Calculus Examples

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

Step 1

Rewrite the equation.

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

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Integrate the left side.

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Step 2.2.1

Split the single integral into multiple integrals.

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

Step 2.2.2

By the Power Rule, the integral of $y^{2}$ with respect to $y$ is $\frac{1}{3} y^{3}$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Apply the constant rule.

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

Step 2.2.6

Simplify.

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Step 2.2.6.1

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

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

Step 2.2.6.2

Simplify.

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

Step 2.2.6.3

Reorder terms.

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

Step 2.2.7

Reorder terms.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

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

Step 2.3.2

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

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

Step 2.3.3

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

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

Step 2.3.4

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

$2 y^{2} + \frac{1}{3} y^{3} - y + C_{4} = \frac{1}{3} x^{3} + C_{5} - 2 (\frac{1}{2} x^{2} + C_{6}) + \int 3 d x$

Step 2.3.5

Apply the constant rule.

$2 y^{2} + \frac{1}{3} y^{3} - y + C_{4} = \frac{1}{3} x^{3} + C_{5} - 2 (\frac{1}{2} x^{2} + C_{6}) + 3 x + C_{7}$

Step 2.3.6

Simplify.

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Step 2.3.6.1

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

$2 y^{2} + \frac{1}{3} y^{3} - y + C_{4} = \frac{1}{3} x^{3} + C_{5} - 2 (\frac{x^{2}}{2} + C_{6}) + 3 x + C_{7}$

Step 2.3.6.2

Simplify.

$2 y^{2} + \frac{1}{3} y^{3} - y + C_{4} = \frac{x^{3}}{3} - x^{2} + 3 x + C_{8}$

Step 2.3.6.3

Reorder terms.

$2 y^{2} + \frac{1}{3} y^{3} - y + C_{4} = \frac{1}{3} x^{3} - x^{2} + 3 x + C_{8}$

Step 2.3.7

Reorder terms.

$2 y^{2} + \frac{1}{3} y^{3} - y + C_{4} = - x^{2} + \frac{1}{3} x^{3} + 3 x + C_{8}$

Step 2.4

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

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