Calculus Examples

$x^{2} d y + (y - 1) d x = 0$

Step 1

Subtract $(y - 1) d x$ from both sides of the equation.

$x^{2} d y = - (y - 1) d x$

Step 2

Multiply both sides by $\frac{1}{x^{2} (y - 1)}$ .

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

Step 3

Simplify.

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Step 3.1

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

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Step 3.1.1

Cancel the common factor.

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

Step 3.1.2

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

Cancel the common factor of $y - 1$ .

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Step 3.3.1

Move the leading negative in $- \frac{1}{x^{2} (y - 1)}$ into the numerator.

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

Step 3.3.2

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

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

Step 3.3.3

Cancel the common factor.

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

Step 3.3.4

Rewrite the expression.

$\frac{1}{y - 1} d y = \frac{- 1}{x^{2}} d x$

Step 3.4

Move the negative in front of the fraction.

$\frac{1}{y - 1} d y = - \frac{1}{x^{2}} d x$

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

Integrate the left side.

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Step 4.2.1

Let $u = y - 1$ . Then $d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.2.1.1

Let $u = y - 1$ . Find $\frac{d u}{d y}$ .

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Step 4.2.1.1.1

Differentiate $y - 1$ .

$\frac{d}{d y} [y - 1]$

Step 4.2.1.1.2

By the Sum Rule, the derivative of $y - 1$ with respect to $y$ is $\frac{d}{d y} [y] + \frac{d}{d y} [- 1]$ .

$\frac{d}{d y} [y] + \frac{d}{d y} [- 1]$

Step 4.2.1.1.3

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

$1 + \frac{d}{d y} [- 1]$

Step 4.2.1.1.4

Since $- 1$ is constant with respect to $y$ , the derivative of $- 1$ with respect to $y$ is $0$ .

$1 + 0$

Step 4.2.1.1.5

Add $1$ and $0$ .

$1$

Step 4.2.1.2

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

$\int \frac{1}{u} d u = \int - \frac{1}{x^{2}} d x$

Step 4.2.2

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

$\ln (| u |) + C_{1} = \int - \frac{1}{x^{2}} d x$

Step 4.2.3

Replace all occurrences of $u$ with $y - 1$ .

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

Step 4.3

Integrate the right side.

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Step 4.3.1

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

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

Step 4.3.2

Apply basic rules of exponents.

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Step 4.3.2.1

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

$\ln (| y - 1 |) + C_{1} = - \int {(x^{2})}^{- 1} d x$

Step 4.3.2.2

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

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Step 4.3.2.2.1

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

$\ln (| y - 1 |) + C_{1} = - \int x^{2 \cdot - 1} d x$

Step 4.3.2.2.2

Multiply $2$ by $- 1$ .

$\ln (| y - 1 |) + C_{1} = - \int x^{- 2} d x$

Step 4.3.3

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

$\ln (| y - 1 |) + C_{1} = - (- x^{- 1} + C_{2})$

Step 4.3.4

Simplify the answer.

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Step 4.3.4.1

Rewrite $- (- x^{- 1} + C_{2})$ as $- - \frac{1}{x} + C_{2}$ .

$\ln (| y - 1 |) + C_{1} = - - \frac{1}{x} + C_{2}$

Step 4.3.4.2

Simplify.

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Step 4.3.4.2.1

Multiply $- 1$ by $- 1$ .

$\ln (| y - 1 |) + C_{1} = 1 \frac{1}{x} + C_{2}$

Step 4.3.4.2.2

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

$\ln (| y - 1 |) + C_{1} = \frac{1}{x} + C_{2}$

Step 4.4

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

$\ln (| y - 1 |) = \frac{1}{x} + C$

Step 5

Solve for $y$ .

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Step 5.1

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

$e^{\ln (| y - 1 |)} = e^{\frac{1}{x} + C}$

Step 5.2

Rewrite $\ln (| y - 1 |) = \frac{1}{x} + C$ 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 $b^{y} = x$ .

$e^{\frac{1}{x} + C} = | y - 1 |$

Step 5.3

Solve for $y$ .

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Step 5.3.1

Rewrite the equation as $| y - 1 | = e^{\frac{1}{x} + C}$ .

$| y - 1 | = e^{\frac{1}{x} + C}$

Step 5.3.2

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

$y - 1 = \pm e^{\frac{1}{x} + C}$

Step 5.3.3

Add $1$ to both sides of the equation.

$y = \pm e^{\frac{1}{x} + C} + 1$

Step 6

Group the constant terms together.

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Step 6.1

Rewrite $e^{\frac{1}{x} + C}$ as $e^{\frac{1}{x}} e^{C}$ .

$y = \pm e^{\frac{1}{x}} e^{C} + 1$

Step 6.2

Reorder $e^{\frac{1}{x}}$ and $e^{C}$ .

$y = \pm e^{C} e^{\frac{1}{x}} + 1$

Step 6.3

Combine constants with the plus or minus.

$y = C e^{\frac{1}{x}} + 1$