Calculus Examples

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

Step 1

Subtract $y d x$ from both sides of the equation.

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

Step 2

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

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

Step 3

Simplify.

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Step 3.1

Cancel the common factor of $1 + x^{2}$ .

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Step 3.1.1

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

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

Step 3.1.2

Cancel the common factor.

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

Step 3.1.3

Rewrite the expression.

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

Step 3.3

Cancel the common factor of $y$ .

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

Cancel the common factor.

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

Step 3.3.4

Rewrite the expression.

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

Step 3.4

Move the negative in front of the fraction.

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

Step 4

Integrate both sides.

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Step 4.1

Set up an integral on each side.

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

Step 4.2

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

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

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.

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

Step 4.3.2

Rewrite $1$ as $1^{2}$ .

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

Step 4.3.3

The integral of $\frac{1}{1^{2} + x^{2}}$ with respect to $x$ is $\arctan (x) + C_{2}$ .

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

Step 4.3.4

Simplify.

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

Step 4.4

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

$\ln (| y |) = - \arctan (x) + C$

Step 5

Solve for $y$ .

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Step 5.1

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

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

Step 5.2

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

$e^{- \arctan (x) + C} = | y |$

Step 5.3

Solve for $y$ .

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Step 5.3.1

Rewrite the equation as $| y | = e^{- \arctan (x) + C}$ .

$| y | = e^{- \arctan (x) + C}$

Step 5.3.2

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

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

Step 6

Group the constant terms together.

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Step 6.1

Rewrite $e^{- \arctan (x) + C}$ as $e^{- \arctan (x)} e^{C}$ .

$y = \pm e^{- \arctan (x)} e^{C}$

Step 6.2

Reorder $e^{- \arctan (x)}$ and $e^{C}$ .

$y = \pm e^{C} e^{- \arctan (x)}$

Step 6.3

Combine constants with the plus or minus.

$y = C e^{- \arctan (x)}$