Calculus Examples

$e^{y} \frac{d y}{d x} = 4 x$

Step 1

Rewrite the equation.

$e^{y} d y = 4 x d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int e^{y} d y = \int 4 x d x$

Step 2.2

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

$e^{y} + C_{1} = \int 4 x d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

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

$e^{y} + C_{1} = 4 \int x d x$

Step 2.3.2

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

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

Step 2.3.3

Simplify the answer.

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Step 2.3.3.1

Rewrite $4 (\frac{1}{2} x^{2} + C_{2})$ as $4 (\frac{1}{2}) x^{2} + C_{2}$ .

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

Step 2.3.3.2

Simplify.

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Step 2.3.3.2.1

Combine $4$ and $\frac{1}{2}$ .

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

Step 2.3.3.2.2

Cancel the common factor of $4$ and $2$ .

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Step 2.3.3.2.2.1

Factor $2$ out of $4$ .

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

Step 2.3.3.2.2.2

Cancel the common factors.

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Step 2.3.3.2.2.2.1

Factor $2$ out of $2$ .

$e^{y} + C_{1} = \frac{2 \cdot 2}{2 (1)} x^{2} + C_{2}$

Step 2.3.3.2.2.2.2

Cancel the common factor.

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

Step 2.3.3.2.2.2.3

Rewrite the expression.

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

Step 2.3.3.2.2.2.4

Divide $2$ by $1$ .

$e^{y} + C_{1} = 2 x^{2} + C_{2}$

Step 2.4

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

$e^{y} = 2 x^{2} + K$

Step 3

Solve for $y$ .

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Step 3.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{y}) = \ln (2 x^{2} + K)$

Step 3.2

Expand the left side.

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Step 3.2.1

Expand $\ln (e^{y})$ by moving $y$ outside the logarithm.

$y \ln (e) = \ln (2 x^{2} + K)$

Step 3.2.2

The natural logarithm of $e$ is $1$ .

$y \cdot 1 = \ln (2 x^{2} + K)$

Step 3.2.3

Multiply $y$ by $1$ .

$y = \ln (2 x^{2} + K)$