Calculus Examples

$y'''' = 2 t e^{- 6 y} - e^{- 6 y}$ and $y (0) = 2$

Step 1

Separate the variables.

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Step 1.1

Factor $e^{- 6 y}$ out of $2 t e^{- 6 y} - e^{- 6 y}$ .

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Step 1.1.1

Factor $e^{- 6 y}$ out of $2 t e^{- 6 y}$ .

$\frac{d y}{d t} = e^{- 6 y} (2 t) - e^{- 6 y}$

Step 1.1.2

Factor $e^{- 6 y}$ out of $- e^{- 6 y}$ .

$\frac{d y}{d t} = e^{- 6 y} (2 t) + e^{- 6 y} \cdot - 1$

Step 1.1.3

Factor $e^{- 6 y}$ out of $e^{- 6 y} (2 t) + e^{- 6 y} \cdot - 1$ .

$\frac{d y}{d t} = e^{- 6 y} (2 t - 1)$

Step 1.2

Multiply both sides by $\frac{1}{e^{- 6 y}}$ .

$\frac{1}{e^{- 6 y}} \frac{d y}{d t} = \frac{1}{e^{- 6 y}} (e^{- 6 y} (2 t - 1))$

Step 1.3

Cancel the common factor of $e^{- 6 y}$ .

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Step 1.3.1

Cancel the common factor.

$\frac{1}{e^{- 6 y}} \frac{d y}{d t} = \frac{1}{e^{- 6 y}} (e^{- 6 y} (2 t - 1))$

Step 1.3.2

Rewrite the expression.

$\frac{1}{e^{- 6 y}} \frac{d y}{d t} = 2 t - 1$

Step 1.4

Rewrite the equation.

$\frac{1}{e^{- 6 y}} d y = (2 t - 1) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{e^{- 6 y}} d y = \int 2 t - 1 d t$

Step 2.2

Integrate the left side.

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Step 2.2.1

Simplify the expression.

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Step 2.2.1.1

Negate the exponent of $e^{- 6 y}$ and move it out of the denominator.

$\int 1 {(e^{- 6 y})}^{- 1} d y = \int 2 t - 1 d t$

Step 2.2.1.2

Simplify.

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Step 2.2.1.2.1

Multiply the exponents in ${(e^{- 6 y})}^{- 1}$ .

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Step 2.2.1.2.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\int 1 e^{- 6 y \cdot - 1} d y = \int 2 t - 1 d t$

Step 2.2.1.2.1.2

Multiply $- 1$ by $- 6$ .

$\int 1 e^{6 y} d y = \int 2 t - 1 d t$

Step 2.2.1.2.2

Multiply $e^{6 y}$ by $1$ .

$\int e^{6 y} d y = \int 2 t - 1 d t$

Step 2.2.2

Let $u = 6 y$ . Then $d u = 6 d y$ , so $\frac{1}{6} d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.2.1

Let $u = 6 y$ . Find $\frac{d u}{d y}$ .

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Step 2.2.2.1.1

Differentiate $6 y$ .

$\frac{d}{d y} [6 y]$

Step 2.2.2.1.2

Since $6$ is constant with respect to $y$ , the derivative of $6 y$ with respect to $y$ is $6 \frac{d}{d y} [y]$ .

$6 \frac{d}{d y} [y]$

Step 2.2.2.1.3

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

$6 \cdot 1$

Step 2.2.2.1.4

Multiply $6$ by $1$ .

$6$

Step 2.2.2.2

Rewrite the problem using $u$ and $d u$ .

$\int e^{u} \frac{1}{6} d u = \int 2 t - 1 d t$

Step 2.2.3

Combine $e^{u}$ and $\frac{1}{6}$ .

$\int \frac{e^{u}}{6} d u = \int 2 t - 1 d t$

Step 2.2.4

Since $\frac{1}{6}$ is constant with respect to $u$ , move $\frac{1}{6}$ out of the integral.

$\frac{1}{6} \int e^{u} d u = \int 2 t - 1 d t$

Step 2.2.5

The integral of $e^{u}$ with respect to $u$ is $e^{u}$ .

$\frac{1}{6} (e^{u} + C_{1}) = \int 2 t - 1 d t$

Step 2.2.6

Simplify.

$\frac{1}{6} e^{u} + C_{1} = \int 2 t - 1 d t$

Step 2.2.7

Replace all occurrences of $u$ with $6 y$ .

$\frac{1}{6} e^{6 y} + C_{1} = \int 2 t - 1 d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$\frac{1}{6} e^{6 y} + C_{1} = \int 2 t d t + \int - 1 d t$

Step 2.3.2

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

$\frac{1}{6} e^{6 y} + C_{1} = 2 \int t d t + \int - 1 d t$

Step 2.3.3

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

$\frac{1}{6} e^{6 y} + C_{1} = 2 (\frac{1}{2} t^{2} + C_{2}) + \int - 1 d t$

Step 2.3.4

Apply the constant rule.

$\frac{1}{6} e^{6 y} + C_{1} = 2 (\frac{1}{2} t^{2} + C_{2}) - t + C_{3}$

Step 2.3.5

Simplify.

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Step 2.3.5.1

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

$\frac{1}{6} e^{6 y} + C_{1} = 2 (\frac{t^{2}}{2} + C_{2}) - t + C_{3}$

Step 2.3.5.2

Simplify.

$\frac{1}{6} e^{6 y} + C_{1} = t^{2} - t + C_{4}$

Step 2.4

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

$\frac{1}{6} e^{6 y} = t^{2} - t + K$

Step 3

Solve for $y$ .

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Step 3.1

Multiply both sides of the equation by $6$ .

$6 (\frac{1}{6} e^{6 y}) = 6 (t^{2} - t + K)$

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $6 (\frac{1}{6} e^{6 y})$ .

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Step 3.2.1.1.1

Combine $\frac{1}{6}$ and $e^{6 y}$ .

$6 \frac{e^{6 y}}{6} = 6 (t^{2} - t + K)$

Step 3.2.1.1.2

Cancel the common factor of $6$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

$6 \frac{e^{6 y}}{6} = 6 (t^{2} - t + K)$

Step 3.2.1.1.2.2

Rewrite the expression.

$e^{6 y} = 6 (t^{2} - t + K)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $6 (t^{2} - t + K)$ .

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Step 3.2.2.1.1

Apply the distributive property.

$e^{6 y} = 6 t^{2} + 6 (- t) + 6 K$

Step 3.2.2.1.2

Multiply $- 1$ by $6$ .

$e^{6 y} = 6 t^{2} - 6 t + 6 K$

Step 3.3

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{6 y}) = \ln (6 t^{2} - 6 t + 6 K)$

Step 3.4

Expand the left side.

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Step 3.4.1

Expand $\ln (e^{6 y})$ by moving $6 y$ outside the logarithm.

$6 y \ln (e) = \ln (6 t^{2} - 6 t + 6 K)$

Step 3.4.2

The natural logarithm of $e$ is $1$ .

$6 y \cdot 1 = \ln (6 t^{2} - 6 t + 6 K)$

Step 3.4.3

Multiply $6$ by $1$ .

$6 y = \ln (6 t^{2} - 6 t + 6 K)$

Step 3.5

Divide each term in $6 y = \ln (6 t^{2} - 6 t + 6 K)$ by $6$ and simplify.

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Step 3.5.1

Divide each term in $6 y = \ln (6 t^{2} - 6 t + 6 K)$ by $6$ .

$\frac{6 y}{6} = \frac{\ln (6 t^{2} - 6 t + 6 K)}{6}$

Step 3.5.2

Simplify the left side.

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Step 3.5.2.1

Cancel the common factor of $6$ .

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Step 3.5.2.1.1

Cancel the common factor.

$\frac{6 y}{6} = \frac{\ln (6 t^{2} - 6 t + 6 K)}{6}$

Step 3.5.2.1.2

Divide $y$ by $1$ .

$y = \frac{\ln (6 t^{2} - 6 t + 6 K)}{6}$

Step 4

Simplify the constant of integration.

$y = \frac{\ln (6 t^{2} - 6 t + K)}{6}$

Step 5

Use the initial condition to find the value of $K$ by substituting $0$ for $t$ and $2$ for $y$ in $y = \frac{\ln (6 t^{2} - 6 t + K)}{6}$ .

$2 = \frac{\ln (6 \cdot 0^{2} - 6 \cdot 0 + K)}{6}$

Step 6

Solve for $K$ .

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Step 6.1

Rewrite the equation as $\frac{\ln (6 \cdot 0^{2} - 6 \cdot 0 + K)}{6} = 2$ .

$\frac{\ln (6 \cdot 0^{2} - 6 \cdot 0 + K)}{6} = 2$

Step 6.2

Multiply both sides of the equation by $6$ .

$6 \frac{\ln (6 \cdot 0^{2} - 6 \cdot 0 + K)}{6} = 6 \cdot 2$

Step 6.3

Simplify both sides of the equation.

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Step 6.3.1

Simplify the left side.

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Step 6.3.1.1

Simplify $6 \frac{\ln (6 \cdot 0^{2} - 6 \cdot 0 + K)}{6}$ .

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Step 6.3.1.1.1

Cancel the common factor of $6$ .

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Step 6.3.1.1.1.1

Cancel the common factor.

$6 \frac{\ln (6 \cdot 0^{2} - 6 \cdot 0 + K)}{6} = 6 \cdot 2$

Step 6.3.1.1.1.2

Rewrite the expression.

$\ln (6 \cdot 0^{2} - 6 \cdot 0 + K) = 6 \cdot 2$

Step 6.3.1.1.2

Simplify each term.

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Step 6.3.1.1.2.1

Raising $0$ to any positive power yields $0$ .

$\ln (6 \cdot 0 - 6 \cdot 0 + K) = 6 \cdot 2$

Step 6.3.1.1.2.2

Multiply $6$ by $0$ .

$\ln (0 - 6 \cdot 0 + K) = 6 \cdot 2$

Step 6.3.1.1.2.3

Multiply $- 6$ by $0$ .

$\ln (0 + 0 + K) = 6 \cdot 2$

Step 6.3.1.1.3

Combine the opposite terms in $0 + 0 + K$ .

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Step 6.3.1.1.3.1

Add $0$ and $0$ .

$\ln (0 + K) = 6 \cdot 2$

Step 6.3.1.1.3.2

Add $0$ and $K$ .

$\ln (K) = 6 \cdot 2$

Step 6.3.2

Simplify the right side.

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Step 6.3.2.1

Multiply $6$ by $2$ .

$\ln (K) = 12$

Step 6.4

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

$e^{\ln (K)} = e^{12}$

Step 6.5

Rewrite $\ln (K) = 12$ 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 $b^{y} = x$ .

$e^{12} = K$

Step 6.6

Rewrite the equation as $K = e^{12}$ .

$K = e^{12}$

Step 7

Substitute $e^{12}$ for $K$ in $y = \frac{\ln (6 t^{2} - 6 t + K)}{6}$ and simplify.

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Step 7.1

Substitute $e^{12}$ for $K$ .

$y = \frac{\ln (6 t^{2} - 6 t + e^{12})}{6}$

Step 7.2

Rewrite $\frac{\ln (6 t^{2} - 6 t + e^{12})}{6}$ as $\frac{1}{6} \ln (6 t^{2} - 6 t + e^{12})$ .

$y = \frac{1}{6} \ln (6 t^{2} - 6 t + e^{12})$

Step 7.3

Simplify $\frac{1}{6} \ln (6 t^{2} - 6 t + e^{12})$ by moving $\frac{1}{6}$ inside the logarithm.

$y = \ln ({(6 t^{2} - 6 t + e^{12})}^{\frac{1}{6}})$