Calculus Examples

$25 \frac{d q}{d t} - \frac{d^{2} q}{d t^{2}} = 0$

Step 1

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

$25 v - \frac{d v}{d t} = 0$

Step 2

Separate the variables.

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Step 2.1

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

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Step 2.1.1

Subtract $25 v$ from both sides of the equation.

$- \frac{d v}{d t} = - 25 v$

Step 2.1.2

Divide each term in $- \frac{d v}{d t} = - 25 v$ by $- 1$ and simplify.

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Step 2.1.2.1

Divide each term in $- \frac{d v}{d t} = - 25 v$ by $- 1$ .

$\frac{- \frac{d v}{d t}}{- 1} = \frac{- 25 v}{- 1}$

Step 2.1.2.2

Simplify the left side.

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Step 2.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{\frac{d v}{d t}}{1} = \frac{- 25 v}{- 1}$

Step 2.1.2.2.2

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

$\frac{d v}{d t} = \frac{- 25 v}{- 1}$

Step 2.1.2.3

Simplify the right side.

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Step 2.1.2.3.1

Move the negative one from the denominator of $\frac{- 25 v}{- 1}$ .

$\frac{d v}{d t} = - 1 \cdot (- 25 v)$

Step 2.1.2.3.2

Rewrite $- 1 \cdot (- 25 v)$ as $- (- 25 v)$ .

$\frac{d v}{d t} = - (- 25 v)$

Step 2.1.2.3.3

Multiply $- 25$ by $- 1$ .

$\frac{d v}{d t} = 25 v$

Step 2.2

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

$\frac{1}{v} \frac{d v}{d t} = \frac{1}{v} (25 v)$

Step 2.3

Simplify.

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Step 2.3.1

Rewrite using the commutative property of multiplication.

$\frac{1}{v} \frac{d v}{d t} = 25 \frac{1}{v} v$

Step 2.3.2

Combine $25$ and $\frac{1}{v}$ .

$\frac{1}{v} \frac{d v}{d t} = \frac{25}{v} v$

Step 2.3.3

Cancel the common factor of $v$ .

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Step 2.3.3.1

Cancel the common factor.

$\frac{1}{v} \frac{d v}{d t} = \frac{25}{v} v$

Step 2.3.3.2

Rewrite the expression.

$\frac{1}{v} \frac{d v}{d t} = 25$

Step 2.4

Rewrite the equation.

$\frac{1}{v} d v = 25 d t$

Step 3

Integrate both sides.

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Step 3.1

Set up an integral on each side.

$\int \frac{1}{v} d v = \int 25 d t$

Step 3.2

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

$\ln (| v |) + C_{1} = \int 25 d t$

Step 3.3

Apply the constant rule.

$\ln (| v |) + C_{1} = 25 t + C_{2}$

Step 3.4

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

$\ln (| v |) = 25 t + C_{3}$

Step 4

Solve for $v$ .

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Step 4.1

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

$e^{\ln (| v |)} = e^{25 t + C_{3}}$

Step 4.2

Rewrite $\ln (| v |) = 25 t + C_{3}$ 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 $b^{y} = x$ .

$e^{25 t + C_{3}} = | v |$

Step 4.3

Solve for $v$ .

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Step 4.3.1

Rewrite the equation as $| v | = e^{25 t + C_{3}}$ .

$| v | = e^{25 t + C_{3}}$

Step 4.3.2

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

$v = \pm e^{25 t + C_{3}}$

Step 5

Group the constant terms together.

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Step 5.1

Rewrite $e^{25 t + C_{3}}$ as $e^{25 t} e^{C_{3}}$ .

$v = \pm e^{25 t} e^{C_{3}}$

Step 5.2

Reorder $e^{25 t}$ and $e^{C_{3}}$ .

$v = \pm e^{C_{3}} e^{25 t}$

Step 5.3

Combine constants with the plus or minus.

$v = C_{4} e^{25 t}$

Step 6

Replace all occurrences of $v$ with $\frac{d q}{d t}$ .

$\frac{d q}{d t} = C_{4} e^{25 t}$

Step 7

Rewrite the equation.

$d q = C_{4} e^{25 t} d t$

Step 8

Integrate both sides.

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Step 8.1

Set up an integral on each side.

$\int d q = \int C_{4} e^{25 t} d t$

Step 8.2

Apply the constant rule.

$q + C_{5} = \int C_{4} e^{25 t} d t$

Step 8.3

Integrate the right side.

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Step 8.3.1

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

$q + C_{5} = C_{4} \int e^{25 t} d t$

Step 8.3.2

Let $u = 25 t$ . Then $d u = 25 d t$ , so $\frac{1}{25} d u = d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 8.3.2.1

Let $u = 25 t$ . Find $\frac{d u}{d t}$ .

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Step 8.3.2.1.1

Differentiate $25 t$ .

$\frac{d}{d t} [25 t]$

Step 8.3.2.1.2

Since $25$ is constant with respect to $t$ , the derivative of $25 t$ with respect to $t$ is $25 \frac{d}{d t} [t]$ .

$25 \frac{d}{d t} [t]$

Step 8.3.2.1.3

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

$25 \cdot 1$

Step 8.3.2.1.4

Multiply $25$ by $1$ .

$25$

Step 8.3.2.2

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

$q + C_{5} = C_{4} \int e^{u} \frac{1}{25} d u$

Step 8.3.3

Combine $e^{u}$ and $\frac{1}{25}$ .

$q + C_{5} = C_{4} \int \frac{e^{u}}{25} d u$

Step 8.3.4

Since $\frac{1}{25}$ is constant with respect to $u$ , move $\frac{1}{25}$ out of the integral.

$q + C_{5} = C_{4} (\frac{1}{25} \int e^{u} d u)$

Step 8.3.5

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

$q + C_{5} = C_{4} \frac{1}{25} (e^{u} + C_{6})$

Step 8.3.6

Simplify.

$q + C_{5} = \frac{C_{4}}{25} e^{u} + C_{7}$

Step 8.3.7

Replace all occurrences of $u$ with $25 t$ .

$q + C_{5} = \frac{C_{4}}{25} e^{25 t} + C_{7}$

Step 8.3.8

Reorder terms.

$q + C_{5} = \frac{1}{25} C_{4} e^{25 t} + C_{7}$

Step 8.3.9

Reorder terms.

$q + C_{5} = \frac{1}{25} e^{25 t} C_{4} + C_{7}$

Step 8.4

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

$q = \frac{1}{25} e^{25 t} C + D$