Calculus Examples

$(4 t e^{2 x}) d y = y e^{2 x} d x$

Step 1

Multiply both sides by $\frac{1}{e^{2 x} y}$ .

$\frac{1}{e^{2 x} y} (4 t e^{2 x}) d y = \frac{1}{e^{2 x} y} (y e^{2 x}) d x$

Step 2

Simplify.

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Step 2.1

Rewrite using the commutative property of multiplication.

$4 \frac{1}{e^{2 x} y} (t e^{2 x}) d y = \frac{1}{e^{2 x} y} (y e^{2 x}) d x$

Step 2.2

Combine $4$ and $\frac{1}{e^{2 x} y}$ .

$\frac{4}{e^{2 x} y} (t e^{2 x}) d y = \frac{1}{e^{2 x} y} (y e^{2 x}) d x$

Step 2.3

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

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Step 2.3.1

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

$\frac{4}{e^{2 x} (y)} (t e^{2 x}) d y = \frac{1}{e^{2 x} y} (y e^{2 x}) d x$

Step 2.3.2

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

$\frac{4}{e^{2 x} (y)} (e^{2 x} t) d y = \frac{1}{e^{2 x} y} (y e^{2 x}) d x$

Step 2.3.3

Cancel the common factor.

$\frac{4}{e^{2 x} y} (e^{2 x} t) d y = \frac{1}{e^{2 x} y} (y e^{2 x}) d x$

Step 2.3.4

Rewrite the expression.

$\frac{4}{y} t d y = \frac{1}{e^{2 x} y} (y e^{2 x}) d x$

Step 2.4

Combine $\frac{4}{y}$ and $t$ .

$\frac{4 t}{y} d y = \frac{1}{e^{2 x} y} (y e^{2 x}) d x$

Step 2.5

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

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Step 2.5.1

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

$\frac{4 t}{y} d y = \frac{1}{y e^{2 x} (1)} (y e^{2 x}) d x$

Step 2.5.2

Cancel the common factor.

$\frac{4 t}{y} d y = \frac{1}{y e^{2 x} \cdot 1} (y e^{2 x}) d x$

Step 2.5.3

Rewrite the expression.

$\frac{4 t}{y} d y = 1 d x$

Step 3

Integrate both sides.

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Step 3.1

Set up an integral on each side.

$\int \frac{4 t}{y} d y = \int d x$

Step 3.2

Integrate the left side.

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Step 3.2.1

Since $4 t$ is constant with respect to $y$ , move $4 t$ out of the integral.

$4 t \int \frac{1}{y} d y = \int d x$

Step 3.2.2

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

$4 t (\ln (| y |) + C_{1}) = \int d x$

Step 3.2.3

Simplify.

$4 t \ln (| y |) + C_{1} = \int d x$

Step 3.3

Apply the constant rule.

$4 t \ln (| y |) + C_{1} = x + C_{2}$

Step 3.4

Group the constant of integration on the right side as $C$ .

$4 t \ln (| y |) = x + C$

Step 4

Solve for $y$ .

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Step 4.1

Divide each term in $4 t \ln (| y |) = x + C$ by $4 t$ and simplify.

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Step 4.1.1

Divide each term in $4 t \ln (| y |) = x + C$ by $4 t$ .

$\frac{4 t \ln (| y |)}{4 t} = \frac{x}{4 t} + \frac{C}{4 t}$

Step 4.1.2

Simplify the left side.

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Step 4.1.2.1

Cancel the common factor of $4$ .

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Step 4.1.2.1.1

Cancel the common factor.

$\frac{4 t \ln (| y |)}{4 t} = \frac{x}{4 t} + \frac{C}{4 t}$

Step 4.1.2.1.2

Rewrite the expression.

$\frac{t \ln (| y |)}{t} = \frac{x}{4 t} + \frac{C}{4 t}$

Step 4.1.2.2

Cancel the common factor of $t$ .

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Step 4.1.2.2.1

Cancel the common factor.

$\frac{t \ln (| y |)}{t} = \frac{x}{4 t} + \frac{C}{4 t}$

Step 4.1.2.2.2

Divide $\ln (| y |)$ by $1$ .

$\ln (| y |) = \frac{x}{4 t} + \frac{C}{4 t}$

Step 4.2

To solve for $y$ , rewrite the equation using properties of logarithms.

$e^{\ln (| y |)} = e^{\frac{x}{4 t} + \frac{C}{4 t}}$

Step 4.3

Rewrite $\ln (| y |) = \frac{x}{4 t} + \frac{C}{4 t}$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{\frac{x}{4 t} + \frac{C}{4 t}} = | y |$

Step 4.4

Solve for $y$ .

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Step 4.4.1

Rewrite the equation as $| y | = e^{\frac{x}{4 t} + \frac{C}{4 t}}$ .

$| y | = e^{\frac{x}{4 t} + \frac{C}{4 t}}$

Step 4.4.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$y = \pm e^{\frac{x}{4 t} + \frac{C}{4 t}}$