Calculus Examples

$\frac{4}{t} d t - \frac{y - 3}{y} d y = 0$

Step 1

Subtract $\frac{4}{t} d t$ from both sides of the equation.

$- \frac{y - 3}{y} d y = - \frac{4}{t} d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int - \frac{y - 3}{y} d y = \int - \frac{4}{t} d t$

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.

$- \int \frac{y - 3}{y} d y = \int - \frac{4}{t} d t$

Step 2.2.2

Divide $y - 3$ by $y$ .

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Step 2.2.2.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$y$

$0$

$y$

$3$

Step 2.2.2.2

Divide the highest order term in the dividend $y$ by the highest order term in divisor $y$ .

					$1$
	$y$	+	$0$		$y$	-	$3$

Step 2.2.2.3

Multiply the new quotient term by the divisor.

				$1$
$y$	+	$0$		$y$	-	$3$
			+	$y$	+	$0$

Step 2.2.2.4

The expression needs to be subtracted from the dividend, so change all the signs in $y + 0$

				$1$
$y$	+	$0$		$y$	-	$3$
			-	$y$	-	$0$

Step 2.2.2.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$1$
$y$	+	$0$		$y$	-	$3$
			-	$y$	-	$0$
					-	$3$

Step 2.2.2.6

The final answer is the quotient plus the remainder over the divisor.

$- \int 1 - \frac{3}{y} d y = \int - \frac{4}{t} d t$

Step 2.2.3

Split the single integral into multiple integrals.

$- (\int d y + \int - \frac{3}{y} d y) = \int - \frac{4}{t} d t$

Step 2.2.4

Apply the constant rule.

$- (y + C_{1} + \int - \frac{3}{y} d y) = \int - \frac{4}{t} d t$

Step 2.2.5

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

$- (y + C_{1} - \int \frac{3}{y} d y) = \int - \frac{4}{t} d t$

Step 2.2.6

Since $3$ is constant with respect to $y$ , move $3$ out of the integral.

$- (y + C_{1} - (3 \int \frac{1}{y} d y)) = \int - \frac{4}{t} d t$

Step 2.2.7

Multiply $3$ by $- 1$ .

$- (y + C_{1} - 3 \int \frac{1}{y} d y) = \int - \frac{4}{t} d t$

Step 2.2.8

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

$- (y + C_{1} - 3 (\ln (| y |) + C_{2})) = \int - \frac{4}{t} d t$

Step 2.2.9

Simplify.

$- (y - 3 \ln (| y |)) + C_{3} = \int - \frac{4}{t} d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

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

$- (y - 3 \ln (| y |)) + C_{3} = - \int \frac{4}{t} d t$

Step 2.3.2

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

$- (y - 3 \ln (| y |)) + C_{3} = - (4 \int \frac{1}{t} d t)$

Step 2.3.3

Multiply $4$ by $- 1$ .

$- (y - 3 \ln (| y |)) + C_{3} = - 4 \int \frac{1}{t} d t$

Step 2.3.4

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

$- (y - 3 \ln (| y |)) + C_{3} = - 4 (\ln (| t |) + C_{4})$

Step 2.3.5

Simplify.

$- (y - 3 \ln (| y |)) + C_{3} = - 4 \ln (| t |) + C_{4}$

Step 2.4

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

$- (y - 3 \ln (| y |)) = - 4 \ln (| t |) + C$