Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

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

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

Step 1.2

Cancel the common factor of $y - 1$ .

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Step 1.2.1

Cancel the common factor.

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

Step 1.2.2

Rewrite the expression.

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

Step 1.3

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

Differentiate $y - 1$ .

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

Step 2.2.1.1.2

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 2.2.1.1.3

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 2.2.1.1.4

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

$1 + 0$

Step 2.2.1.1.5

Add $1$ and $0$ .

$1$

Step 2.2.1.2

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

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

Step 2.3.2

Apply the constant rule.

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

Simplify.

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Step 2.3.5.1

Simplify.

$\ln (| y - 1 |) + C_{1} = x - 2 (\frac{1}{2}) x^{2} + C_{4}$

Step 2.3.5.2

Simplify.

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Step 2.3.5.2.1

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

$\ln (| y - 1 |) + C_{1} = x + \frac{- 2}{2} x^{2} + C_{4}$

Step 2.3.5.2.2

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

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Step 2.3.5.2.2.1

Factor $2$ out of $- 2$ .

$\ln (| y - 1 |) + C_{1} = x + \frac{2 \cdot - 1}{2} x^{2} + C_{4}$

Step 2.3.5.2.2.2

Cancel the common factors.

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Step 2.3.5.2.2.2.1

Factor $2$ out of $2$ .

$\ln (| y - 1 |) + C_{1} = x + \frac{2 \cdot - 1}{2 (1)} x^{2} + C_{4}$

Step 2.3.5.2.2.2.2

Cancel the common factor.

$\ln (| y - 1 |) + C_{1} = x + \frac{2 \cdot - 1}{2 \cdot 1} x^{2} + C_{4}$

Step 2.3.5.2.2.2.3

Rewrite the expression.

$\ln (| y - 1 |) + C_{1} = x + \frac{- 1}{1} x^{2} + C_{4}$

Step 2.3.5.2.2.2.4

Divide $- 1$ by $1$ .

$\ln (| y - 1 |) + C_{1} = x - x^{2} + C_{4}$

Step 2.3.6

Reorder terms.

$\ln (| y - 1 |) + C_{1} = - x^{2} + x + C_{4}$

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Rewrite $\ln (| y - 1 |) = - x^{2} + x + C$ 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 $b^{y} = x$ .

$e^{- x^{2} + x + C} = | y - 1 |$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Rewrite the equation as $| y - 1 | = e^{- x^{2} + x + C}$ .

$| y - 1 | = e^{- x^{2} + x + C}$

Step 3.3.2

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

$y - 1 = \pm e^{- x^{2} + x + C}$

Step 3.3.3

Add $1$ to both sides of the equation.

$y = \pm e^{- x^{2} + x + C} + 1$

Step 4

Group the constant terms together.

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Step 4.1

Rewrite $e^{- x^{2} + x + C}$ as $e^{- x^{2} + x} e^{C}$ .

$y = \pm e^{- x^{2} + x} e^{C} + 1$

Step 4.2

Reorder $e^{- x^{2} + x}$ and $e^{C}$ .

$y = \pm e^{C} e^{- x^{2} + x} + 1$

Step 4.3

Combine constants with the plus or minus.

$y = C e^{- x^{2} + x} + 1$