Calculus Examples

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

Step 1

Rewrite the differential equation as a function of $\frac{x}{t}$ .

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Step 1.1

Split $\frac{4 t^{2} + 3 x^{2}}{2 t x}$ and simplify.

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Step 1.1.1

Split the fraction $\frac{4 t^{2} + 3 x^{2}}{2 t x}$ into two fractions.

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

Step 1.1.2

Simplify each term.

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Step 1.1.2.1

Cancel the common factor of $4$ and $2$ .

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Step 1.1.2.1.1

Factor $2$ out of $4 t^{2}$ .

$\frac{d x}{d t} = \frac{2 (2 t^{2})}{2 t x} + \frac{3 x^{2}}{2 t x}$

Step 1.1.2.1.2

Cancel the common factors.

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Step 1.1.2.1.2.1

Factor $2$ out of $2 t x$ .

$\frac{d x}{d t} = \frac{2 (2 t^{2})}{2 (t x)} + \frac{3 x^{2}}{2 t x}$

Step 1.1.2.1.2.2

Cancel the common factor.

$\frac{d x}{d t} = \frac{2 (2 t^{2})}{2 (t x)} + \frac{3 x^{2}}{2 t x}$

Step 1.1.2.1.2.3

Rewrite the expression.

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

Step 1.1.2.2

Cancel the common factor of $t^{2}$ and $t$ .

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Step 1.1.2.2.1

Factor $t$ out of $2 t^{2}$ .

$\frac{d x}{d t} = \frac{t (2 t)}{t x} + \frac{3 x^{2}}{2 t x}$

Step 1.1.2.2.2

Cancel the common factors.

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Step 1.1.2.2.2.1

Factor $t$ out of $t x$ .

$\frac{d x}{d t} = \frac{t (2 t)}{t (x)} + \frac{3 x^{2}}{2 t x}$

Step 1.1.2.2.2.2

Cancel the common factor.

$\frac{d x}{d t} = \frac{t (2 t)}{t x} + \frac{3 x^{2}}{2 t x}$

Step 1.1.2.2.2.3

Rewrite the expression.

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

Step 1.1.2.3

Cancel the common factor of $x^{2}$ and $x$ .

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Step 1.1.2.3.1

Factor $x$ out of $3 x^{2}$ .

$\frac{d x}{d t} = \frac{2 t}{x} + \frac{x (3 x)}{2 t x}$

Step 1.1.2.3.2

Cancel the common factors.

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Step 1.1.2.3.2.1

Factor $x$ out of $2 t x$ .

$\frac{d x}{d t} = \frac{2 t}{x} + \frac{x (3 x)}{x (2 t)}$

Step 1.1.2.3.2.2

Cancel the common factor.

$\frac{d x}{d t} = \frac{2 t}{x} + \frac{x (3 x)}{x (2 t)}$

Step 1.1.2.3.2.3

Rewrite the expression.

$\frac{d x}{d t} = \frac{2 t}{x} + \frac{3 x}{2 t}$

Step 1.2

Rewrite the differential equation as $f (\frac{x}{t})$ .

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Step 1.2.1

Factor out $\frac{t}{x}$ from $\frac{2 t}{x}$ .

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Step 1.2.1.1

Factor $\frac{t}{x}$ out of $\frac{2 t}{x}$ .

$\frac{d x}{d t} = \frac{t}{x} \cdot 2 + \frac{3 x}{2 t}$

Step 1.2.1.2

Reorder $\frac{t}{x}$ and $2$ .

$\frac{d x}{d t} = 2 \cdot \frac{t}{x} + \frac{3 x}{2 t}$

Step 1.2.2

Rewrite $\frac{t}{x}$ as ${(\frac{x}{t})}^{- 1}$ .

$\frac{d x}{d t} = 2 \cdot {(\frac{x}{t})}^{- 1} + \frac{3 x}{2 t}$

Step 1.3

Factor out $\frac{x}{t}$ from $\frac{3 x}{2 t}$ .

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Step 1.3.1

Factor $\frac{x}{t}$ out of $\frac{3 x}{2 t}$ .

$\frac{d x}{d t} = 2 \cdot {(\frac{x}{t})}^{- 1} + \frac{x}{t} \cdot \frac{3}{2}$

Step 1.3.2

Reorder $\frac{x}{t}$ and $\frac{3}{2}$ .

$\frac{d x}{d t} = 2 \cdot {(\frac{x}{t})}^{- 1} + \frac{3}{2} \cdot \frac{x}{t}$

Step 2

Let $V = \frac{x}{t}$ . Substitute $V$ for $\frac{x}{t}$ .

$\frac{d x}{d t} = 2 \cdot V^{- 1} + \frac{3}{2} V$

Step 3

Solve $V = \frac{x}{t}$ for $x$ .

$x = V t$

Step 4

Use the product rule to find the derivative of $x = V t$ with respect to $t$ .

$\frac{d x}{d t} = t \frac{d V}{d t} + V$

Step 5

Substitute $t \frac{d V}{d t} + V$ for $\frac{d x}{d t}$ .

$t \frac{d V}{d t} + V = 2 \cdot V^{- 1} + \frac{3}{2} V$

Step 6

Solve the substituted differential equation.

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Step 6.1

Separate the variables.

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Step 6.1.1

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

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Step 6.1.1.1

Simplify each term.

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Step 6.1.1.1.1

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

$t \frac{d V}{d t} + V = 2 \cdot \frac{1}{V} + \frac{3}{2} V$

Step 6.1.1.1.2

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

$t \frac{d V}{d t} + V = \frac{2}{V} + \frac{3}{2} V$

Step 6.1.1.1.3

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

$t \frac{d V}{d t} + V = \frac{2}{V} + \frac{3 V}{2}$

Step 6.1.1.2

Subtract $V$ from both sides of the equation.

$t \frac{d V}{d t} = \frac{2}{V} + \frac{3 V}{2} - V$

Step 6.1.1.3

Divide each term in $t \frac{d V}{d t} = \frac{2}{V} + \frac{3 V}{2} - V$ by $t$ and simplify.

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Step 6.1.1.3.1

Divide each term in $t \frac{d V}{d t} = \frac{2}{V} + \frac{3 V}{2} - V$ by $t$ .

$\frac{t \frac{d V}{d t}}{t} = \frac{\frac{2}{V}}{t} + \frac{\frac{3 V}{2}}{t} + \frac{- V}{t}$

Step 6.1.1.3.2

Simplify the left side.

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Step 6.1.1.3.2.1

Cancel the common factor of $t$ .

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Step 6.1.1.3.2.1.1

Cancel the common factor.

$\frac{t \frac{d V}{d t}}{t} = \frac{\frac{2}{V}}{t} + \frac{\frac{3 V}{2}}{t} + \frac{- V}{t}$

Step 6.1.1.3.2.1.2

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

$\frac{d V}{d t} = \frac{\frac{2}{V}}{t} + \frac{\frac{3 V}{2}}{t} + \frac{- V}{t}$

Step 6.1.1.3.3

Simplify the right side.

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Step 6.1.1.3.3.1

Combine the numerators over the common denominator.

$\frac{d V}{d t} = \frac{\frac{2}{V} + \frac{3 V}{2} - V}{t}$

Step 6.1.1.3.3.2

To write $- V$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{d V}{d t} = \frac{\frac{2}{V} + \frac{3 V}{2} - V \cdot \frac{2}{2}}{t}$

Step 6.1.1.3.3.3

Simplify terms.

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Step 6.1.1.3.3.3.1

Combine $- V$ and $\frac{2}{2}$ .

$\frac{d V}{d t} = \frac{\frac{2}{V} + \frac{3 V}{2} + \frac{- V \cdot 2}{2}}{t}$

Step 6.1.1.3.3.3.2

Combine the numerators over the common denominator.

$\frac{d V}{d t} = \frac{\frac{2}{V} + \frac{3 V - V \cdot 2}{2}}{t}$

Step 6.1.1.3.3.3.3

Simplify each term.

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Step 6.1.1.3.3.3.3.1

Simplify the numerator.

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Step 6.1.1.3.3.3.3.1.1

Factor $V$ out of $3 V - V \cdot 2$ .

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Step 6.1.1.3.3.3.3.1.1.1

Factor $V$ out of $3 V$ .

$\frac{d V}{d t} = \frac{\frac{2}{V} + \frac{V \cdot 3 - V \cdot 2}{2}}{t}$

Step 6.1.1.3.3.3.3.1.1.2

Factor $V$ out of $- V \cdot 2$ .

$\frac{d V}{d t} = \frac{\frac{2}{V} + \frac{V \cdot 3 + V (- 1 \cdot 2)}{2}}{t}$

Step 6.1.1.3.3.3.3.1.1.3

Factor $V$ out of $V \cdot 3 + V (- 1 \cdot 2)$ .

$\frac{d V}{d t} = \frac{\frac{2}{V} + \frac{V (3 - 1 \cdot 2)}{2}}{t}$

Step 6.1.1.3.3.3.3.1.2

Multiply $- 1$ by $2$ .

$\frac{d V}{d t} = \frac{\frac{2}{V} + \frac{V (3 - 2)}{2}}{t}$

Step 6.1.1.3.3.3.3.1.3

Subtract $2$ from $3$ .

$\frac{d V}{d t} = \frac{\frac{2}{V} + \frac{V \cdot 1}{2}}{t}$

Step 6.1.1.3.3.3.3.2

Multiply $V$ by $1$ .

$\frac{d V}{d t} = \frac{\frac{2}{V} + \frac{V}{2}}{t}$

Step 6.1.1.3.3.4

Simplify the numerator.

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Step 6.1.1.3.3.4.1

To write $\frac{2}{V}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{d V}{d t} = \frac{\frac{2}{V} \cdot \frac{2}{2} + \frac{V}{2}}{t}$

Step 6.1.1.3.3.4.2

To write $\frac{V}{2}$ as a fraction with a common denominator, multiply by $\frac{V}{V}$ .

$\frac{d V}{d t} = \frac{\frac{2}{V} \cdot \frac{2}{2} + \frac{V}{2} \cdot \frac{V}{V}}{t}$

Step 6.1.1.3.3.4.3

Write each expression with a common denominator of $V \cdot 2$ , by multiplying each by an appropriate factor of $1$ .

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Step 6.1.1.3.3.4.3.1

Multiply $\frac{2}{V}$ by $\frac{2}{2}$ .

$\frac{d V}{d t} = \frac{\frac{2 \cdot 2}{V \cdot 2} + \frac{V}{2} \cdot \frac{V}{V}}{t}$

Step 6.1.1.3.3.4.3.2

Multiply $\frac{V}{2}$ by $\frac{V}{V}$ .

$\frac{d V}{d t} = \frac{\frac{2 \cdot 2}{V \cdot 2} + \frac{V \cdot V}{2 V}}{t}$

Step 6.1.1.3.3.4.3.3

Reorder the factors of $V \cdot 2$ .

$\frac{d V}{d t} = \frac{\frac{2 \cdot 2}{2 V} + \frac{V \cdot V}{2 V}}{t}$

Step 6.1.1.3.3.4.4

Combine the numerators over the common denominator.

$\frac{d V}{d t} = \frac{\frac{2 \cdot 2 + V \cdot V}{2 V}}{t}$

Step 6.1.1.3.3.4.5

Simplify the numerator.

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Step 6.1.1.3.3.4.5.1

Multiply $2$ by $2$ .

$\frac{d V}{d t} = \frac{\frac{4 + V \cdot V}{2 V}}{t}$

Step 6.1.1.3.3.4.5.2

Multiply $V$ by $V$ .

$\frac{d V}{d t} = \frac{\frac{4 + V^{2}}{2 V}}{t}$

Step 6.1.1.3.3.5

Multiply the numerator by the reciprocal of the denominator.

$\frac{d V}{d t} = \frac{4 + V^{2}}{2 V} \cdot \frac{1}{t}$

Step 6.1.1.3.3.6

Multiply $\frac{4 + V^{2}}{2 V}$ by $\frac{1}{t}$ .

$\frac{d V}{d t} = \frac{4 + V^{2}}{2 V t}$

Step 6.1.2

Regroup factors.

$\frac{d V}{d t} = \frac{4 + V^{2}}{2 V} \cdot \frac{1}{t}$

Step 6.1.3

Multiply both sides by $\frac{2 V}{4 + V^{2}}$ .

$\frac{2 V}{4 + V^{2}} \frac{d V}{d t} = \frac{2 V}{4 + V^{2}} (\frac{4 + V^{2}}{2 V} \cdot \frac{1}{t})$

Step 6.1.4

Simplify.

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Step 6.1.4.1

Multiply $\frac{4 + V^{2}}{2 V}$ by $\frac{1}{t}$ .

$\frac{2 V}{4 + V^{2}} \frac{d V}{d t} = \frac{2 V}{4 + V^{2}} \cdot \frac{4 + V^{2}}{2 V t}$

Step 6.1.4.2

Cancel the common factor of $2 V$ .

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Step 6.1.4.2.1

Factor $2 V$ out of $2 V t$ .

$\frac{2 V}{4 + V^{2}} \frac{d V}{d t} = \frac{2 V}{4 + V^{2}} \cdot \frac{4 + V^{2}}{2 V (t)}$

Step 6.1.4.2.2

Cancel the common factor.

$\frac{2 V}{4 + V^{2}} \frac{d V}{d t} = \frac{2 V}{4 + V^{2}} \cdot \frac{4 + V^{2}}{2 V t}$

Step 6.1.4.2.3

Rewrite the expression.

$\frac{2 V}{4 + V^{2}} \frac{d V}{d t} = \frac{1}{4 + V^{2}} \cdot \frac{4 + V^{2}}{t}$

Step 6.1.4.3

Cancel the common factor of $4 + V^{2}$ .

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Step 6.1.4.3.1

Cancel the common factor.

$\frac{2 V}{4 + V^{2}} \frac{d V}{d t} = \frac{1}{4 + V^{2}} \cdot \frac{4 + V^{2}}{t}$

Step 6.1.4.3.2

Rewrite the expression.

$\frac{2 V}{4 + V^{2}} \frac{d V}{d t} = \frac{1}{t}$

Step 6.1.5

Rewrite the equation.

$\frac{2 V}{4 + V^{2}} d V = \frac{1}{t} d t$

Step 6.2

Integrate both sides.

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Step 6.2.1

Set up an integral on each side.

$\int \frac{2 V}{4 + V^{2}} d V = \int \frac{1}{t} d t$

Step 6.2.2

Integrate the left side.

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Step 6.2.2.1

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

$2 \int \frac{V}{4 + V^{2}} d V = \int \frac{1}{t} d t$

Step 6.2.2.2

Let $u = 4 + V^{2}$ . Then $d u = 2 V d V$ , so $\frac{1}{2} d u = V d V$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.2.2.2.1

Let $u = 4 + V^{2}$ . Find $\frac{d u}{d V}$ .

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Step 6.2.2.2.1.1

Differentiate $4 + V^{2}$ .

$\frac{d}{d V} [4 + V^{2}]$

Step 6.2.2.2.1.2

By the Sum Rule, the derivative of $4 + V^{2}$ with respect to $V$ is $\frac{d}{d V} [4] + \frac{d}{d V} [V^{2}]$ .

$\frac{d}{d V} [4] + \frac{d}{d V} [V^{2}]$

Step 6.2.2.2.1.3

Since $4$ is constant with respect to $V$ , the derivative of $4$ with respect to $V$ is $0$ .

$0 + \frac{d}{d V} [V^{2}]$

Step 6.2.2.2.1.4

Differentiate using the Power Rule which states that $\frac{d}{d V} [V^{n}]$ is $n V^{n - 1}$ where $n = 2$ .

$0 + 2 V$

Step 6.2.2.2.1.5

Add $0$ and $2 V$ .

$2 V$

Step 6.2.2.2.2

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

$2 \int \frac{1}{u} \cdot \frac{1}{2} d u = \int \frac{1}{t} d t$

Step 6.2.2.3

Simplify.

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Step 6.2.2.3.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$2 \int \frac{1}{u \cdot 2} d u = \int \frac{1}{t} d t$

Step 6.2.2.3.2

Move $2$ to the left of $u$ .

$2 \int \frac{1}{2 u} d u = \int \frac{1}{t} d t$

Step 6.2.2.4

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$2 (\frac{1}{2} \int \frac{1}{u} d u) = \int \frac{1}{t} d t$

Step 6.2.2.5

Simplify.

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Step 6.2.2.5.1

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

$\frac{2}{2} \int \frac{1}{u} d u = \int \frac{1}{t} d t$

Step 6.2.2.5.2

Cancel the common factor of $2$ .

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Step 6.2.2.5.2.1

Cancel the common factor.

$\frac{2}{2} \int \frac{1}{u} d u = \int \frac{1}{t} d t$

Step 6.2.2.5.2.2

Rewrite the expression.

$1 \int \frac{1}{u} d u = \int \frac{1}{t} d t$

Step 6.2.2.5.3

Multiply $\int \frac{1}{u} d u$ by $1$ .

$\int \frac{1}{u} d u = \int \frac{1}{t} d t$

Step 6.2.2.6

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

$\ln (| u |) + C_{1} = \int \frac{1}{t} d t$

Step 6.2.2.7

Replace all occurrences of $u$ with $4 + V^{2}$ .

$\ln (| 4 + V^{2} |) + C_{1} = \int \frac{1}{t} d t$

Step 6.2.3

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

$\ln (| 4 + V^{2} |) + C_{1} = \ln (| t |) + C_{2}$

Step 6.2.4

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

$\ln (| 4 + V^{2} |) = \ln (| t |) + C$

Step 6.3

Solve for $V$ .

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Step 6.3.1

Move all the terms containing a logarithm to the left side of the equation.

$\ln (| 4 + V^{2} |) - \ln (| t |) = C$

Step 6.3.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| 4 + V^{2} |}{| t |}) = C$

Step 6.3.3

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

$e^{\ln (\frac{| 4 + V^{2} |}{| t |})} = e^{C}$

Step 6.3.4

Rewrite $\ln (\frac{| 4 + V^{2} |}{| t |}) = C$ 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 $b^{y} = x$ .

$e^{C} = \frac{| 4 + V^{2} |}{| t |}$

Step 6.3.5

Solve for $V$ .

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Step 6.3.5.1

Rewrite the equation as $\frac{| 4 + V^{2} |}{| t |} = e^{C}$ .

$\frac{| 4 + V^{2} |}{| t |} = e^{C}$

Step 6.3.5.2

Multiply both sides by $| t |$ .

$\frac{| 4 + V^{2} |}{| t |} | t | = e^{C} | t |$

Step 6.3.5.3

Simplify the left side.

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Step 6.3.5.3.1

Cancel the common factor of $| t |$ .

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Step 6.3.5.3.1.1

Cancel the common factor.

$\frac{| 4 + V^{2} |}{| t |} | t | = e^{C} | t |$

Step 6.3.5.3.1.2

Rewrite the expression.

$| 4 + V^{2} | = e^{C} | t |$

Step 6.3.5.4

Solve for $V$ .

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Step 6.3.5.4.1

Reorder factors in $e^{C} | t |$ .

$| 4 + V^{2} | = | t | e^{C}$

Step 6.3.5.4.2

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

$4 + V^{2} = \pm | t | e^{C}$

Step 6.3.5.4.3

Reorder factors in $\pm | t | e^{C}$ .

$4 + V^{2} = \pm e^{C} | t |$

Step 6.3.5.4.4

Subtract $4$ from both sides of the equation.

$V^{2} = \pm e^{C} | t | - 4$

Step 6.3.5.4.5

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$V = \pm \sqrt{\pm e^{C} | t | - 4}$

Step 6.4

Group the constant terms together.

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Step 6.4.1

Simplify the constant of integration.

$V = \pm \sqrt{\pm C | t | - 4}$

Step 6.4.2

Combine constants with the plus or minus.

$V = \pm \sqrt{C | t | - 4}$

Step 7

Substitute $\frac{x}{t}$ for $V$ .

$\frac{x}{t} = \pm \sqrt{C | t | - 4}$

Step 8

Solve $\frac{x}{t} = \pm \sqrt{C | t | - 4}$ for $x$ .

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Step 8.1

Multiply both sides by $t$ .

$\frac{x}{t} t = (\pm \sqrt{C | t | - 4}) t$

Step 8.2

Simplify the left side.

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Step 8.2.1

Cancel the common factor of $t$ .

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Step 8.2.1.1

Cancel the common factor.

$\frac{x}{t} t = (\pm \sqrt{C | t | - 4}) t$

Step 8.2.1.2

Rewrite the expression.

$x = (\pm \sqrt{C | t | - 4}) t$