Calculus Examples

$\frac{d y}{d x} = 2 x - 3$ , $y (1) = 4$

Step 1

Rewrite the equation.

$d y = (2 x - 3) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 2 x - 3 d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int 2 x - 3 d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int 2 x d x + \int - 3 d x$

Step 2.3.2

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

$y + C_{1} = 2 \int x d x + \int - 3 d x$

Step 2.3.3

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

$y + C_{1} = 2 (\frac{1}{2} x^{2} + C_{2}) + \int - 3 d x$

Step 2.3.4

Apply the constant rule.

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

Step 2.3.5

Simplify.

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Step 2.3.5.1

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

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

Step 2.3.5.2

Simplify.

$y + C_{1} = x^{2} - 3 x + C_{4}$

Step 2.4

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

$y = x^{2} - 3 x + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $1$ for $x$ and $4$ for $y$ in $y = x^{2} - 3 x + K$ .

$4 = 1^{2} - 3 \cdot 1 + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $1^{2} - 3 \cdot 1 + K = 4$ .

$1^{2} - 3 \cdot 1 + K = 4$

Step 4.2

Simplify $1^{2} - 3 \cdot 1 + K$ .

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

One to any power is one.

$1 - 3 \cdot 1 + K = 4$

Step 4.2.1.2

Multiply $- 3$ by $1$ .

$1 - 3 + K = 4$

Step 4.2.2

Subtract $3$ from $1$ .

$- 2 + K = 4$

Step 4.3

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

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Step 4.3.1

Add $2$ to both sides of the equation.

$K = 4 + 2$

Step 4.3.2

Add $4$ and $2$ .

$K = 6$

Step 5

Substitute $6$ for $K$ in $y = x^{2} - 3 x + K$ and simplify.

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Step 5.1

Substitute $6$ for $K$ .

$y = x^{2} - 3 x + 6$