Calculus Examples

$(2 x + 3) d x + (2 y - 2) d y = 0$

Step 1

Subtract $(2 x + 3) d x$ from both sides of the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int 2 y - 2 d y = \int - (2 x + 3) d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Split the single integral into multiple integrals.

$\int 2 y d y + \int - 2 d y = \int - (2 x + 3) d x$

Step 2.2.2

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

$2 \int y d y + \int - 2 d y = \int - (2 x + 3) d x$

Step 2.2.3

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

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

Step 2.2.4

Apply the constant rule.

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

Step 2.2.5

Simplify.

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Step 2.2.5.1

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

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

Step 2.2.5.2

Simplify.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Multiply $- (2 x + 3)$ .

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

Step 2.3.2

Simplify.

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Step 2.3.2.1

Multiply $2$ by $- 1$ .

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

Step 2.3.2.2

Multiply $- 1$ by $3$ .

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

Step 2.3.3

Split the single integral into multiple integrals.

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

Step 2.3.4

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

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

Step 2.3.5

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

$y^{2} - 2 y + C_{3} = - 2 (\frac{1}{2} x^{2} + C_{4}) + \int - 3 d x$

Step 2.3.6

Apply the constant rule.

$y^{2} - 2 y + C_{3} = - 2 (\frac{1}{2} x^{2} + C_{4}) - 3 x + C_{5}$

Step 2.3.7

Simplify.

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Step 2.3.7.1

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

$y^{2} - 2 y + C_{3} = - 2 (\frac{x^{2}}{2} + C_{4}) - 3 x + C_{5}$

Step 2.3.7.2

Simplify.

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

Step 2.4

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

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

Step 3

Solve for $y$ .

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Step 3.1

Move all the expressions to the left side of the equation.

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Step 3.1.1

Add $x^{2}$ to both sides of the equation.

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

Step 3.1.2

Add $3 x$ to both sides of the equation.

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

Step 3.1.3

Subtract $K$ from both sides of the equation.

$y^{2} - 2 y + x^{2} + 3 x - K = 0$

Step 3.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.3

Substitute the values $a = 1$ , $b = - 2$ , and $c = x^{2} + 3 x - K$ into the quadratic formula and solve for $y$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot (x^{2} + 3 x - K))}}{2 \cdot 1}$

Step 3.4

Simplify.

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Step 3.4.1

Simplify the numerator.

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Step 3.4.1.1

Raise $- 2$ to the power of $2$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot (x^{2} + 3 x - K)}}{2 \cdot 1}$

Step 3.4.1.2

Multiply $- 4$ by $1$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot (x^{2} + 3 x - K)}}{2 \cdot 1}$

Step 3.4.1.3

Apply the distributive property.

$y = \frac{2 \pm \sqrt{4 - 4 x^{2} - 4 (3 x) - 4 (- K)}}{2 \cdot 1}$

Step 3.4.1.4

Simplify.

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Step 3.4.1.4.1

Multiply $3$ by $- 4$ .

$y = \frac{2 \pm \sqrt{4 - 4 x^{2} - 12 x - 4 (- K)}}{2 \cdot 1}$

Step 3.4.1.4.2

Multiply $- 1$ by $- 4$ .

$y = \frac{2 \pm \sqrt{4 - 4 x^{2} - 12 x + 4 K}}{2 \cdot 1}$

Step 3.4.1.5

Factor $4$ out of $4 - 4 x^{2} - 12 x + 4 K$ .

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Step 3.4.1.5.1

Factor $4$ out of $4$ .

$y = \frac{2 \pm \sqrt{4 \cdot 1 - 4 x^{2} - 12 x + 4 K}}{2 \cdot 1}$

Step 3.4.1.5.2

Factor $4$ out of $- 4 x^{2}$ .

$y = \frac{2 \pm \sqrt{4 \cdot 1 + 4 (- x^{2}) - 12 x + 4 K}}{2 \cdot 1}$

Step 3.4.1.5.3

Factor $4$ out of $- 12 x$ .

$y = \frac{2 \pm \sqrt{4 \cdot 1 + 4 (- x^{2}) + 4 (- 3 x) + 4 K}}{2 \cdot 1}$

Step 3.4.1.5.4

Factor $4$ out of $4 \cdot 1 + 4 (- x^{2})$ .

$y = \frac{2 \pm \sqrt{4 \cdot (1 - x^{2}) + 4 (- 3 x) + 4 K}}{2 \cdot 1}$

Step 3.4.1.5.5

Factor $4$ out of $4 \cdot (1 - x^{2}) + 4 (- 3 x)$ .

$y = \frac{2 \pm \sqrt{4 \cdot (1 - x^{2} - 3 x) + 4 K}}{2 \cdot 1}$

Step 3.4.1.5.6

Factor $4$ out of $4 \cdot (1 - x^{2} - 3 x) + 4 K$ .

$y = \frac{2 \pm \sqrt{4 (1 - x^{2} - 3 x + K)}}{2 \cdot 1}$

Step 3.4.1.6

Rewrite $4 (1 - x^{2} - 3 x + K)$ as $2^{2} (1^{2} - x^{2} - 3 x + K)$ .

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Step 3.4.1.6.1

Rewrite $4$ as $2^{2}$ .

$y = \frac{2 \pm \sqrt{2^{2} (1 - x^{2} - 3 x + K)}}{2 \cdot 1}$

Step 3.4.1.6.2

Rewrite $1$ as $1^{2}$ .

$y = \frac{2 \pm \sqrt{2^{2} (1^{2} - x^{2} - 3 x + K)}}{2 \cdot 1}$

Step 3.4.1.7

Pull terms out from under the radical.

$y = \frac{2 \pm 2 \sqrt{1^{2} - x^{2} - 3 x + K}}{2 \cdot 1}$

Step 3.4.1.8

One to any power is one.

$y = \frac{2 \pm 2 \sqrt{1 - x^{2} - 3 x + K}}{2 \cdot 1}$

Step 3.4.2

Multiply $2$ by $1$ .

$y = \frac{2 \pm 2 \sqrt{1 - x^{2} - 3 x + K}}{2}$

Step 3.4.3

Simplify $\frac{2 \pm 2 \sqrt{1 - x^{2} - 3 x + K}}{2}$ .

$y = 1 \pm \sqrt{1 - x^{2} - 3 x + K}$

Step 3.5

The final answer is the combination of both solutions.

$y = 1 + \sqrt{1 - x^{2} - 3 x + K}$

$y = 1 - \sqrt{1 - x^{2} - 3 x + K}$

$y = 1 + \sqrt{1 - x^{2} - 3 x + K}$

$y = 1 - \sqrt{1 - x^{2} - 3 x + K}$