Calculus Examples

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

Separate the variables.

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Divide each term in $x \frac{d y}{d x} = 4 y - 3$ by $x$ and simplify.

$\frac{d y}{d x} = \frac{4 y}{x} - \frac{3}{x}$

Combine the numerators over the common denominator.

$\frac{d y}{d x} = \frac{4 y - 3}{x}$

Multiply both sides by $\frac{1}{4 y - 3}$ .

$\frac{1}{4 y - 3} \frac{d y}{d x} = \frac{1}{4 y - 3} \cdot \frac{4 y - 3}{x}$

Cancel the common factor of $4 y - 3$ .

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Cancel the common factor.

$\frac{1}{4 y - 3} \frac{d y}{d x} = \frac{1}{4 y - 3} \cdot \frac{4 y - 3}{x}$

Rewrite the expression.

$\frac{1}{4 y - 3} \frac{d y}{d x} = \frac{1}{x}$

Rewrite the equation.

$\frac{1}{4 y - 3} d y = \frac{1}{x} d x$

Integrate both sides.

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Set up an integral on each side.

$\int \frac{1}{4 y - 3} d y = \int \frac{1}{x} d x$

Integrate the left side.

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Let $u = 4 y - 3$ . Then $d u = 4 d y$ , so $\frac{1}{4} d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Let $u = 4 y - 3$ . Find $\frac{d u}{d y}$ .

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Differentiate $4 y - 3$ .

$\frac{d}{d y} [4 y - 3]$

By the Sum Rule, the derivative of $4 y - 3$ with respect to $y$ is $\frac{d}{d y} [4 y] + \frac{d}{d y} [- 3]$ .

$\frac{d}{d y} [4 y] + \frac{d}{d y} [- 3]$

Evaluate $\frac{d}{d y} [4 y]$ .

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Since $4$ is constant with respect to $y$ , the derivative of $4 y$ with respect to $y$ is $4 \frac{d}{d y} [y]$ .

$4 \frac{d}{d y} [y] + \frac{d}{d y} [- 3]$

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 1$ .

$4 \cdot 1 + \frac{d}{d y} [- 3]$

Multiply $4$ by $1$ .

$4 + \frac{d}{d y} [- 3]$

Since $- 3$ is constant with respect to $y$ , the derivative of $- 3$ with respect to $y$ is $0$ .

$4 + 0$

Add $4$ and $0$ .

$4$

Rewrite the problem using $u$ and $d u$ .

$\int \frac{1}{u} \cdot \frac{1}{4} d u = \int \frac{1}{x} d x$

Simplify.

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Multiply $\frac{1}{u}$ by $\frac{1}{4}$ .

$\int \frac{1}{u \cdot 4} d u = \int \frac{1}{x} d x$

Move $4$ to the left of $u$ .

$\int \frac{1}{4 u} d u = \int \frac{1}{x} d x$

Since $\frac{1}{4}$ is constant with respect to $u$ , move $\frac{1}{4}$ out of the integral.

$\frac{1}{4} \int \frac{1}{u} d u = \int \frac{1}{x} d x$

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$\frac{1}{4} (\ln (| u |) + C_{1}) = \int \frac{1}{x} d x$

Simplify.

$\frac{1}{4} \ln (| u |) + C_{1} = \int \frac{1}{x} d x$

Replace all occurrences of $u$ with $4 y - 3$ .

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

The integral of $\frac{1}{x}$ with respect to $x$ is $\ln (| x |)$ .

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

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

$\frac{1}{4} \ln (| 4 y - 3 |) = \ln (| x |) + C$

Solve for $y$ .

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Multiply both sides of the equation by $4$ .

$4 (\frac{1}{4} \ln (| 4 y - 3 |)) = 4 (\ln (| x |) + C)$

Simplify both sides of the equation.

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Simplify the left side.

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Simplify $4 (\frac{1}{4} \ln (| 4 y - 3 |))$ .

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Combine $\frac{1}{4}$ and $\ln (| 4 y - 3 |)$ .

$4 \frac{\ln (| 4 y - 3 |)}{4} = 4 (\ln (| x |) + C)$

Cancel the common factor of $4$ .

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Cancel the common factor.

$4 \frac{\ln (| 4 y - 3 |)}{4} = 4 (\ln (| x |) + C)$

Rewrite the expression.

$\ln (| 4 y - 3 |) = 4 (\ln (| x |) + C)$

Simplify the right side.

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Apply the distributive property.

$\ln (| 4 y - 3 |) = 4 \ln (| x |) + 4 C$

Move all the terms containing a logarithm to the left side of the equation.

$\ln (| 4 y - 3 |) - 4 \ln (| x |) = 4 C$

Simplify the left side.

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Simplify $\ln (| 4 y - 3 |) - 4 \ln (| x |)$ .

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Simplify each term.

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Simplify $- 4 \ln (| x |)$ by moving $4$ inside the logarithm.

$\ln (| 4 y - 3 |) - \ln ({| x |}^{4}) = 4 C$

Remove the absolute value in ${| x |}^{4}$ because exponentiations with even powers are always positive.

$\ln (| 4 y - 3 |) - \ln (x^{4}) = 4 C$

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| 4 y - 3 |}{x^{4}}) = 4 C$

To solve for $y$ , rewrite the equation using properties of logarithms.

$e^{\ln (\frac{| 4 y - 3 |}{x^{4}})} = e^{4 C}$

Rewrite $\ln (\frac{| 4 y - 3 |}{x^{4}}) = 4 C$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{4 C} = \frac{| 4 y - 3 |}{x^{4}}$

Solve for $y$ .

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Rewrite the equation as $\frac{| 4 y - 3 |}{x^{4}} = e^{4 C}$ .

$\frac{| 4 y - 3 |}{x^{4}} = e^{4 C}$

Multiply both sides by $x^{4}$ .

$\frac{| 4 y - 3 |}{x^{4}} x^{4} = e^{4 C} x^{4}$

Simplify.

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Simplify the left side.

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Cancel the common factor of $x^{4}$ .

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Cancel the common factor.

$\frac{| 4 y - 3 |}{x^{4}} x^{4} = e^{4 C} x^{4}$

Rewrite the expression.

$| 4 y - 3 | = e^{4 C} x^{4}$

Simplify the right side.

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Reorder factors in $e^{4 C} x^{4}$ .

$| 4 y - 3 | = x^{4} e^{4 C}$

Solve for $y$ .

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Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$4 y - 3 = \pm x^{4} e^{4 C}$

Add $3$ to both sides of the equation.

$4 y = \pm x^{4} e^{4 C} + 3$

Divide each term in $4 y = \pm x^{4} e^{4 C} + 3$ by $4$ and simplify.

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Divide each term in $4 y = \pm x^{4} e^{4 C} + 3$ by $4$ .

$\frac{4 y}{4} = \frac{\pm x^{4} e^{4 C}}{4} + \frac{3}{4}$

Simplify the left side.

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Cancel the common factor of $4$ .

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Cancel the common factor.

$\frac{4 y}{4} = \frac{\pm x^{4} e^{4 C}}{4} + \frac{3}{4}$

Divide $y$ by $1$ .

$y = \frac{\pm x^{4} e^{4 C}}{4} + \frac{3}{4}$

Group the constant terms together.

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Simplify the constant of integration.

$y = \frac{\pm x^{4} C}{4} + \frac{3}{4}$

Combine constants with the plus or minus.

$y = \frac{C x^{4}}{4} + \frac{3}{4}$

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

$1 = \frac{C \cdot 1^{4}}{4} + \frac{3}{4}$

Solve for $C$ .

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Rewrite the equation as $\frac{C \cdot 1^{4}}{4} + \frac{3}{4} = 1$ .

$\frac{C \cdot 1^{4}}{4} + \frac{3}{4} = 1$

Simplify each term.

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One to any power is one.

$\frac{C \cdot 1}{4} + \frac{3}{4} = 1$

Multiply $C$ by $1$ .

$\frac{C}{4} + \frac{3}{4} = 1$

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

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Subtract $\frac{3}{4}$ from both sides of the equation.

$\frac{C}{4} = 1 - \frac{3}{4}$

Write $1$ as a fraction with a common denominator.

$\frac{C}{4} = \frac{4}{4} - \frac{3}{4}$

Combine the numerators over the common denominator.

$\frac{C}{4} = \frac{4 - 3}{4}$

Subtract $3$ from $4$ .

$\frac{C}{4} = \frac{1}{4}$

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$C = 1$

Substitute $1$ for $C$ in $y = \frac{C x^{4}}{4} + \frac{3}{4}$ and simplify.

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Substitute.

$y = \frac{1 x^{4}}{4} + \frac{3}{4}$

Multiply $x^{4}$ by $1$ .

$y = \frac{x^{4}}{4} + \frac{3}{4}$