Calculus Examples

$(2 y + t) d y + y d t = 0$

Step 1

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

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Step 1.1

Rewrite.

$y d t + (2 y + t) d y = 0$

Step 2

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

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Step 2.1

Differentiate $M$ with respect to $y$ .

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

Step 2.2

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

$\frac{\partial M}{\partial y} = 1$

Step 3

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

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Step 3.1

Differentiate $N$ with respect to $x$ .

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

Step 3.2

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

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

Step 3.3

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

$\frac{\partial N}{\partial x} = 0 + \frac{d}{d x} [t]$

Step 3.4

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

$\frac{\partial N}{\partial x} = 0 + 0$

Step 3.5

Add $0$ and $0$ .

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

Step 4

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

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Step 4.1

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

$1 = 0$

Step 4.2

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

$1 = 0$ is not an identity.

Step 5

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

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Step 5.1

Substitute $0$ for $\frac{\partial N}{\partial x}$ .

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

Step 5.2

Substitute $1$ for $\frac{\partial M}{\partial y}$ .

$\frac{0 - 1}{M}$

Step 5.3

Substitute $y$ for $M$ .

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Step 5.3.1

Substitute $y$ for $M$ .

$\frac{0 - 1}{y}$

Step 5.3.2

Subtract $1$ from $0$ .

$\frac{- 1}{y}$

Step 5.3.3

Substitute $y$ for $M$ .

$- \frac{1}{y}$

Step 5.4

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

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

Step 6

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

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Step 6.1

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

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

Step 6.2

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

$μ (x, y) = e^{- (\ln (| y |) + C)}$

Step 6.3

Simplify.

$μ (x, y) = e^{- \ln (| y |)} + C$

Step 6.4

Simplify each term.

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Step 6.4.1

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

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

Step 6.4.2

Exponentiation and log are inverse functions.

$μ (x, y) = {| y |}^{- 1} + C$

Step 6.4.3

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

$μ (x, y) = \frac{1}{| y |} + C$

Step 7

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

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Step 7.1

Multiply $y$ by $\frac{1}{y}$ .

$y \frac{1}{y} d t + (2 y + t) d y = 0$

Step 7.2

Cancel the common factor of $y$ .

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Step 7.2.1

Cancel the common factor.

$y \frac{1}{y} d t + (2 y + t) d y = 0$

Step 7.2.2

Rewrite the expression.

$1 d t + (2 y + t) d y = 0$

Step 7.3

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

$1 d t + (2 y + t) \frac{1}{y} d y = 0$

Step 7.4

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

$1 d t + \frac{2 y + t}{y} d y = 0$

Step 8

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

$f (x, y) = \int d x$

Step 9

Integrate $M (x, y) = 1$ to find $f (x, y)$ .

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Step 9.1

Apply the constant rule.

$f (x, y) = x + C$

Step 10

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

$f (x, y) = x + g (y)$

Step 11

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

$\frac{\partial f}{\partial y} = \frac{2 y + t}{y}$

Step 12

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

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Step 12.1

Differentiate $f$ with respect to $y$ .

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

Step 12.2

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

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

Step 12.3

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

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

Step 12.4

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

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

Step 12.5

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

$\frac{d g}{d y} = \frac{2 y + t}{y}$

Step 13

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

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Step 13.1

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

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

Step 13.2

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

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

Step 13.3

Split the fraction into multiple fractions.

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

Step 13.4

Split the single integral into multiple integrals.

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

Step 13.5

Cancel the common factor of $y$ .

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Step 13.5.1

Cancel the common factor.

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

Step 13.5.2

Divide $2$ by $1$ .

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

Step 13.6

Apply the constant rule.

$g (y) = 2 y + C + \int \frac{t}{y} d y$

Step 13.7

Since $t$ is constant with respect to $y$ , move $t$ out of the integral.

$g (y) = 2 y + C + t \int \frac{1}{y} d y$

Step 13.8

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

$g (y) = 2 y + C + t (\ln (| y |) + C)$

Step 13.9

Simplify.

$g (y) = 2 y + t \ln (| y |) + C$

Step 14

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

$f (x, y) = x + 2 y + t \ln (| y |) + C$