Calculus Examples

$x d y + (2 y - 10 x^{2}) d x = 0$

Step 1

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

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Step 1.1

Rewrite.

$(2 y - 10 x^{2}) d x + x d y = 0$

Step 2

Find $\frac{\partial M}{\partial y}$ where $M (x, y) = 2 y - 10 x^{2}$ .

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Step 2.1

Differentiate $M$ with respect to $y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 y - 10 x^{2}]$

Step 2.2

By the Sum Rule, the derivative of $2 y - 10 x^{2}$ with respect to $y$ is $\frac{d}{d y} [2 y] + \frac{d}{d y} [- 10 x^{2}]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [2 y] + \frac{d}{d y} [- 10 x^{2}]$

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

$\frac{\partial M}{\partial y} = 2 \frac{d}{d y} [y] + \frac{d}{d y} [- 10 x^{2}]$

Step 2.3.2

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

$\frac{\partial M}{\partial y} = 2 \cdot 1 + \frac{d}{d y} [- 10 x^{2}]$

Step 2.3.3

Multiply $2$ by $1$ .

$\frac{\partial M}{\partial y} = 2 + \frac{d}{d y} [- 10 x^{2}]$

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

Since $- 10 x^{2}$ is constant with respect to $y$ , the derivative of $- 10 x^{2}$ with respect to $y$ is $0$ .

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

Step 2.4.2

Add $2$ and $0$ .

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

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 $2$ .

$\frac{1}{x}$

Step 5.4

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

$μ (x, y) = e^{\int \frac{1}{x} d x}$

Step 6

Evaluate the integral $e^{\int \frac{1}{x} d x}$ .

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Step 6.1

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

$μ (x, y) = e^{\ln (| x |) + C}$

Step 6.2

Simplify the answer.

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Step 6.2.1

Simplify.

$μ (x, y) = e^{\ln (| x |)} + C$

Step 6.2.2

Exponentiation and log are inverse functions.

$μ (x, y) = | x | + C$

Step 7

Multiply both sides of $(2 y - 10 x^{2}) d x + x d y = 0$ by the integration factor $x$ .

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Step 7.1

Multiply $2 y - 10 x^{2}$ by $x$ .

$(2 y - 10 x^{2}) x d x + x d y = 0$

Step 7.2

Apply the distributive property.

$(2 y x - 10 x^{2} x) d x + x d y = 0$

Step 7.3

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 7.3.1

Move $x$ .

$(2 y x - 10 (x \cdot x^{2})) d x + x d y = 0$

Step 7.3.2

Multiply $x$ by $x^{2}$ .

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Step 7.3.2.1

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

$(2 y x - 10 (x^{1} x^{2})) d x + x d y = 0$

Step 7.3.2.2

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

$(2 y x - 10 x^{1 + 2}) d x + x d y = 0$

Step 7.3.3

Add $1$ and $2$ .

$(2 y x - 10 x^{3}) d x + x d y = 0$

Step 7.4

Multiply $x$ by $x$ .

$(2 y x - 10 x^{3}) d x + x \cdot x d y = 0$

Step 7.5

Multiply $x$ by $x$ .

$(2 y x - 10 x^{3}) d x + x^{2} d y = 0$

Step 8

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

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

Step 9

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

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Step 9.1

Apply the constant rule.

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

Step 10

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

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

Step 11

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

$\frac{\partial f}{\partial x} = 2 y x - 10 x^{3}$

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $x$ .

$\frac{d}{d x} [x^{2} y + g (x)] = 2 y x - 10 x^{3}$

Step 12.2

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

$\frac{d}{d x} [x^{2} y] + \frac{d}{d x} [g (x)] = 2 y x - 10 x^{3}$

Step 12.3

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

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Step 12.3.1

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

$y \frac{d}{d x} [x^{2}] + \frac{d}{d x} [g (x)] = 2 y x - 10 x^{3}$

Step 12.3.2

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

$y (2 x) + \frac{d}{d x} [g (x)] = 2 y x - 10 x^{3}$

Step 12.3.3

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

$2 y x + \frac{d}{d x} [g (x)] = 2 y x - 10 x^{3}$

Step 12.4

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

$2 y x + \frac{d g}{d x} = 2 y x - 10 x^{3}$

Step 12.5

Reorder terms.

$\frac{d g}{d x} + 2 x y = 2 y x - 10 x^{3}$

Step 13

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

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Step 13.1

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

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Step 13.1.1

Subtract $2 x y$ from both sides of the equation.

$\frac{d g}{d x} = 2 y x - 10 x^{3} - 2 x y$

Step 13.1.2

Combine the opposite terms in $2 y x - 10 x^{3} - 2 x y$ .

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Step 13.1.2.1

Reorder the factors in the terms $2 y x$ and $- 2 x y$ .

$\frac{d g}{d x} = 2 x y - 10 x^{3} - 2 x y$

Step 13.1.2.2

Subtract $2 x y$ from $2 x y$ .

$\frac{d g}{d x} = - 10 x^{3} + 0$

Step 13.1.2.3

Add $- 10 x^{3}$ and $0$ .

$\frac{d g}{d x} = - 10 x^{3}$

Step 14

Find the antiderivative of $- 10 x^{3}$ to find $g (x)$ .

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Step 14.1

Integrate both sides of $\frac{d g}{d x} = - 10 x^{3}$ .

$\int \frac{d g}{d x} d x = \int - 10 x^{3} d x$

Step 14.2

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

$g (x) = \int - 10 x^{3} d x$

Step 14.3

Since $- 10$ is constant with respect to $x$ , move $- 10$ out of the integral.

$g (x) = - 10 \int x^{3} d x$

Step 14.4

By the Power Rule, the integral of $x^{3}$ with respect to $x$ is $\frac{1}{4} x^{4}$ .

$g (x) = - 10 (\frac{1}{4} x^{4} + C)$

Step 14.5

Simplify the answer.

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Step 14.5.1

Rewrite $- 10 (\frac{1}{4} x^{4} + C)$ as $- 10 (\frac{1}{4}) x^{4} + C$ .

$g (x) = - 10 (\frac{1}{4}) x^{4} + C$

Step 14.5.2

Simplify.

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Step 14.5.2.1

Combine $- 10$ and $\frac{1}{4}$ .

$g (x) = \frac{- 10}{4} x^{4} + C$

Step 14.5.2.2

Cancel the common factor of $- 10$ and $4$ .

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Step 14.5.2.2.1

Factor $2$ out of $- 10$ .

$g (x) = \frac{2 (- 5)}{4} x^{4} + C$

Step 14.5.2.2.2

Cancel the common factors.

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Step 14.5.2.2.2.1

Factor $2$ out of $4$ .

$g (x) = \frac{2 \cdot - 5}{2 \cdot 2} x^{4} + C$

Step 14.5.2.2.2.2

Cancel the common factor.

$g (x) = \frac{2 \cdot - 5}{2 \cdot 2} x^{4} + C$

Step 14.5.2.2.2.3

Rewrite the expression.

$g (x) = \frac{- 5}{2} x^{4} + C$

Step 14.5.2.3

Move the negative in front of the fraction.

$g (x) = - \frac{5}{2} x^{4} + C$

Step 15

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

$f (x, y) = x^{2} y - \frac{5}{2} x^{4} + C$

Step 16

Simplify each term.

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Step 16.1

Combine $x^{4}$ and $\frac{5}{2}$ .

$f (x, y) = x^{2} y - \frac{x^{4} \cdot 5}{2} + C$

Step 16.2

Move $5$ to the left of $x^{4}$ .

$f (x, y) = x^{2} y - \frac{5 x^{4}}{2} + C$