Calculus Examples

$4 \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = 0$

Step 1

Let $v = \frac{d y}{d x}$ . Then $\frac{d v}{d x} = \frac{d^{2} y}{d x^{2}}$ . Substitute $v$ for $\frac{d y}{d x}$ and $\frac{d v}{d x}$ for $\frac{d^{2} y}{d x^{2}}$ to get a differential equation with dependent variable $v$ and independent variable $x$ .

$4 \frac{d v}{d x} + v = 0$

Step 2

Separate the variables.

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Step 2.1

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

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Step 2.1.1

Subtract $v$ from both sides of the equation.

$4 \frac{d v}{d x} = - v$

Step 2.1.2

Divide each term in $4 \frac{d v}{d x} = - v$ by $4$ and simplify.

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Step 2.1.2.1

Divide each term in $4 \frac{d v}{d x} = - v$ by $4$ .

$\frac{4 \frac{d v}{d x}}{4} = \frac{- v}{4}$

Step 2.1.2.2

Simplify the left side.

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Step 2.1.2.2.1

Cancel the common factor of $4$ .

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Step 2.1.2.2.1.1

Cancel the common factor.

$\frac{4 \frac{d v}{d x}}{4} = \frac{- v}{4}$

Step 2.1.2.2.1.2

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

$\frac{d v}{d x} = \frac{- v}{4}$

Step 2.1.2.3

Simplify the right side.

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Step 2.1.2.3.1

Move the negative in front of the fraction.

$\frac{d v}{d x} = - \frac{v}{4}$

Step 2.2

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

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

Step 2.3

Simplify.

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Step 2.3.1

Rewrite using the commutative property of multiplication.

$\frac{1}{v} \frac{d v}{d x} = - \frac{1}{v} \cdot \frac{v}{4}$

Step 2.3.2

Cancel the common factor of $v$ .

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Step 2.3.2.1

Move the leading negative in $- \frac{1}{v}$ into the numerator.

$\frac{1}{v} \frac{d v}{d x} = \frac{- 1}{v} \cdot \frac{v}{4}$

Step 2.3.2.2

Cancel the common factor.

$\frac{1}{v} \frac{d v}{d x} = \frac{- 1}{v} \cdot \frac{v}{4}$

Step 2.3.2.3

Rewrite the expression.

$\frac{1}{v} \frac{d v}{d x} = - \frac{1}{4}$

Step 2.4

Rewrite the equation.

$\frac{1}{v} d v = - \frac{1}{4} d x$

Step 3

Integrate both sides.

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Step 3.1

Set up an integral on each side.

$\int \frac{1}{v} d v = \int - \frac{1}{4} d x$

Step 3.2

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

$\ln (| v |) + C_{1} = \int - \frac{1}{4} d x$

Step 3.3

Apply the constant rule.

$\ln (| v |) + C_{1} = - \frac{1}{4} x + C_{2}$

Step 3.4

Group the constant of integration on the right side as $C_{3}$ .

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

Step 4

Solve for $v$ .

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Step 4.1

To solve for $v$ , rewrite the equation using properties of logarithms.

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

Step 4.2

Rewrite $\ln (| v |) = - \frac{1}{4} x + C_{3}$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{- \frac{1}{4} x + C_{3}} = | v |$

Step 4.3

Solve for $v$ .

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Step 4.3.1

Rewrite the equation as $| v | = e^{- \frac{1}{4} x + C_{3}}$ .

$| v | = e^{- \frac{1}{4} x + C_{3}}$

Step 4.3.2

Combine $x$ and $\frac{1}{4}$ .

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

Step 4.3.3

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

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

Step 5

Group the constant terms together.

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Step 5.1

Rewrite $e^{- \frac{x}{4} + C_{3}}$ as $e^{- \frac{x}{4}} e^{C_{3}}$ .

$v = \pm e^{- \frac{x}{4}} e^{C_{3}}$

Step 5.2

Reorder $e^{- \frac{x}{4}}$ and $e^{C_{3}}$ .

$v = \pm e^{C_{3}} e^{- \frac{x}{4}}$

Step 5.3

Combine constants with the plus or minus.

$v = C_{4} e^{- \frac{x}{4}}$

Step 6

Replace all occurrences of $v$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = C_{4} e^{- \frac{x}{4}}$

Step 7

Rewrite the equation.

$d y = C_{4} e^{- \frac{x}{4}} d x$

Step 8

Integrate both sides.

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Step 8.1

Set up an integral on each side.

$\int d y = \int C_{4} e^{- \frac{x}{4}} d x$

Step 8.2

Apply the constant rule.

$y + C_{5} = \int C_{4} e^{- \frac{x}{4}} d x$

Step 8.3

Integrate the right side.

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Step 8.3.1

Since $C_{4}$ is constant with respect to $x$ , move $C_{4}$ out of the integral.

$y + C_{5} = C_{4} \int e^{- \frac{x}{4}} d x$

Step 8.3.2

Let $u = - \frac{x}{4}$ . Then $d u = - \frac{1}{4} d x$ , so $- 4 d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 8.3.2.1

Let $u = - \frac{x}{4}$ . Find $\frac{d u}{d x}$ .

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Step 8.3.2.1.1

Differentiate $- \frac{x}{4}$ .

$\frac{d}{d x} [- \frac{x}{4}]$

Step 8.3.2.1.2

Since $- \frac{1}{4}$ is constant with respect to $x$ , the derivative of $- \frac{x}{4}$ with respect to $x$ is $- \frac{1}{4} \frac{d}{d x} [x]$ .

$- \frac{1}{4} \frac{d}{d x} [x]$

Step 8.3.2.1.3

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

$- \frac{1}{4} \cdot 1$

Step 8.3.2.1.4

Multiply $- 1$ by $1$ .

$- \frac{1}{4}$

Step 8.3.2.2

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

$y + C_{5} = C_{4} \int - e^{u} \frac{- 1}{- \frac{1}{4}} d u$

Step 8.3.3

Simplify.

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Step 8.3.3.1

Dividing two negative values results in a positive value.

$y + C_{5} = C_{4} \int - e^{u} \frac{1}{\frac{1}{4}} d u$

Step 8.3.3.2

Multiply by the reciprocal of the fraction to divide by $\frac{1}{4}$ .

$y + C_{5} = C_{4} \int - e^{u} (1 \cdot 4) d u$

Step 8.3.3.3

Multiply $4$ by $1$ .

$y + C_{5} = C_{4} \int - e^{u} \cdot 4 d u$

Step 8.3.3.4

Multiply $4$ by $- 1$ .

$y + C_{5} = C_{4} \int - 4 e^{u} d u$

Step 8.3.4

Since $- 4$ is constant with respect to $u$ , move $- 4$ out of the integral.

$y + C_{5} = C_{4} (- 4 \int e^{u} d u)$

Step 8.3.5

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

$y + C_{5} = C_{4} (- 4 (e^{u} + C_{6}))$

Step 8.3.6

Simplify.

$y + C_{5} = C_{4} (- 4 e^{u}) + C_{7}$

Step 8.3.7

Replace all occurrences of $u$ with $- \frac{x}{4}$ .

$y + C_{5} = C_{4} (- 4 e^{- \frac{x}{4}}) + C_{7}$

Step 8.3.8

Reorder terms.

$y + C_{5} = C_{4} (- 4 e^{- \frac{1}{4} x}) + C_{7}$

Step 8.3.9

Reorder terms.

$y + C_{5} = - 4 e^{- \frac{1}{4} x} C_{4} + C_{7}$

Step 8.4

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

$y = - 4 e^{- \frac{1}{4} x} C + D$