Calculus Examples

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

Step 1

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

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Step 1.1

Differentiate $M$ with respect to $y$ .

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

Step 1.2

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

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

Step 1.3

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

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

Step 1.4

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

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

Step 1.5

Add $0$ and $0$ .

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

Step 2

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

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Step 2.1

Differentiate $N$ with respect to $x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [2 t x - x]$

Step 2.2

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

$\frac{\partial N}{\partial x} = \frac{d}{d x} [2 t x] + \frac{d}{d x} [- x]$

Step 2.3

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

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Step 2.3.1

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

$\frac{\partial N}{\partial x} = 2 t \frac{d}{d x} [x] + \frac{d}{d x} [- x]$

Step 2.3.2

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

$\frac{\partial N}{\partial x} = 2 t \cdot 1 + \frac{d}{d x} [- x]$

Step 2.3.3

Multiply $2$ by $1$ .

$\frac{\partial N}{\partial x} = 2 t + \frac{d}{d x} [- x]$

Step 2.4

Evaluate $\frac{d}{d x} [- x]$ .

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Step 2.4.1

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

$\frac{\partial N}{\partial x} = 2 t - \frac{d}{d x} [x]$

Step 2.4.2

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

$\frac{\partial N}{\partial x} = 2 t - 1 \cdot 1$

Step 2.4.3

Multiply $- 1$ by $1$ .

$\frac{\partial N}{\partial x} = 2 t - 1$

Step 3

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

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Step 3.1

Substitute $0$ for $\frac{\partial M}{\partial y}$ and $2 t - 1$ for $\frac{\partial N}{\partial x}$ .

$0 = 2 t - 1$

Step 3.2

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

$0 = 2 t - 1$ is not an identity.

Step 4

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 4.1

Substitute $0$ for $\frac{\partial M}{\partial y}$ .

$\frac{0 - \frac{\partial N}{\partial x}}{N}$

Step 4.2

Substitute $2 t - 1$ for $\frac{\partial N}{\partial x}$ .

$\frac{0 - (2 t - 1)}{N}$

Step 4.3

Substitute $2 t x - x$ for $N$ .

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Step 4.3.1

Substitute $2 t x - x$ for $N$ .

$\frac{0 - (2 t - 1)}{2 t x - x}$

Step 4.3.2

Simplify the numerator.

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Step 4.3.2.1

Apply the distributive property.

$\frac{0 - (2 t) - - 1}{2 t x - x}$

Step 4.3.2.2

Multiply $2$ by $- 1$ .

$\frac{0 - 2 t - - 1}{2 t x - x}$

Step 4.3.2.3

Multiply $- 1$ by $- 1$ .

$\frac{0 - 2 t + 1}{2 t x - x}$

Step 4.3.2.4

Subtract $- (- 2 t + 1)$ from $0$ .

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

Step 4.3.3

Factor $x$ out of $2 t x - x$ .

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Step 4.3.3.1

Factor $x$ out of $2 t x$ .

$\frac{- 2 t + 1}{x (2 t) - x}$

Step 4.3.3.2

Factor $x$ out of $- x$ .

$\frac{- 2 t + 1}{x (2 t) + x \cdot - 1}$

Step 4.3.3.3

Factor $x$ out of $x (2 t) + x \cdot - 1$ .

$\frac{- 2 t + 1}{x (2 t - 1)}$

Step 4.3.4

Cancel the common factor of $- 2 t + 1$ and $2 t - 1$ .

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Step 4.3.4.1

Factor $- 1$ out of $- 2 t$ .

$\frac{- (2 t) + 1}{x (2 t - 1)}$

Step 4.3.4.2

Rewrite $1$ as $- 1 (- 1)$ .

$\frac{- (2 t) - 1 (- 1)}{x (2 t - 1)}$

Step 4.3.4.3

Factor $- 1$ out of $- (2 t) - 1 (- 1)$ .

$\frac{- (2 t - 1)}{x (2 t - 1)}$

Step 4.3.4.4

Rewrite $- (2 t - 1)$ as $- 1 (2 t - 1)$ .

$\frac{- 1 (2 t - 1)}{x (2 t - 1)}$

Step 4.3.4.5

Cancel the common factor.

$\frac{- 1 (2 t - 1)}{x (2 t - 1)}$

Step 4.3.4.6

Rewrite the expression.

$\frac{- 1}{x}$

Step 4.3.5

Move the negative in front of the fraction.

$- \frac{1}{x}$

Step 4.4

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 5

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

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Step 5.1

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

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

Step 5.2

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

$μ (x, y) = e^{- (\ln (| x |) + C)}$

Step 5.3

Simplify.

$μ (x, y) = e^{- \ln (| x |)} + C$

Step 5.4

Simplify each term.

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Step 5.4.1

Simplify $- \ln (| x |)$ by moving $- 1$ inside the logarithm.

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

Step 5.4.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| x |}^{- 1} + C$

Step 5.4.3

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

$μ (x, y) = \frac{1}{| x |} + C$

Step 6

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

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Step 6.1

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

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

Step 6.2

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

$\frac{t^{2} + x^{2}}{x} d t + (2 t x - x) d x = 0$

Step 6.3

Multiply $2 t x - x$ by $\frac{1}{x}$ .

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

Step 6.4

Multiply $2 t x - x$ by $\frac{1}{x}$ .

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

Step 6.5

Factor $x$ out of $2 t x - x$ .

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Step 6.5.1

Factor $x$ out of $2 t x$ .

$\frac{t^{2} + x^{2}}{x} d t + \frac{x (2 t) - x}{x} d x = 0$

Step 6.5.2

Factor $x$ out of $- x$ .

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

Step 6.5.3

Factor $x$ out of $x (2 t) + x \cdot - 1$ .

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

Step 6.6

Cancel the common factor of $x$ .

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Step 6.6.1

Cancel the common factor.

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

Step 6.6.2

Divide $2 t - 1$ by $1$ .

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

Step 7

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

$f (x, y) = \int 2 t - 1 d y$

Step 8

Integrate $N (x, y) = 2 t - 1$ to find $f (x, y)$ .

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Step 8.1

Split the single integral into multiple integrals.

$f (x, y) = \int 2 t d y + \int - 1 d y$

Step 8.2

Apply the constant rule.

$f (x, y) = 2 t y + C + \int - 1 d y$

Step 8.3

Apply the constant rule.

$f (x, y) = 2 t y + C - y + C$

Step 8.4

Simplify.

$f (x, y) = 2 t y - y + C$

Step 9

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

$f (x, y) = 2 t y - y + g (x)$

Step 10

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

$\frac{\partial f}{\partial x} = \frac{t^{2} + x^{2}}{x}$

Step 11

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

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Step 11.1

Differentiate $f$ with respect to $x$ .

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

Step 11.2

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

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

Step 11.3

Since $2 t y$ is constant with respect to $x$ , the derivative of $2 t y$ with respect to $x$ is $0$ .

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

Step 11.4

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

$0 + 0 + \frac{d}{d x} [g (x)] = \frac{t^{2} + x^{2}}{x}$

Step 11.5

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

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

Step 11.6

Combine terms.

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Step 11.6.1

Add $0$ and $0$ .

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

Step 11.6.2

Add $0$ and $\frac{d g}{d x}$ .

$\frac{d g}{d x} = \frac{t^{2} + x^{2}}{x}$

Step 12

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

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Step 12.1

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

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

Step 12.2

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

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

Step 12.3

Split the fraction into multiple fractions.

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

Step 12.4

Split the single integral into multiple integrals.

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

Step 12.5

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

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Step 12.5.1

Factor $x$ out of $x^{2}$ .

$g (x) = \int \frac{t^{2}}{x} d x + \int \frac{x \cdot x}{x} d x$

Step 12.5.2

Cancel the common factors.

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Step 12.5.2.1

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

$g (x) = \int \frac{t^{2}}{x} d x + \int \frac{x \cdot x}{x^{1}} d x$

Step 12.5.2.2

Factor $x$ out of $x^{1}$ .

$g (x) = \int \frac{t^{2}}{x} d x + \int \frac{x \cdot x}{x \cdot 1} d x$

Step 12.5.2.3

Cancel the common factor.

$g (x) = \int \frac{t^{2}}{x} d x + \int \frac{x \cdot x}{x \cdot 1} d x$

Step 12.5.2.4

Rewrite the expression.

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

Step 12.5.2.5

Divide $x$ by $1$ .

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

Step 12.6

Since $t^{2}$ is constant with respect to $x$ , move $t^{2}$ out of the integral.

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

Step 12.7

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

$g (x) = t^{2} (\ln (| x |) + C) + \int x d x$

Step 12.8

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

$g (x) = t^{2} (\ln (| x |) + C) + \frac{1}{2} x^{2} + C$

Step 12.9

Simplify.

$g (x) = t^{2} \ln (| x |) + \frac{1}{2} x^{2} + C$

Step 13

Substitute for $g (x)$ in $f (x, y) = 2 t y - y + g (x)$ .

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

Step 14

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

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