Calculus Examples

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

Step 1

Separate the variables.

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Step 1.1

Solve for $\frac{d y}{d x}$ .

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Step 1.1.1

Subtract $\frac{1}{x^{2}}$ from both sides of the equation.

$\frac{d y}{d x} + \frac{y}{x^{2}} - \frac{1}{x^{2}} = 0$

Step 1.1.2

Move all terms not containing $\frac{d y}{d x}$ to the right side of the equation.

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Step 1.1.2.1

Subtract $\frac{y}{x^{2}}$ from both sides of the equation.

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

Step 1.1.2.2

Add $\frac{1}{x^{2}}$ to both sides of the equation.

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

Step 1.2

Combine the numerators over the common denominator.

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

Step 1.3

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

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

Step 1.4

Simplify.

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Step 1.4.1

Multiply $\frac{1}{- y + 1}$ by $\frac{- y + 1}{x^{2}}$ .

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

Step 1.4.2

Cancel the common factor of $- y + 1$ .

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Step 1.4.2.1

Cancel the common factor.

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

Step 1.4.2.2

Rewrite the expression.

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

Step 1.5

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = - y + 1$ . Then $d u = - d y$ , so $- d u = d y$ . 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

Rewrite.

$\frac{- 1}{1}$

Step 2.2.1.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 2.2.1.2

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

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

Step 2.2.2

Split the fraction into multiple fractions.

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

Step 2.2.3

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

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

Step 2.2.4

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 2.2.5

Simplify.

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

Step 2.2.6

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

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Apply basic rules of exponents.

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Step 2.3.1.1

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 2.3.1.2

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

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Step 2.3.1.2.1

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 2.3.1.2.2

Multiply $2$ by $- 1$ .

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

Step 2.3.2

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 2.3.3

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

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Divide each term in $- \ln (| - y + 1 |) = - \frac{1}{x} + C$ by $- 1$ and simplify.

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Step 3.1.1

Divide each term in $- \ln (| - y + 1 |) = - \frac{1}{x} + C$ by $- 1$ .

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

Step 3.1.2

Simplify the left side.

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Step 3.1.2.1

Dividing two negative values results in a positive value.

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

Step 3.1.2.2

Divide $\ln (| - y + 1 |)$ by $1$ .

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

Step 3.1.3

Simplify the right side.

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Step 3.1.3.1

Simplify each term.

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Step 3.1.3.1.1

Dividing two negative values results in a positive value.

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

Step 3.1.3.1.2

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

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

Step 3.1.3.1.3

Move the negative one from the denominator of $\frac{C}{- 1}$ .

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

Step 3.1.3.1.4

Rewrite $- 1 \cdot C$ as $- C$ .

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

Step 3.2

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

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

Step 3.3

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 3.4

Solve for $y$ .

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Step 3.4.1

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

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

Step 3.4.2

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 3.4.3

Subtract $1$ from both sides of the equation.

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

Step 3.4.4

Divide each term in $- y = \pm e^{\frac{1}{x} - C} - 1$ by $- 1$ and simplify.

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Step 3.4.4.1

Divide each term in $- y = \pm e^{\frac{1}{x} - C} - 1$ by $- 1$ .

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

Step 3.4.4.2

Simplify the left side.

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Step 3.4.4.2.1

Dividing two negative values results in a positive value.

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

Step 3.4.4.2.2

Divide $y$ by $1$ .

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

Step 3.4.4.3

Simplify the right side.

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Step 3.4.4.3.1

Simplify each term.

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Step 3.4.4.3.1.1

Move the negative one from the denominator of $\frac{\pm e^{\frac{1}{x} - C}}{- 1}$ .

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

Step 3.4.4.3.1.2

Rewrite $- 1 \cdot (\pm e^{\frac{1}{x} - C})$ as $- (\pm e^{\frac{1}{x} - C})$ .

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

Step 3.4.4.3.1.3

Divide $- 1$ by $- 1$ .

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

Step 4

Group the constant terms together.

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Step 4.1

Simplify the constant of integration.

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

Step 4.2

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

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

Step 4.3

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

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

Step 4.4

Combine constants with the plus or minus.

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