Calculus Examples

$(t + 1) d t - \frac{1}{y^{2}} d y = 0$

Step 1

Subtract $(t + 1) d t$ from both sides of the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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

Step 2.2.2

Apply basic rules of exponents.

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Step 2.2.2.1

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

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

Step 2.2.2.2

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

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Step 2.2.2.2.1

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

$- \int y^{2 \cdot - 1} d y = \int - (t + 1) d t$

Step 2.2.2.2.2

Multiply $2$ by $- 1$ .

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

Step 2.2.3

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

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

Step 2.2.4

Simplify the answer.

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Step 2.2.4.1

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

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

Step 2.2.4.2

Simplify.

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Step 2.2.4.2.1

Multiply $- 1$ by $- 1$ .

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

Step 2.2.4.2.2

Multiply $\frac{1}{y}$ by $1$ .

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Multiply $- (t + 1)$ .

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

Step 2.3.2

Multiply $- 1$ by $1$ .

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

Step 2.3.3

Split the single integral into multiple integrals.

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

Step 2.3.4

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

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

Step 2.3.5

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

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

Step 2.3.6

Apply the constant rule.

$\frac{1}{y} + C_{1} = - (\frac{1}{2} t^{2} + C_{2}) - t + C_{3}$

Step 2.3.7

Simplify.

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Step 2.3.7.1

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

$\frac{1}{y} + C_{1} = - (\frac{t^{2}}{2} + C_{2}) - t + C_{3}$

Step 2.3.7.2

Simplify.

$\frac{1}{y} + C_{1} = - \frac{1}{2} t^{2} - t + C_{4}$

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Find the LCD of the terms in the equation.

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Step 3.2.1

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

$y, 2, 1, 1$

Step 3.2.2

Since $y, 2, 1, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 2, 1, 1$ then find LCM for the variable part $y^{1}$ .

Step 3.2.3

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 3.2.4

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

Not prime

Step 3.2.5

Since $2$ has no factors besides $1$ and $2$ .

$2$ is a prime number

Step 3.2.6

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

Not prime

Step 3.2.7

The LCM of $1, 2, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2$

Step 3.2.8

The factor for $y^{1}$ is $y$ itself.

$y^{1} = y$

$y$ occurs $1$ time.

Step 3.2.9

The LCM of $y^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$y$

Step 3.2.10

The LCM for $y, 2, 1, 1$ is the numeric part $2$ multiplied by the variable part.

$2 y$

Step 3.3

Multiply each term in $\frac{1}{y} = - \frac{t^{2}}{2} - t + K$ by $2 y$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $\frac{1}{y} = - \frac{t^{2}}{2} - t + K$ by $2 y$ .

$\frac{1}{y} (2 y) = - \frac{t^{2}}{2} (2 y) - t (2 y) + K (2 y)$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Rewrite using the commutative property of multiplication.

$2 \frac{1}{y} y = - \frac{t^{2}}{2} (2 y) - t (2 y) + K (2 y)$

Step 3.3.2.2

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

$\frac{2}{y} y = - \frac{t^{2}}{2} (2 y) - t (2 y) + K (2 y)$

Step 3.3.2.3

Cancel the common factor of $y$ .

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Step 3.3.2.3.1

Cancel the common factor.

$\frac{2}{y} y = - \frac{t^{2}}{2} (2 y) - t (2 y) + K (2 y)$

Step 3.3.2.3.2

Rewrite the expression.

$2 = - \frac{t^{2}}{2} (2 y) - t (2 y) + K (2 y)$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Simplify each term.

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Step 3.3.3.1.1

Cancel the common factor of $2$ .

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Step 3.3.3.1.1.1

Move the leading negative in $- \frac{t^{2}}{2}$ into the numerator.

$2 = \frac{- t^{2}}{2} (2 y) - t (2 y) + K (2 y)$

Step 3.3.3.1.1.2

Factor $2$ out of $2 y$ .

$2 = \frac{- t^{2}}{2} (2 (y)) - t (2 y) + K (2 y)$

Step 3.3.3.1.1.3

Cancel the common factor.

$2 = \frac{- t^{2}}{2} (2 y) - t (2 y) + K (2 y)$

Step 3.3.3.1.1.4

Rewrite the expression.

$2 = - t^{2} y - t (2 y) + K (2 y)$

Step 3.3.3.1.2

Rewrite using the commutative property of multiplication.

$2 = - t^{2} y - 1 \cdot 2 t y + K (2 y)$

Step 3.3.3.1.3

Multiply $- 1$ by $2$ .

$2 = - t^{2} y - 2 t y + K (2 y)$

Step 3.3.3.1.4

Rewrite using the commutative property of multiplication.

$2 = - t^{2} y - 2 t y + 2 K y$

Step 3.4

Solve the equation.

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Step 3.4.1

Rewrite the equation as $- t^{2} y - 2 t y + 2 K y = 2$ .

$- t^{2} y - 2 t y + 2 K y = 2$

Step 3.4.2

Factor $y$ out of $- t^{2} y - 2 t y + 2 K y$ .

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Step 3.4.2.1

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

$y (- t^{2}) - 2 t y + 2 K y = 2$

Step 3.4.2.2

Factor $y$ out of $- 2 t y$ .

$y (- t^{2}) + y (- 2 t) + 2 K y = 2$

Step 3.4.2.3

Factor $y$ out of $2 K y$ .

$y (- t^{2}) + y (- 2 t) + y (2 K) = 2$

Step 3.4.2.4

Factor $y$ out of $y (- t^{2}) + y (- 2 t)$ .

$y (- t^{2} - 2 t) + y (2 K) = 2$

Step 3.4.2.5

Factor $y$ out of $y (- t^{2} - 2 t) + y (2 K)$ .

$y (- t^{2} - 2 t + 2 K) = 2$

Step 3.4.3

Divide each term in $y (- t^{2} - 2 t + 2 K) = 2$ by $- t^{2} - 2 t + 2 K$ and simplify.

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Step 3.4.3.1

Divide each term in $y (- t^{2} - 2 t + 2 K) = 2$ by $- t^{2} - 2 t + 2 K$ .

$\frac{y (- t^{2} - 2 t + 2 K)}{- t^{2} - 2 t + 2 K} = \frac{2}{- t^{2} - 2 t + 2 K}$

Step 3.4.3.2

Simplify the left side.

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Step 3.4.3.2.1

Cancel the common factor of $- t^{2} - 2 t + 2 K$ .

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Step 3.4.3.2.1.1

Cancel the common factor.

$\frac{y (- t^{2} - 2 t + 2 K)}{- t^{2} - 2 t + 2 K} = \frac{2}{- t^{2} - 2 t + 2 K}$

Step 3.4.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{2}{- t^{2} - 2 t + 2 K}$

Step 3.4.3.3

Simplify the right side.

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Step 3.4.3.3.1

Factor $- 1$ out of $- t^{2}$ .

$y = \frac{2}{- (t^{2}) - 2 t + 2 K}$

Step 3.4.3.3.2

Factor $- 1$ out of $- 2 t$ .

$y = \frac{2}{- (t^{2}) - (2 t) + 2 K}$

Step 3.4.3.3.3

Factor $- 1$ out of $- (t^{2}) - (2 t)$ .

$y = \frac{2}{- (t^{2} + 2 t) + 2 K}$

Step 3.4.3.3.4

Factor $- 1$ out of $2 K$ .

$y = \frac{2}{- (t^{2} + 2 t) - (- 2 K)}$

Step 3.4.3.3.5

Factor $- 1$ out of $- (t^{2} + 2 t) - (- 2 K)$ .

$y = \frac{2}{- (t^{2} + 2 t - 2 K)}$

Step 3.4.3.3.6

Rewrite negatives.

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Step 3.4.3.3.6.1

Rewrite $- (t^{2} + 2 t - 2 K)$ as $- 1 (t^{2} + 2 t - 2 K)$ .

$y = \frac{2}{- 1 (t^{2} + 2 t - 2 K)}$

Step 3.4.3.3.6.2

Move the negative in front of the fraction.

$y = - \frac{2}{t^{2} + 2 t - 2 K}$

Step 4

Simplify the constant of integration.

$y = - \frac{2}{t^{2} + 2 t + K}$