Calculus Examples

$x d y + (3 x - 2 y) d x = 0$

Step 1

Rewrite the differential equation to fit the Exact differential equation technique.

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Step 1.1

Rewrite.

$(3 x - 2 y) d x + x d y = 0$

Step 2

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 3 x - 2 y$ .

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Step 2.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [3 x - 2 y]$

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of $3 x - 2 y$ with respect to $y$ is $\frac{d}{d y} [3 x] + \frac{d}{d y} [- 2 y]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [3 x] + \frac{d}{d y} [- 2 y]$

Step 2.2.2

Since $3 x$ is constant with respect to $y$ , the derivative of $3 x$ with respect to $y$ is $0$ .

$\frac{\partial M}{\partial y} = 0 + \frac{d}{d y} [- 2 y]$

Step 2.3

Evaluate $\frac{d}{d y} [- 2 y]$ .

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Step 2.3.1

Since $- 2$ is constant with respect to $y$ , the derivative of $- 2 y$ with respect to $y$ is $- 2 \frac{d}{d y} [y]$ .

$\frac{\partial M}{\partial y} = 0 - 2 \frac{d}{d y} [y]$

Step 2.3.2

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

$\frac{\partial M}{\partial y} = 0 - 2 \cdot 1$

Step 2.3.3

Multiply $- 2$ by $1$ .

$\frac{\partial M}{\partial y} = 0 - 2$

Step 2.4

Subtract $2$ from $0$ .

$\frac{\partial M}{\partial y} = - 2$

Step 3

Find $\frac{\partial N}{\partial x}$ where $N (x, y) = x$ .

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Step 3.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [x]$

Step 3.2

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

$\frac{\partial N}{\partial x} = 1$

Step 4

Check that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .

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Step 4.1

Substitute $- 2$ for $\frac{\partial M}{\partial y}$ and $1$ for $\frac{\partial N}{\partial x}$ .

$- 2 = 1$

Step 4.2

Since the left side does not equal the right side, the equation is not an identity.

$- 2 = 1$ is not an identity.

Step 5

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} d x}$ .

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Step 5.1

Substitute $- 2$ for $\frac{\partial M}{\partial y}$ .

$\frac{- 2 - \frac{\partial N}{\partial x}}{N}$

Step 5.2

Substitute $1$ for $\frac{\partial N}{\partial x}$ .

$\frac{- 2 - 1}{N}$

Step 5.3

Substitute $x$ for $N$ .

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Step 5.3.1

Substitute $x$ for $N$ .

$\frac{- 2 - 1}{x}$

Step 5.3.2

Subtract $1$ from $- 2$ .

$\frac{- 3}{x}$

Step 5.3.3

Move the negative in front of the fraction.

$- \frac{3}{x}$

Step 5.4

Find the integration factor $μ (x, y) = e^{\int \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} d x}$ .

$μ (x, y) = e^{\int - \frac{3}{x} d x}$

Step 6

Evaluate the integral $e^{\int - \frac{3}{x} d x}$ .

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Step 6.1

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

$μ (x, y) = e^{- \int \frac{3}{x} d x}$

Step 6.2

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

$μ (x, y) = e^{- (3 \int \frac{1}{x} d x)}$

Step 6.3

Multiply $3$ by $- 1$ .

$μ (x, y) = e^{- 3 \int \frac{1}{x} d x}$

Step 6.4

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

$μ (x, y) = e^{- 3 (\ln (| x |) + C)}$

Step 6.5

Simplify.

$μ (x, y) = e^{- 3 \ln (| x |)} + C$

Step 6.6

Simplify each term.

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Step 6.6.1

Simplify $- 3 \ln (| x |)$ by moving $- 3$ inside the logarithm.

$μ (x, y) = e^{\ln ({| x |}^{- 3})} + C$

Step 6.6.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x |}^{- 3} + C$

Step 6.6.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$μ (x, y) = \frac{1}{{| x |}^{3}} + C$

Step 7

Multiply both sides of $(3 x - 2 y) d x + x d y = 0$ by the integration factor $\frac{1}{x^{3}}$ .

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Step 7.1

Multiply $3 x - 2 y$ by $\frac{1}{x^{3}}$ .

$(3 x - 2 y) \frac{1}{x^{3}} d x + x d y = 0$

Step 7.2

Multiply $3 x - 2 y$ by $\frac{1}{x^{3}}$ .

$\frac{3 x - 2 y}{x^{3}} d x + x d y = 0$

Step 7.3

Multiply $x$ by $\frac{1}{x^{3}}$ .

$\frac{3 x - 2 y}{x^{3}} d x + x \frac{1}{x^{3}} d y = 0$

Step 7.4

Cancel the common factor of $x$ .

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Step 7.4.1

Factor $x$ out of $x^{3}$ .

$\frac{3 x - 2 y}{x^{3}} d x + x \frac{1}{x \cdot x^{2}} d y = 0$

Step 7.4.2

Cancel the common factor.

$\frac{3 x - 2 y}{x^{3}} d x + x \frac{1}{x \cdot x^{2}} d y = 0$

Step 7.4.3

Rewrite the expression.

$\frac{3 x - 2 y}{x^{3}} d x + \frac{1}{x^{2}} d y = 0$

Step 8

Set $f (x, y)$ equal to the integral of $N (x, y)$ .

$f (x, y) = \int \frac{1}{x^{2}} d y$

Step 9

Integrate $N (x, y) = \frac{1}{x^{2}}$ to find $f (x, y)$ .

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Step 9.1

Apply the constant rule.

$f (x, y) = \frac{1}{x^{2}} y + C$

Step 9.2

Combine $\frac{1}{x^{2}}$ and $y$ .

$f (x, y) = \frac{y}{x^{2}} + C$

Step 10

Since the integral of $g (x)$ will contain an integration constant, we can replace $C$ with $g (x)$ .

$f (x, y) = \frac{y}{x^{2}} + g (x)$

Step 11

Set $\frac{\partial f}{\partial x} = M (x, y)$ .

$\frac{\partial f}{\partial x} = \frac{3 x - 2 y}{x^{3}}$

Step 12

Find $\frac{\partial f}{\partial x}$ .

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Step 12.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [\frac{y}{x^{2}} + g (x)] = \frac{3 x - 2 y}{x^{3}}$

Step 12.2

By the Sum Rule, the derivative of $\frac{y}{x^{2}} + g (x)$ with respect to $x$ is $\frac{d}{d x} [\frac{y}{x^{2}}] + \frac{d}{d x} [g (x)]$ .

$\frac{d}{d x} [\frac{y}{x^{2}}] + \frac{d}{d x} [g (x)] = \frac{3 x - 2 y}{x^{3}}$

Step 12.3

Evaluate $\frac{d}{d x} [\frac{y}{x^{2}}]$ .

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Step 12.3.1

Since $y$ is constant with respect to $x$ , the derivative of $\frac{y}{x^{2}}$ with respect to $x$ is $y \frac{d}{d x} [\frac{1}{x^{2}}]$ .

$y \frac{d}{d x} [\frac{1}{x^{2}}] + \frac{d}{d x} [g (x)] = \frac{3 x - 2 y}{x^{3}}$

Step 12.3.2

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

$y \frac{d}{d x} [{(x^{2})}^{- 1}] + \frac{d}{d x} [g (x)] = \frac{3 x - 2 y}{x^{3}}$

Step 12.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{2}$ .

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Step 12.3.3.1

To apply the Chain Rule, set $u$ as $x^{2}$ .

$y (\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [g (x)] = \frac{3 x - 2 y}{x^{3}}$

Step 12.3.3.2

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

$y (- u^{- 2} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [g (x)] = \frac{3 x - 2 y}{x^{3}}$

Step 12.3.3.3

Replace all occurrences of $u$ with $x^{2}$ .

$y (- {(x^{2})}^{- 2} \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [g (x)] = \frac{3 x - 2 y}{x^{3}}$

Step 12.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$y (- {(x^{2})}^{- 2} (2 x)) + \frac{d}{d x} [g (x)] = \frac{3 x - 2 y}{x^{3}}$

Step 12.3.5

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

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Step 12.3.5.1

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

$y (- x^{2 \cdot - 2} (2 x)) + \frac{d}{d x} [g (x)] = \frac{3 x - 2 y}{x^{3}}$

Step 12.3.5.2

Multiply $2$ by $- 2$ .

$y (- x^{- 4} (2 x)) + \frac{d}{d x} [g (x)] = \frac{3 x - 2 y}{x^{3}}$

Step 12.3.6

Multiply $2$ by $- 1$ .

$y (- 2 x^{- 4} x) + \frac{d}{d x} [g (x)] = \frac{3 x - 2 y}{x^{3}}$

Step 12.3.7

Raise $x$ to the power of $1$ .

$y (- 2 (x^{1} x^{- 4})) + \frac{d}{d x} [g (x)] = \frac{3 x - 2 y}{x^{3}}$

Step 12.3.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y (- 2 x^{1 - 4}) + \frac{d}{d x} [g (x)] = \frac{3 x - 2 y}{x^{3}}$

Step 12.3.9

Subtract $4$ from $1$ .

$y (- 2 x^{- 3}) + \frac{d}{d x} [g (x)] = \frac{3 x - 2 y}{x^{3}}$

Step 12.4

Differentiate using the function rule which states that the derivative of $g (x)$ is $\frac{d g}{d x}$ .

$y (- 2 x^{- 3}) + \frac{d g}{d x} = \frac{3 x - 2 y}{x^{3}}$

Step 12.5

Simplify.

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Step 12.5.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$y (- 2 \frac{1}{x^{3}}) + \frac{d g}{d x} = \frac{3 x - 2 y}{x^{3}}$

Step 12.5.2

Combine terms.

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Step 12.5.2.1

Combine $- 2$ and $\frac{1}{x^{3}}$ .

$y \frac{- 2}{x^{3}} + \frac{d g}{d x} = \frac{3 x - 2 y}{x^{3}}$

Step 12.5.2.2

Move the negative in front of the fraction.

$y (- \frac{2}{x^{3}}) + \frac{d g}{d x} = \frac{3 x - 2 y}{x^{3}}$

Step 12.5.2.3

Combine $y$ and $\frac{2}{x^{3}}$ .

$- \frac{y \cdot 2}{x^{3}} + \frac{d g}{d x} = \frac{3 x - 2 y}{x^{3}}$

Step 12.5.2.4

Move $2$ to the left of $y$ .

$- \frac{2 y}{x^{3}} + \frac{d g}{d x} = \frac{3 x - 2 y}{x^{3}}$

Step 12.5.3

Reorder terms.

$\frac{d g}{d x} - \frac{2 y}{x^{3}} = \frac{3 x - 2 y}{x^{3}}$

Step 13

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

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Step 13.1

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

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Step 13.1.1

Move all terms containing variables to the left side of the equation.

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Step 13.1.1.1

Subtract $\frac{3 x - 2 y}{x^{3}}$ from both sides of the equation.

$\frac{d g}{d x} - \frac{2 y}{x^{3}} - \frac{3 x - 2 y}{x^{3}} = 0$

Step 13.1.1.2

Combine the numerators over the common denominator.

$\frac{d g}{d x} + \frac{- 2 y - (3 x - 2 y)}{x^{3}} = 0$

Step 13.1.1.3

Simplify each term.

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Step 13.1.1.3.1

Apply the distributive property.

$\frac{d g}{d x} + \frac{- 2 y - (3 x) - (- 2 y)}{x^{3}} = 0$

Step 13.1.1.3.2

Multiply $3$ by $- 1$ .

$\frac{d g}{d x} + \frac{- 2 y - 3 x - (- 2 y)}{x^{3}} = 0$

Step 13.1.1.3.3

Multiply $- 2$ by $- 1$ .

$\frac{d g}{d x} + \frac{- 2 y - 3 x + 2 y}{x^{3}} = 0$

Step 13.1.1.4

Combine the opposite terms in $- 2 y - 3 x + 2 y$ .

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Step 13.1.1.4.1

Add $- 2 y$ and $2 y$ .

$\frac{d g}{d x} + \frac{- 3 x + 0}{x^{3}} = 0$

Step 13.1.1.4.2

Add $- 3 x$ and $0$ .

$\frac{d g}{d x} + \frac{- 3 x}{x^{3}} = 0$

Step 13.1.1.5

Simplify each term.

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Step 13.1.1.5.1

Cancel the common factor of $x$ and $x^{3}$ .

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Step 13.1.1.5.1.1

Factor $x$ out of $- 3 x$ .

$\frac{d g}{d x} + \frac{x \cdot - 3}{x^{3}} = 0$

Step 13.1.1.5.1.2

Cancel the common factors.

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Step 13.1.1.5.1.2.1

Factor $x$ out of $x^{3}$ .

$\frac{d g}{d x} + \frac{x \cdot - 3}{x \cdot x^{2}} = 0$

Step 13.1.1.5.1.2.2

Cancel the common factor.

$\frac{d g}{d x} + \frac{x \cdot - 3}{x \cdot x^{2}} = 0$

Step 13.1.1.5.1.2.3

Rewrite the expression.

$\frac{d g}{d x} + \frac{- 3}{x^{2}} = 0$

Step 13.1.1.5.2

Move the negative in front of the fraction.

$\frac{d g}{d x} - \frac{3}{x^{2}} = 0$

Step 13.1.2

Add $\frac{3}{x^{2}}$ to both sides of the equation.

$\frac{d g}{d x} = \frac{3}{x^{2}}$

Step 14

Find the antiderivative of $\frac{3}{x^{2}}$ to find $g (x)$ .

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Step 14.1

Integrate both sides of $\frac{d g}{d x} = \frac{3}{x^{2}}$ .

$\int \frac{d g}{d x} d x = \int \frac{3}{x^{2}} d x$

Step 14.2

Evaluate $\int \frac{d g}{d x} d x$ .

$g (x) = \int \frac{3}{x^{2}} d x$

Step 14.3

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

$g (x) = 3 \int \frac{1}{x^{2}} d x$

Step 14.4

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

$g (x) = 3 \int {(x^{2})}^{- 1} d x$

Step 14.5

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

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Step 14.5.1

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

$g (x) = 3 \int x^{2 \cdot - 1} d x$

Step 14.5.2

Multiply $2$ by $- 1$ .

$g (x) = 3 \int x^{- 2} d x$

Step 14.6

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

$g (x) = 3 (- x^{- 1} + C)$

Step 14.7

Simplify the answer.

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Step 14.7.1

Rewrite $3 (- x^{- 1} + C)$ as $3 (- \frac{1}{x}) + C$ .

$g (x) = 3 (- \frac{1}{x}) + C$

Step 14.7.2

Simplify.

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Step 14.7.2.1

Multiply $- 1$ by $3$ .

$g (x) = - 3 \frac{1}{x} + C$

Step 14.7.2.2

Combine $- 3$ and $\frac{1}{x}$ .

$g (x) = \frac{- 3}{x} + C$

Step 14.7.2.3

Move the negative in front of the fraction.

$g (x) = - \frac{3}{x} + C$

Step 15

Substitute for $g (x)$ in $f (x, y) = \frac{y}{x^{2}} + g (x)$ .

$f (x, y) = \frac{y}{x^{2}} - \frac{3}{x} + C$