.2.2

Multiply $x$ by $1$ .

$\frac{d v}{d x} - v = - 3 (x + \frac{1}{3}) {(v^{- \frac{1}{3}})}^{4} v^{\frac{4}{3}}$

Step 7.1.3.3

Simplify terms.

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Step 7.1.3.3.1

Apply the distributive property.

$\frac{d v}{d x} - v = (- 3 x - 3 (\frac{1}{3})) {(v^{- \frac{1}{3}})}^{4} v^{\frac{4}{3}}$

Step 7.1.3.3.2

Cancel the common factor of $3$ .

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Step 7.1.3.3.2.1

Factor $3$ out of $- 3$ .

$\frac{d v}{d x} - v = (- 3 x + 3 (- 1) \frac{1}{3}) {(v^{- \frac{1}{3}})}^{4} v^{\frac{4}{3}}$

Step 7.1.3.3.2.2

Cancel the common factor.

$\frac{d v}{d x} - v = (- 3 x + 3 \cdot - 1 \frac{1}{3}) {(v^{- \frac{1}{3}})}^{4} v^{\frac{4}{3}}$

Step 7.1.3.3.2.3

Rewrite the expression.

$\frac{d v}{d x} - v = (- 3 x - 1) {(v^{- \frac{1}{3}})}^{4} v^{\frac{4}{3}}$

Step 7.1.3.3.3

Multiply the exponents in ${(v^{- \frac{1}{3}})}^{4}$ .

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Step 7.1.3.3.3.1

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

$\frac{d v}{d x} - v = (- 3 x - 1) v^{- \frac{1}{3} \cdot 4} v^{\frac{4}{3}}$

Step 7.1.3.3.3.2

Multiply $- \frac{1}{3} \cdot 4$ .

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Step 7.1.3.3.3.2.1

Multiply $4$ by $- 1$ .

$\frac{d v}{d x} - v = (- 3 x - 1) v^{- 4 (\frac{1}{3})} v^{\frac{4}{3}}$

Step 7.1.3.3.3.2.2

Combine $- 4$ and $\frac{1}{3}$ .

$\frac{d v}{d x} - v = (- 3 x - 1) v^{\frac{- 4}{3}} v^{\frac{4}{3}}$

Step 7.1.3.3.3.3

Move the negative in front of the fraction.

$\frac{d v}{d x} - v = (- 3 x - 1) v^{- \frac{4}{3}} v^{\frac{4}{3}}$

Step 7.1.3.4

Multiply $v^{- \frac{4}{3}}$ by $v^{\frac{4}{3}}$ by adding the exponents.

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Step 7.1.3.4.1

Move $v^{\frac{4}{3}}$ .

$\frac{d v}{d x} - v = (- 3 x - 1) (v^{\frac{4}{3}} v^{- \frac{4}{3}})$

Step 7.1.3.4.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{d v}{d x} - v = (- 3 x - 1) v^{\frac{4}{3} - \frac{4}{3}}$

Step 7.1.3.4.3

Combine the numerators over the common denominator.

$\frac{d v}{d x} - v = (- 3 x - 1) v^{\frac{4 - 4}{3}}$

Step 7.1.3.4.4

Subtract $4$ from $4$ .

$\frac{d v}{d x} - v = (- 3 x - 1) v^{\frac{0}{3}}$

Step 7.1.3.4.5

Divide $0$ by $3$ .

$\frac{d v}{d x} - v = (- 3 x - 1) v^{0}$

Step 7.1.3.5

Simplify $(- 3 x - 1) v^{0}$ .

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

Step 7.2

The integrating factor is defined by the formula $e^{\int P (x) d x}$ , where $P (x) = - 1$ .

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Step 7.2.1

Set up the integration.

$e^{\int - 1 d x}$

Step 7.2.2

Apply the constant rule.

$e^{- x + C}$

Step 7.2.3

Remove the constant of integration.

$e^{- x}$

Step 7.3

Multiply each term by the integrating factor $e^{- x}$ .

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Step 7.3.1

Multiply each term by $e^{- x}$ .

$e^{- x} \frac{d v}{d x} + e^{- x} (- v) = e^{- x} (- 3 x) + e^{- x} \cdot - 1$

Step 7.3.2

Rewrite using the commutative property of multiplication.

$e^{- x} \frac{d v}{d x} - e^{- x} v = e^{- x} (- 3 x) + e^{- x} \cdot - 1$

Step 7.3.3

Simplify each term.

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Step 7.3.3.1

Rewrite using the commutative property of multiplication.

$e^{- x} \frac{d v}{d x} - e^{- x} v = - 3 e^{- x} x + e^{- x} \cdot - 1$

Step 7.3.3.2

Move $- 1$ to the left of $e^{- x}$ .

$e^{- x} \frac{d v}{d x} - e^{- x} v = - 3 e^{- x} x - 1 \cdot e^{- x}$

Step 7.3.3.3

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$e^{- x} \frac{d v}{d x} - e^{- x} v = - 3 e^{- x} x - e^{- x}$

Step 7.3.4

Reorder factors in $e^{- x} \frac{d v}{d x} - e^{- x} v = - 3 e^{- x} x - e^{- x}$ .

$e^{- x} \frac{d v}{d x} - v e^{- x} = - 3 x e^{- x} - e^{- x}$

Step 7.4

Rewrite the left side as a result of differentiating a product.

$\frac{d}{d x} [e^{- x} v] = - 3 x e^{- x} - e^{- x}$

Step 7.5

Set up an integral on each side.

$\int \frac{d}{d x} [e^{- x} v] d x = \int - 3 x e^{- x} - e^{- x} d x$

Step 7.6

Integrate the left side.

$e^{- x} v = \int - 3 x e^{- x} - e^{- x} d x$

Step 7.7

Integrate the right side.

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Step 7.7.1

Split the single integral into multiple integrals.

$e^{- x} v = \int - 3 x e^{- x} d x + \int - e^{- x} d x$

Step 7.7.2

Since $- 3$ is constant with respect to $x$ , move $- 3$ out of the integral.

$e^{- x} v = - 3 \int x e^{- x} d x + \int - e^{- x} d x$

Step 7.7.3

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x$ and $d v = e^{- x}$ .

$e^{- x} v = - 3 (x (- e^{- x}) - \int - e^{- x} d x) + \int - e^{- x} d x$

Step 7.7.4

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

$e^{- x} v = - 3 (x (- e^{- x}) - - \int e^{- x} d x) + \int - e^{- x} d x$

Step 7.7.5

Simplify.

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Step 7.7.5.1

Multiply $- 1$ by $- 1$ .

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

Step 7.7.5.2

Multiply $\int e^{- x} d x$ by $1$ .

$e^{- x} v = - 3 (x (- e^{- x}) + \int e^{- x} d x) + \int - e^{- x} d x$

Step 7.7.6

Let $u_{1} = - x$ . Then $d u_{1} = - d x$ , so $- d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 7.7.6.1

Let $u_{1} = - x$ . Find $\frac{d u_{1}}{d x}$ .

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Step 7.7.6.1.1

Differentiate $- x$ .

$\frac{d}{d x} [- x]$

Step 7.7.6.1.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$- \frac{d}{d x} [x]$

Step 7.7.6.1.3

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

$- 1 \cdot 1$

Step 7.7.6.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 7.7.6.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$e^{- x} v = - 3 (x (- e^{- x}) + \int - e^{u_{1}} d u_{1}) + \int - e^{- x} d x$

Step 7.7.7

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

$e^{- x} v = - 3 (x (- e^{- x}) - \int e^{u_{1}} d u_{1}) + \int - e^{- x} d x$

Step 7.7.8

The integral of $e^{u_{1}}$ with respect to $u_{1}$ is $e^{u_{1}}$ .

$e^{- x} v = - 3 (x (- e^{- x}) - (e^{u_{1}} + C)) + \int - e^{- x} d x$

Step 7.7.9

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

$e^{- x} v = - 3 (x (- e^{- x}) - (e^{u_{1}} + C)) - \int e^{- x} d x$

Step 7.7.10

Let $u_{2} = - x$ . Then $d u_{2} = - d x$ , so $- d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 7.7.10.1

Let $u_{2} = - x$ . Find $\frac{d u_{2}}{d x}$ .

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Step 7.7.10.1.1

Differentiate $- x$ .

$\frac{d}{d x} [- x]$

Step 7.7.10.1.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$- \frac{d}{d x} [x]$

Step 7.7.10.1.3

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

$- 1 \cdot 1$

Step 7.7.10.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 7.7.10.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$e^{- x} v = - 3 (x (- e^{- x}) - (e^{u_{1}} + C)) - \int - e^{u_{2}} d u_{2}$

Step 7.7.11

Since $- 1$ is constant with respect to $u_{2}$ , move $- 1$ out of the integral.

$e^{- x} v = - 3 (x (- e^{- x}) - (e^{u_{1}} + C)) - - \int e^{u_{2}} d u_{2}$

Step 7.7.12

Simplify.

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Step 7.7.12.1

Multiply $- 1$ by $- 1$ .

$e^{- x} v = - 3 (x (- e^{- x}) - (e^{u_{1}} + C)) + 1 \int e^{u_{2}} d u_{2}$

Step 7.7.12.2

Multiply $\int e^{u_{2}} d u_{2}$ by $1$ .

$e^{- x} v = - 3 (x (- e^{- x}) - (e^{u_{1}} + C)) + \int e^{u_{2}} d u_{2}$

Step 7.7.13

The integral of $e^{u_{2}}$ with respect to $u_{2}$ is $e^{u_{2}}$ .

$e^{- x} v = - 3 (x (- e^{- x}) - (e^{u_{1}} + C)) + e^{u_{2}} + C$

Step 7.7.14

Simplify.

$e^{- x} v = - 3 (- x e^{- x} - e^{u_{1}}) + e^{u_{2}} + C$

Step 7.7.15

Substitute back in for each integration substitution variable.

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Step 7.7.15.1

Replace all occurrences of $u_{1}$ with $- x$ .

$e^{- x} v = - 3 (- x e^{- x} - e^{- x}) + e^{u_{2}} + C$

Step 7.7.15.2

Replace all occurrences of $u_{2}$ with $- x$ .

$e^{- x} v = - 3 (- x e^{- x} - e^{- x}) + e^{- x} + C$

Step 7.8

Divide each term in $e^{- x} v = - 3 (- x e^{- x} - e^{- x}) + e^{- x} + C$ by $e^{- x}$ and simplify.

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Step 7.8.1

Divide each term in $e^{- x} v = - 3 (- x e^{- x} - e^{- x}) + e^{- x} + C$ by $e^{- x}$ .

$\frac{e^{- x} v}{e^{- x}} = \frac{- 3 (- x e^{- x} - e^{- x})}{e^{- x}} + \frac{e^{- x}}{e^{- x}} + \frac{C}{e^{- x}}$

Step 7.8.2

Simplify the left side.

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Step 7.8.2.1

Cancel the common factor of $e^{- x}$ .

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Step 7.8.2.1.1

Cancel the common factor.

$\frac{e^{- x} v}{e^{- x}} = \frac{- 3 (- x e^{- x} - e^{- x})}{e^{- x}} + \frac{e^{- x}}{e^{- x}} + \frac{C}{e^{- x}}$

Step 7.8.2.1.2

Divide $v$ by $1$ .

$v = \frac{- 3 (- x e^{- x} - e^{- x})}{e^{- x}} + \frac{e^{- x}}{e^{- x}} + \frac{C}{e^{- x}}$

Step 7.8.3

Simplify the right side.

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Step 7.8.3.1

Combine the numerators over the common denominator.

$v = \frac{- 3 (- x e^{- x} - e^{- x}) + e^{- x} + C}{e^{- x}}$

Step 7.8.3.2

Simplify each term.

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Step 7.8.3.2.1

Apply the distributive property.

$v = \frac{- 3 (- x e^{- x}) - 3 (- e^{- x}) + e^{- x} + C}{e^{- x}}$

Step 7.8.3.2.2

Multiply $- 1$ by $- 3$ .

$v = \frac{3 (x e^{- x}) - 3 (- e^{- x}) + e^{- x} + C}{e^{- x}}$

Step 7.8.3.2.3

Multiply $- 1$ by $- 3$ .

$v = \frac{3 x e^{- x} + 3 e^{- x} + e^{- x} + C}{e^{- x}}$

Step 7.8.3.3

Add $3 e^{- x}$ and $e^{- x}$ .

$v = \frac{3 x e^{- x} + 4 e^{- x} + C}{e^{- x}}$

Step 8

Substitute $y^{- 3}$ for $v$ .

$y^{- 3} = \frac{3 x e^{- x} + 4 e^{- x} + C}{e^{- x}}$