Calculus Examples

3 y^{2} t d y + (y^{3} + 2 t) d t = 0

$3 y^{2} t d y + (y^{3} + 2 t) d t = 0$

Step 1

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

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Step 1.1

Rewrite.

$(y^{3} + 2 t) d t + 3 y^{2} t d y = 0$

Step 2

Find

$\frac{\partial M}{\partial y}$ where

$M (x, y) = y^{3} + 2 t$ .

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Step 2.1

Differentiate

$M$ with respect to

$y$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [y^{3} + 2 t]$

Step 2.2

By the Sum Rule, the derivative of

$y^{3} + 2 t$ with respect to

$y$ is

$\frac{d}{d y} [y^{3}] + \frac{d}{d y} [2 t]$ .

$\frac{\partial M}{\partial y} = \frac{d}{d y} [y^{3}] + \frac{d}{d y} [2 t]$

Step 2.3

Differentiate using the Power Rule which states that

$\frac{d}{d y} [y^{n}]$ is

$n y^{n - 1}$ where

$n = 3$ .

$\frac{\partial M}{\partial y} = 3 y^{2} + \frac{d}{d y} [2 t]$

Step 2.4

Since

$2 t$ is constant with respect to

$y$ , the derivative of

$2 t$ with respect to

$y$ is

$0$ .

$\frac{\partial M}{\partial y} = 3 y^{2} + 0$

Step 2.5

Add

$3 y^{2}$ and

$0$ .

$\frac{\partial M}{\partial y} = 3 y^{2}$

Step 3

Find

$\frac{\partial N}{\partial x}$ where

$N (x, y) = 3 y^{2} t$ .

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Step 3.1

Differentiate

$N$ with respect to

$x$ .

$\frac{\partial N}{\partial x} = \frac{d}{d x} [3 y^{2} t]$

Step 3.2

Since

$3 y^{2} t$ is constant with respect to

$x$ , the derivative of

$3 y^{2} t$ with respect to

$x$ is

$0$ .

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

Step 4

Check that

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

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Step 4.1

Substitute

$3 y^{2}$ for

$\frac{\partial M}{\partial y}$ and

$0$ for

$\frac{\partial N}{\partial x}$ .

$3 y^{2} = 0$

Step 4.2

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

$3 y^{2} = 0$ 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

$3 y^{2}$ for

$\frac{\partial M}{\partial y}$ .

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

Step 5.2

Substitute

$0$ for

$\frac{\partial N}{\partial x}$ .

$\frac{3 y^{2} + 0}{N}$

Step 5.3

Substitute

$3 y^{2} t$ for

$N$ .

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Step 5.3.1

Substitute

$3 y^{2} t$ for

$N$ .

$\frac{3 y^{2} + 0}{3 y^{2} t}$

Step 5.3.2

Cancel the common factor of

$3 y^{2} + 0$ and

$3$ .

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Step 5.3.2.1

Factor

$3$ out of

$3 y^{2}$ .

$\frac{3 (y^{2}) + 0}{3 y^{2} t}$

Step 5.3.2.2

Factor

$3$ out of

$0$ .

$\frac{3 (y^{2}) + 3 \cdot 0}{3 y^{2} t}$

Step 5.3.2.3

Factor

$3$ out of

$3 (y^{2}) + 3 (0)$ .

$\frac{3 (y^{2} + 0)}{3 y^{2} t}$

Step 5.3.2.4

Cancel the common factors.

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Step 5.3.2.4.1

Factor

$3$ out of

$3 y^{2} t$ .

$\frac{3 (y^{2} + 0)}{3 (y^{2} t)}$

Step 5.3.2.4.2

Cancel the common factor.

$\frac{3 (y^{2} + 0)}{3 (y^{2} t)}$

Step 5.3.2.4.3

Rewrite the expression.

$\frac{y^{2} + 0}{y^{2} t}$

Step 5.3.3

Add

$y^{2}$ and

$0$ .

$\frac{y^{2}}{y^{2} t}$

Step 5.3.4

Cancel the common factor of

$y^{2}$ .

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Step 5.3.4.1

Cancel the common factor.

$\frac{y^{2}}{y^{2} t}$

Step 5.3.4.2

Rewrite the expression.

$\frac{1}{t}$

Step 5.4

Find the integration factor

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

$μ (x, y) = e^{\int \frac{1}{t} d x}$

Step 6

Evaluate the integral

$e^{\int \frac{1}{t} d x}$ .

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Step 6.1

Apply the constant rule.

$μ (x, y) = e^{\frac{1}{t} x + C}$

Step 6.2

Simplify the answer.

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Step 6.2.1

Combine

$\frac{1}{t}$ and

$x$ .

$μ (x, y) = e^{\frac{x}{t} + C}$

Step 6.2.2

Simplify.

$μ (x, y) = e^{\frac{x}{t}} + C$

Step 7

Multiply both sides of

$(y^{3} + 2 t) d t + 3 y^{2} t d y = 0$ by the integration factor

$e^{\frac{x}{t}}$ .

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Step 7.1

Multiply

$y^{3} + 2 t$ by

$e^{\frac{x}{t}}$ .

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

Step 7.2

Apply the distributive property.

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

Step 7.3

Multiply

$3 y^{2} t$ by

$e^{\frac{x}{t}}$ .

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

Step 8

Set

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

$N (x, y)$ .

$f (x, y) = \int 3 y^{2} t e^{\frac{x}{t}} d y$

Step 9

Integrate

$N (x, y) = 3 y^{2} t e^{\frac{x}{t}}$ to find

$f (x, y)$ .

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Step 9.1

Since

$3 t e^{\frac{x}{t}}$ is constant with respect to

$y$ , move

$3 t e^{\frac{x}{t}}$ out of the integral.

$f (x, y) = 3 t e^{\frac{x}{t}} \int y^{2} d y$

Step 9.2

By the Power Rule, the integral of

$y^{2}$ with respect to

$y$ is

$\frac{1}{3} y^{3}$ .

$f (x, y) = 3 t e^{\frac{x}{t}} (\frac{1}{3} y^{3} + C)$

Step 9.3

Simplify the answer.

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Step 9.3.1

Rewrite

$3 t e^{\frac{x}{t}} (\frac{1}{3} y^{3} + C)$ as

$3 t e^{\frac{x}{t}} \frac{1}{3} y^{3} + C$ .

$f (x, y) = 3 t e^{\frac{x}{t}} \frac{1}{3} y^{3} + C$

Step 9.3.2

Simplify.

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Step 9.3.2.1

Combine

$3$ and

$\frac{1}{3}$ .

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

Step 9.3.2.2

Combine

$\frac{3}{3}$ and

$t$ .

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

Step 9.3.2.3

Combine

$\frac{3 t}{3}$ and

$e^{\frac{x}{t}}$ .

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

Step 9.3.2.4

Cancel the common factor of

$3$ .

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Step 9.3.2.4.1

Cancel the common factor.

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

Step 9.3.2.4.2

Divide

$t e^{\frac{x}{t}}$ by

$1$ .

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

Step 10

Since the integral of

$g (x)$ will contain an integration constant, we can replace

$C$ with

$g (x)$ .

$f (x, y) = t e^{\frac{x}{t}} y^{3} + g (x)$

Step 11

Set

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

$\frac{\partial f}{\partial x} = y^{3} e^{\frac{x}{t}} + 2 t e^{\frac{x}{t}}$

Step 12

Find

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

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Step 12.1

Differentiate

$f$ with respect to

$x$ .

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

Step 12.2

By the Sum Rule, the derivative of

$t e^{\frac{x}{t}} y^{3} + g (x)$ with respect to

$x$ is

$\frac{d}{d x} [t e^{\frac{x}{t}} y^{3}] + \frac{d}{d x} [g (x)]$ .

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

Step 12.3

Evaluate

$\frac{d}{d x} [t e^{\frac{x}{t}} y^{3}]$ .

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Step 12.3.1

Since

$t y^{3}$ is constant with respect to

$x$ , the derivative of

$t e^{\frac{x}{t}} y^{3}$ with respect to

$x$ is

$t y^{3} \frac{d}{d x} [e^{\frac{x}{t}}]$ .

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

Step 12.3.2

Differentiate using the chain rule, which states that

$\frac{d}{d x} [f (g (x))]$ is

$f' (g (x)) g' (x)$ where

$f (x) = e^{x}$ and

$g (x) = \frac{x}{t}$ .

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Step 12.3.2.1

To apply the Chain Rule, set

$u$ as

$\frac{x}{t}$ .

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

Step 12.3.2.2

Differentiate using the Exponential Rule which states that

$\frac{d}{d u} [a^{u}]$ is

$a^{u} \ln (a)$ where

$a$ =

$e$ .

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

Step 12.3.2.3

Replace all occurrences of

$u$ with

$\frac{x}{t}$ .

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

Step 12.3.3

Since

$\frac{1}{t}$ is constant with respect to

$x$ , the derivative of

$\frac{x}{t}$ with respect to

$x$ is

$\frac{1}{t} \frac{d}{d x} [x]$ .

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

Step 12.3.4

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

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

Step 12.3.5

Multiply

$\frac{1}{t}$ by

$1$ .

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

Step 12.3.6

Combine

$e^{\frac{x}{t}}$ and

$\frac{1}{t}$ .

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

Step 12.3.7

Combine

$t$ and

$\frac{e^{\frac{x}{t}}}{t}$ .

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

Step 12.3.8

Combine

$y^{3}$ and

$\frac{t e^{\frac{x}{t}}}{t}$ .

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

Step 12.3.9

Cancel the common factor of

$t$ .

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Step 12.3.9.1

Cancel the common factor.

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

Step 12.3.9.2

Divide

$y^{3} e^{\frac{x}{t}}$ by

$1$ .

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

Step 12.4

Differentiate using the function rule which states that the derivative of

$g (x)$ is

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

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

Step 12.5

Simplify.

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Step 12.5.1

Reorder terms.

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

Step 12.5.2

Reorder factors in

$\frac{d g}{d x} + e^{\frac{x}{t}} y^{3}$ .

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

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

$y^{3} e^{\frac{x}{t}}$ from both sides of the equation.

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

Step 13.1.2

Combine the opposite terms in

$y^{3} e^{\frac{x}{t}} + 2 t e^{\frac{x}{t}} - y^{3} e^{\frac{x}{t}}$ .

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Step 13.1.2.1

Subtract

$y^{3} e^{\frac{x}{t}}$ from

$y^{3} e^{\frac{x}{t}}$ .

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

Step 13.1.2.2

Add

$2 t e^{\frac{x}{t}}$ and

$0$ .

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

Step 14

Find the antiderivative of

$2 t e^{\frac{x}{t}}$ to find

$g (x)$ .

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Step 14.1

Integrate both sides of

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

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

Step 14.2

Evaluate

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

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

Step 14.3

Since

$2 t$ is constant with respect to

$x$ , move

$2 t$ out of the integral.

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

Step 14.4

Let

$u = \frac{x}{t}$ . Then

$d u = \frac{1}{t} d x$ , so

$t d u = d x$ . Rewrite using

$u$ and

$d$

$u$ .

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Step 14.4.1

Let

$u = \frac{x}{t}$ . Find

$\frac{d u}{d x}$ .

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Step 14.4.1.1

Differentiate

$\frac{x}{t}$ .

$\frac{d}{d x} [\frac{x}{t}]$

Step 14.4.1.2

Since

$\frac{1}{t}$ is constant with respect to

$x$ , the derivative of

$\frac{x}{t}$ with respect to

$x$ is

$\frac{1}{t} \frac{d}{d x} [x]$ .

$\frac{1}{t} \frac{d}{d x} [x]$

Step 14.4.1.3

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

$\frac{1}{t} \cdot 1$

Step 14.4.1.4

Multiply

$\frac{1}{t}$ by

$1$ .

$\frac{1}{t}$

Step 14.4.2

Rewrite the problem using

$u$ and

$d u$ .

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

Step 14.5

Simplify.

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Step 14.5.1

Multiply by the reciprocal of the fraction to divide by

$\frac{1}{t}$ .

$g (x) = 2 t \int e^{u} (1 t) d u$

Step 14.5.2

Multiply

$t$ by

$1$ .

$g (x) = 2 t \int e^{u} t d u$

Step 14.6

Since

$t$ is constant with respect to

$u$ , move

$t$ out of the integral.

$g (x) = 2 t (t \int e^{u} d u)$

Step 14.7

Simplify.

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Step 14.7.1

Raise

$t$ to the power of

$1$ .

$g (x) = 2 (t^{1} t) \int e^{u} d u$

Step 14.7.2

Raise

$t$ to the power of

$1$ .

$g (x) = 2 (t^{1} t^{1}) \int e^{u} d u$

Step 14.7.3

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 14.7.4

Add

$1$ and

$1$ .

$g (x) = 2 t^{2} \int e^{u} d u$

Step 14.8

The integral of

$e^{u}$ with respect to

$u$ is

$e^{u}$ .

$g (x) = 2 t^{2} (e^{u} + C)$

Step 14.9

Simplify.

$g (x) = 2 t^{2} e^{u} + C$

Step 14.10

Replace all occurrences of

$u$ with

$\frac{x}{t}$ .

$g (x) = 2 t^{2} e^{\frac{x}{t}} + C$

Step 15

Substitute for

$g (x)$ in

$f (x, y) = t e^{\frac{x}{t}} y^{3} + g (x)$ .

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

Step 16

Reorder factors in

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

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