Calculus Examples

$4 e^{4 y} \frac{d y}{d x} = 2 x e^{3 x} + 3 e^{4 y}$

Step 1

Let $v = e^{4 y}$ . Substitute $v$ for all occurrences of $e^{4 y}$ .

$4 v \frac{d y}{d x} = 2 x e^{3 x} + 3 v$

Step 2

Find $\frac{d v}{d x}$ by differentiating $e^{4 y}$ .

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Step 2.1

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 $g (x) = 4 y$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $4 y$ .

$\frac{d v}{d x} = \frac{d}{d u} [e^{u}] \frac{d}{d x} [4 y]$

Step 2.1.2

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

$\frac{d v}{d x} = e^{u} \frac{d}{d x} [4 y]$

Step 2.1.3

Replace all occurrences of $u$ with $4 y$ .

$\frac{d v}{d x} = e^{4 y} \frac{d}{d x} [4 y]$

Step 2.2

Since $4$ is constant with respect to $x$ , the derivative of $4 y$ with respect to $x$ is $4 \frac{d}{d x} [y]$ .

$\frac{d v}{d x} = e^{4 y} (4 \frac{d}{d x} [y])$

Step 2.3

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

$\frac{d v}{d x} = e^{4 y} (4 y')$

Step 2.4

Reorder the factors of $e^{4 y} (4 y')$ .

$\frac{d v}{d x} = 4 e^{4 y} y'$

Step 3

Substitute $v$ for $e^{4 y}$ .

$\frac{d v}{d x} = 4 v y'$

Step 4

Substitute the derivative back in to the differential equation.

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Step 4.1

Cancel the common factor.

$4 v \frac{\frac{d v}{d x}}{4 v} = 2 x e^{3 x} + 3 v$

Step 4.2

Rewrite the expression.

$\frac{d v}{d x} = 2 x e^{3 x} + 3 v$

Step 5

Subtract $3 v$ from both sides of the equation.

$\frac{d v}{d x} - 3 v = 2 x e^{3 x}$

Step 6

The integrating factor is defined by the formula $e^{\int P (x) d x}$ , where $P (x) = - 3$ .

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Step 6.1

Set up the integration.

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

Step 6.2

Apply the constant rule.

$e^{- 3 x + C}$

Step 6.3

Remove the constant of integration.

$e^{- 3 x}$

Step 7

Multiply each term by the integrating factor $e^{- 3 x}$ .

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Step 7.1

Multiply each term by $e^{- 3 x}$ .

$e^{- 3 x} \frac{d v}{d x} + e^{- 3 x} (- 3 v) = e^{- 3 x} (2 x e^{3 x})$

Step 7.2

Rewrite using the commutative property of multiplication.

$e^{- 3 x} \frac{d v}{d x} - 3 e^{- 3 x} v = e^{- 3 x} (2 x e^{3 x})$

Step 7.3

Rewrite using the commutative property of multiplication.

$e^{- 3 x} \frac{d v}{d x} - 3 e^{- 3 x} v = 2 e^{- 3 x} (x e^{3 x})$

Step 7.4

Multiply $e^{- 3 x}$ by $e^{3 x}$ by adding the exponents.

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Step 7.4.1

Move $e^{3 x}$ .

$e^{- 3 x} \frac{d v}{d x} - 3 e^{- 3 x} v = 2 (e^{3 x} e^{- 3 x}) x$

Step 7.4.2

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

$e^{- 3 x} \frac{d v}{d x} - 3 e^{- 3 x} v = 2 e^{3 x - 3 x} x$

Step 7.4.3

Subtract $3 x$ from $3 x$ .

$e^{- 3 x} \frac{d v}{d x} - 3 e^{- 3 x} v = 2 e^{0} x$

Step 7.5

Simplify $2 e^{0} x$ .

$e^{- 3 x} \frac{d v}{d x} - 3 e^{- 3 x} v = 2 x$

Step 7.6

Reorder factors in $e^{- 3 x} \frac{d v}{d x} - 3 e^{- 3 x} v = 2 x$ .

$e^{- 3 x} \frac{d v}{d x} - 3 v e^{- 3 x} = 2 x$

Step 8

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

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

Step 9

Set up an integral on each side.

$\int \frac{d}{d x} [e^{- 3 x} v] d x = \int 2 x d x$

Step 10

Integrate the left side.

$e^{- 3 x} v = \int 2 x d x$

Step 11

Integrate the right side.

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Step 11.1

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

$e^{- 3 x} v = 2 \int x d x$

Step 11.2

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

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

Step 11.3

Simplify the answer.

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Step 11.3.1

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

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

Step 11.3.2

Simplify.

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Step 11.3.2.1

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

$e^{- 3 x} v = \frac{2}{2} x^{2} + C$

Step 11.3.2.2

Cancel the common factor of $2$ .

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Step 11.3.2.2.1

Cancel the common factor.

$e^{- 3 x} v = \frac{2}{2} x^{2} + C$

Step 11.3.2.2.2

Rewrite the expression.

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

Step 11.3.2.3

Multiply $x^{2}$ by $1$ .

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

Step 12

Divide each term in $e^{- 3 x} v = x^{2} + C$ by $e^{- 3 x}$ and simplify.

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Step 12.1

Divide each term in $e^{- 3 x} v = x^{2} + C$ by $e^{- 3 x}$ .

$\frac{e^{- 3 x} v}{e^{- 3 x}} = \frac{x^{2}}{e^{- 3 x}} + \frac{C}{e^{- 3 x}}$

Step 12.2

Simplify the left side.

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Step 12.2.1

Cancel the common factor of $e^{- 3 x}$ .

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Step 12.2.1.1

Cancel the common factor.

$\frac{e^{- 3 x} v}{e^{- 3 x}} = \frac{x^{2}}{e^{- 3 x}} + \frac{C}{e^{- 3 x}}$

Step 12.2.1.2

Divide $v$ by $1$ .

$v = \frac{x^{2}}{e^{- 3 x}} + \frac{C}{e^{- 3 x}}$

Step 13

Replace all occurrences of $v$ with $e^{4 y}$ .

$e^{4 y} = \frac{x^{2}}{e^{- 3 x}} + \frac{C}{e^{- 3 x}}$

Step 14

Solve for $y$ .

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Step 14.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{4 y}) = \ln (\frac{x^{2}}{e^{- 3 x}} + \frac{C}{e^{- 3 x}})$

Step 14.2

Expand the left side.

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Step 14.2.1

Expand $\ln (e^{4 y})$ by moving $4 y$ outside the logarithm.

$4 y \ln (e) = \ln (\frac{x^{2}}{e^{- 3 x}} + \frac{C}{e^{- 3 x}})$

Step 14.2.2

The natural logarithm of $e$ is $1$ .

$4 y \cdot 1 = \ln (\frac{x^{2}}{e^{- 3 x}} + \frac{C}{e^{- 3 x}})$

Step 14.2.3

Multiply $4$ by $1$ .

$4 y = \ln (\frac{x^{2}}{e^{- 3 x}} + \frac{C}{e^{- 3 x}})$

Step 14.3

Divide each term in $4 y = \ln (\frac{x^{2}}{e^{- 3 x}} + \frac{C}{e^{- 3 x}})$ by $4$ and simplify.

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Step 14.3.1

Divide each term in $4 y = \ln (\frac{x^{2}}{e^{- 3 x}} + \frac{C}{e^{- 3 x}})$ by $4$ .

$\frac{4 y}{4} = \frac{\ln (\frac{x^{2}}{e^{- 3 x}} + \frac{C}{e^{- 3 x}})}{4}$

Step 14.3.2

Simplify the left side.

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Step 14.3.2.1

Cancel the common factor of $4$ .

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Step 14.3.2.1.1

Cancel the common factor.

$\frac{4 y}{4} = \frac{\ln (\frac{x^{2}}{e^{- 3 x}} + \frac{C}{e^{- 3 x}})}{4}$

Step 14.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{\ln (\frac{x^{2}}{e^{- 3 x}} + \frac{C}{e^{- 3 x}})}{4}$