Calculus Examples

$e^{t} \frac{d y}{d t} = y^{2}$ , $y (0) = \frac{1}{4}$

Step 1

Separate the variables.

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Step 1.1

Divide each term in $e^{t} \frac{d y}{d t} = y^{2}$ by $e^{t}$ and simplify.

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Step 1.1.1

Divide each term in $e^{t} \frac{d y}{d t} = y^{2}$ by $e^{t}$ .

$\frac{e^{t} \frac{d y}{d t}}{e^{t}} = \frac{y^{2}}{e^{t}}$

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of $e^{t}$ .

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Step 1.1.2.1.1

Cancel the common factor.

$\frac{e^{t} \frac{d y}{d t}}{e^{t}} = \frac{y^{2}}{e^{t}}$

Step 1.1.2.1.2

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

$\frac{d y}{d t} = \frac{y^{2}}{e^{t}}$

Step 1.2

Multiply both sides by $\frac{1}{y^{2}}$ .

$\frac{1}{y^{2}} \frac{d y}{d t} = \frac{1}{y^{2}} \cdot \frac{y^{2}}{e^{t}}$

Step 1.3

Simplify.

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Step 1.3.1

Combine.

$\frac{1}{y^{2}} \frac{d y}{d t} = \frac{1 y^{2}}{y^{2} e^{t}}$

Step 1.3.2

Cancel the common factor of $y^{2}$ .

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Step 1.3.2.1

Cancel the common factor.

$\frac{1}{y^{2}} \frac{d y}{d t} = \frac{1 y^{2}}{y^{2} e^{t}}$

Step 1.3.2.2

Rewrite the expression.

$\frac{1}{y^{2}} \frac{d y}{d t} = \frac{1}{e^{t}}$

Step 1.4

Rewrite the equation.

$\frac{1}{y^{2}} d y = \frac{1}{e^{t}} d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y^{2}} d y = \int \frac{1}{e^{t}} d t$

Step 2.2

Integrate the left side.

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Step 2.2.1

Apply basic rules of exponents.

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Step 2.2.1.1

Move $y^{2}$ out of the denominator by raising it to the $- 1$ power.

$\int {(y^{2})}^{- 1} d y = \int \frac{1}{e^{t}} d t$

Step 2.2.1.2

Multiply the exponents in ${(y^{2})}^{- 1}$ .

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Step 2.2.1.2.1

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

$\int y^{2 \cdot - 1} d y = \int \frac{1}{e^{t}} d t$

Step 2.2.1.2.2

Multiply $2$ by $- 1$ .

$\int y^{- 2} d y = \int \frac{1}{e^{t}} d t$

Step 2.2.2

By the Power Rule, the integral of $y^{- 2}$ with respect to $y$ is $- y^{- 1}$ .

$- y^{- 1} + C_{1} = \int \frac{1}{e^{t}} d t$

Step 2.2.3

Rewrite $- y^{- 1} + C_{1}$ as $- \frac{1}{y} + C_{1}$ .

$- \frac{1}{y} + C_{1} = \int \frac{1}{e^{t}} d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Simplify the expression.

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Step 2.3.1.1

Negate the exponent of $e^{t}$ and move it out of the denominator.

$- \frac{1}{y} + C_{1} = \int 1 {(e^{t})}^{- 1} d t$

Step 2.3.1.2

Simplify.

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Step 2.3.1.2.1

Multiply the exponents in ${(e^{t})}^{- 1}$ .

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Step 2.3.1.2.1.1

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

$- \frac{1}{y} + C_{1} = \int 1 e^{t \cdot - 1} d t$

Step 2.3.1.2.1.2

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

$- \frac{1}{y} + C_{1} = \int 1 e^{- 1 \cdot t} d t$

Step 2.3.1.2.1.3

Rewrite $- 1 t$ as $- t$ .

$- \frac{1}{y} + C_{1} = \int 1 e^{- t} d t$

Step 2.3.1.2.2

Multiply $e^{- t}$ by $1$ .

$- \frac{1}{y} + C_{1} = \int e^{- t} d t$

Step 2.3.2

Let $u = - t$ . Then $d u = - d t$ , so $- d u = d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.3.2.1

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

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Step 2.3.2.1.1

Differentiate $- t$ .

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

Step 2.3.2.1.2

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

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

Step 2.3.2.1.3

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

$- 1 \cdot 1$

Step 2.3.2.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 2.3.2.2

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

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

Step 2.3.3

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

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

Step 2.3.4

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

$- \frac{1}{y} + C_{1} = - (e^{u} + C_{2})$

Step 2.3.5

Simplify.

$- \frac{1}{y} + C_{1} = - e^{u} + C_{2}$

Step 2.3.6

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

$- \frac{1}{y} + C_{1} = - e^{- t} + C_{2}$

Step 2.4

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

$- \frac{1}{y} = - e^{- t} + K$

Step 3

Solve for $y$ .

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Step 3.1

Find the LCD of the terms in the equation.

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Step 3.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y, 1, 1$

Step 3.1.2

The LCM of one and any expression is the expression.

$y$

Step 3.2

Multiply each term in $- \frac{1}{y} = - e^{- t} + K$ by $y$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $- \frac{1}{y} = - e^{- t} + K$ by $y$ .

$- \frac{1}{y} y = - e^{- t} y + K y$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $y$ .

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Step 3.2.2.1.1

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

$\frac{- 1}{y} y = - e^{- t} y + K y$

Step 3.2.2.1.2

Cancel the common factor.

$\frac{- 1}{y} y = - e^{- t} y + K y$

Step 3.2.2.1.3

Rewrite the expression.

$- 1 = - e^{- t} y + K y$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Reorder factors in $- e^{- t} y + K y$ .

$- 1 = - y e^{- t} + K y$

Step 3.3

Solve the equation.

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Step 3.3.1

Rewrite the equation as $- y e^{- t} + K y = - 1$ .

$- y e^{- t} + K y = - 1$

Step 3.3.2

Factor $y$ out of $- y e^{- t} + K y$ .

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Step 3.3.2.1

Factor $y$ out of $- y e^{- t}$ .

$y (- 1 e^{- t}) + K y = - 1$

Step 3.3.2.2

Factor $y$ out of $K y$ .

$y (- 1 e^{- t}) + y K = - 1$

Step 3.3.2.3

Factor $y$ out of $y (- 1 e^{- t}) + y K$ .

$y (- 1 e^{- t} + K) = - 1$

Step 3.3.3

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

$y (- e^{- t} + K) = - 1$

Step 3.3.4

Divide each term in $y (- e^{- t} + K) = - 1$ by $- e^{- t} + K$ and simplify.

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Step 3.3.4.1

Divide each term in $y (- e^{- t} + K) = - 1$ by $- e^{- t} + K$ .

$\frac{y (- e^{- t} + K)}{- e^{- t} + K} = \frac{- 1}{- e^{- t} + K}$

Step 3.3.4.2

Simplify the left side.

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Step 3.3.4.2.1

Cancel the common factor of $- e^{- t} + K$ .

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Step 3.3.4.2.1.1

Cancel the common factor.

$\frac{y (- e^{- t} + K)}{- e^{- t} + K} = \frac{- 1}{- e^{- t} + K}$

Step 3.3.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 1}{- e^{- t} + K}$

Step 3.3.4.3

Simplify the right side.

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Step 3.3.4.3.1

Move the negative in front of the fraction.

$y = - \frac{1}{- e^{- t} + K}$

Step 3.3.4.3.2

Factor $- 1$ out of $- e^{- t}$ .

$y = - \frac{1}{- (e^{- t}) + K}$

Step 3.3.4.3.3

Factor $- 1$ out of $K$ .

$y = - \frac{1}{- (e^{- t}) - 1 (- K)}$

Step 3.3.4.3.4

Factor $- 1$ out of $- (e^{- t}) - 1 (- K)$ .

$y = - \frac{1}{- (e^{- t} - K)}$

Step 3.3.4.3.5

Simplify the expression.

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Step 3.3.4.3.5.1

Rewrite $- (e^{- t} - K)$ as $- 1 (e^{- t} - K)$ .

$y = - \frac{1}{- 1 (e^{- t} - K)}$

Step 3.3.4.3.5.2

Move the negative in front of the fraction.

$y = - - \frac{1}{e^{- t} - K}$

Step 3.3.4.3.5.3

Multiply $- 1$ by $- 1$ .

$y = 1 \frac{1}{e^{- t} - K}$

Step 3.3.4.3.5.4

Multiply $\frac{1}{e^{- t} - K}$ by $1$ .

$y = \frac{1}{e^{- t} - K}$

Step 4

Simplify the constant of integration.

$y = \frac{1}{e^{- t} + K}$

Step 5

Use the initial condition to find the value of $K$ by substituting $0$ for $t$ and $\frac{1}{4}$ for $y$ in $y = \frac{1}{e^{- t} + K}$ .

$\frac{1}{4} = \frac{1}{e^{0} + K}$

Step 6

Solve for $K$ .

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Step 6.1

Rewrite the equation as $\frac{1}{e^{0} + K} = \frac{1}{4}$ .

$\frac{1}{e^{0} + K} = \frac{1}{4}$

Step 6.2

Anything raised to $0$ is $1$ .

$\frac{1}{1 + K} = \frac{1}{4}$

Step 6.3

Find the LCD of the terms in the equation.

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Step 6.3.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1 + K, 4$

Step 6.3.2

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 6.3.3

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 6.3.4

$4$ has factors of $2$ and $2$ .

$2 \cdot 2$

Step 6.3.5

Multiply $2$ by $2$ .

$4$

Step 6.3.6

The factor for $1 + K$ is $1 + K$ itself.

$(1 + K) = 1 + K$

$(1 + K)$ occurs $1$ time.

Step 6.3.7

The LCM of $1 + K$ is the result of multiplying all factors the greatest number of times they occur in either term.

$1 + K$

Step 6.3.8

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$4 (1 + K)$

Step 6.4

Multiply each term in $\frac{1}{1 + K} = \frac{1}{4}$ by $4 (1 + K)$ to eliminate the fractions.

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Step 6.4.1

Multiply each term in $\frac{1}{1 + K} = \frac{1}{4}$ by $4 (1 + K)$ .

$\frac{1}{1 + K} (4 (1 + K)) = \frac{1}{4} (4 (1 + K))$

Step 6.4.2

Simplify the left side.

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Step 6.4.2.1

Rewrite using the commutative property of multiplication.

$4 \frac{1}{1 + K} (1 + K) = \frac{1}{4} (4 (1 + K))$

Step 6.4.2.2

Combine $4$ and $\frac{1}{1 + K}$ .

$\frac{4}{1 + K} (1 + K) = \frac{1}{4} (4 (1 + K))$

Step 6.4.2.3

Cancel the common factor of $1 + K$ .

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Step 6.4.2.3.1

Cancel the common factor.

$\frac{4}{1 + K} (1 + K) = \frac{1}{4} (4 (1 + K))$

Step 6.4.2.3.2

Rewrite the expression.

$4 = \frac{1}{4} (4 (1 + K))$

Step 6.4.3

Simplify the right side.

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Step 6.4.3.1

Cancel the common factor of $4$ .

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Step 6.4.3.1.1

Cancel the common factor.

$4 = \frac{1}{4} (4 (1 + K))$

Step 6.4.3.1.2

Rewrite the expression.

$4 = 1 + K$

Step 6.5

Solve the equation.

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Step 6.5.1

Rewrite the equation as $1 + K = 4$ .

$1 + K = 4$

Step 6.5.2

Move all terms not containing $K$ to the right side of the equation.

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Step 6.5.2.1

Subtract $1$ from both sides of the equation.

$K = 4 - 1$

Step 6.5.2.2

Subtract $1$ from $4$ .

$K = 3$

Step 7

Substitute $3$ for $K$ in $y = \frac{1}{e^{- t} + K}$ and simplify.

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Step 7.1

Substitute $3$ for $K$ .

$y = \frac{1}{e^{- t} + 3}$