Calculus Examples

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

Step 1

Rewrite the equation.

$d y = (x + 1) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int x + 1 d x$

Step 2.2

Apply the constant rule.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

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

Step 2.3.2

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

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

Step 2.3.3

Apply the constant rule.

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

Step 2.3.4

Simplify.

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

Step 2.4

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

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

Step 3

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

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

Step 4

Solve for $K$ .

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Step 4.1

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

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

Step 4.2

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

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Step 4.2.1

Cancel the common factor of $2$ .

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Step 4.2.1.1

Factor $2$ out of $2^{2}$ .

$\frac{1}{2} \cdot (2 \cdot 2) + 2 + K = 3$

Step 4.2.1.2

Cancel the common factor.

$\frac{1}{2} \cdot (2 \cdot 2) + 2 + K = 3$

Step 4.2.1.3

Rewrite the expression.

$2 + 2 + K = 3$

Step 4.2.2

Add $2$ and $2$ .

$4 + K = 3$

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.

$K = 3 - 4$

Step 4.3.2

Subtract $4$ from $3$ .

$K = - 1$

Step 5

Substitute $- 1$ for $K$ in $y = \frac{1}{2} x^{2} + x + K$ and simplify.

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Step 5.1

Substitute $- 1$ for $K$ .

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

Step 5.2

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

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