Calculus Examples

$\frac{d x}{d t} = 4 e^{t} - 6$ ; , $x (0) = 2$

Step 1

Rewrite the equation.

$d x = (4 e^{t} - 6) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d x = \int 4 e^{t} - 6 d t$

Step 2.2

Apply the constant rule.

$x + C_{1} = \int 4 e^{t} - 6 d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$x + C_{1} = \int 4 e^{t} d t + \int - 6 d t$

Step 2.3.2

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

$x + C_{1} = 4 \int e^{t} d t + \int - 6 d t$

Step 2.3.3

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

$x + C_{1} = 4 (e^{t} + C_{2}) + \int - 6 d t$

Step 2.3.4

Apply the constant rule.

$x + C_{1} = 4 (e^{t} + C_{2}) - 6 t + C_{3}$

Step 2.3.5

Simplify.

$x + C_{1} = 4 e^{t} - 6 t + C_{4}$

Step 2.3.6

Reorder terms.

$x + C_{1} = - 6 t + 4 e^{t} + C_{4}$

Step 2.4

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

$x = - 6 t + 4 e^{t} + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $0$ for $t$ and $2$ for $x$ in $x = - 6 t + 4 e^{t} + K$ .

$2 = - 6 \cdot 0 + 4 e^{0} + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $- 6 \cdot 0 + 4 e^{0} + K = 2$ .

$- 6 \cdot 0 + 4 e^{0} + K = 2$

Step 4.2

Simplify $- 6 \cdot 0 + 4 e^{0} + K$ .

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Multiply $- 6$ by $0$ .

$0 + 4 e^{0} + K = 2$

Step 4.2.1.2

Anything raised to $0$ is $1$ .

$0 + 4 \cdot 1 + K = 2$

Step 4.2.1.3

Multiply $4$ by $1$ .

$0 + 4 + K = 2$

Step 4.2.2

Add $0$ and $4$ .

$4 + K = 2$

Step 4.3

Move all terms not containing $K$ to the right side of the equation.

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Step 4.3.1

Subtract $4$ from both sides of the equation.

$K = 2 - 4$

Step 4.3.2

Subtract $4$ from $2$ .

$K = - 2$

Step 5

Substitute $- 2$ for $K$ in $x = - 6 t + 4 e^{t} + K$ and simplify.

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Step 5.1

Substitute $- 2$ for $K$ .

$x = - 6 t + 4 e^{t} - 2$