Calculus Examples

$(t + 1) \frac{d y}{d t} + y = \ln (t)$

Step 1

Check if the left side of the equation is the result of the derivative of the term $(t + 1) y$ .

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Step 1.1

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = t + 1$ and $g (t) = y$ .

$(t + 1) \frac{d}{d t} [y] + y \frac{d}{d t} [t + 1]$

Step 1.2

Rewrite $\frac{d}{d t} [y]$ as $y'$ .

$(t + 1) y' + y \frac{d}{d t} [t + 1]$

Step 1.3

By the Sum Rule, the derivative of $t + 1$ with respect to $t$ is $\frac{d}{d t} [t] + \frac{d}{d t} [1]$ .

$(t + 1) y' + y (\frac{d}{d t} [t] + \frac{d}{d t} [1])$

Step 1.4

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

$(t + 1) y' + y (1 + \frac{d}{d t} [1])$

Step 1.5

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

$(t + 1) y' + y (1 + 0)$

Step 1.6

Add $1$ and $0$ .

$(t + 1) y' + y \cdot 1$

Step 1.7

Substitute $\frac{d y}{d t}$ for $y'$ .

$(t + 1) \frac{d y}{d t} + y \cdot 1$

Step 1.8

Reorder $y$ and $1$ .

$(t + 1) \frac{d y}{d t} + 1 \cdot y$

Step 1.9

Multiply $y$ by $1$ .

$(t + 1) \frac{d y}{d t} + y$

Step 2

Rewrite the left side as a result of differentiating a product.

$\frac{d}{d t} [(t + 1) y] = \ln (t)$

Step 3

Set up an integral on each side.

$\int \frac{d}{d t} [(t + 1) y] d t = \int \ln (t) d t$

Step 4

Integrate the left side.

$(t + 1) y = \int \ln (t) d t$

Step 5

Integrate the right side.

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Step 5.1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (t)$ and $d v = 1$ .

$(t + 1) y = \ln (t) t - \int t \frac{1}{t} d t$

Step 5.2

Simplify.

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Step 5.2.1

Combine $t$ and $\frac{1}{t}$ .

$(t + 1) y = \ln (t) t - \int \frac{t}{t} d t$

Step 5.2.2

Cancel the common factor of $t$ .

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Step 5.2.2.1

Cancel the common factor.

$(t + 1) y = \ln (t) t - \int \frac{t}{t} d t$

Step 5.2.2.2

Rewrite the expression.

$(t + 1) y = \ln (t) t - \int d t$

Step 5.3

Apply the constant rule.

$(t + 1) y = \ln (t) t - (t + C)$

Step 5.4

Simplify.

$(t + 1) y = \ln (t) t - t + C$

Step 6

Divide each term in $(t + 1) y = \ln (t) t - t + C$ by $t + 1$ and simplify.

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Step 6.1

Divide each term in $(t + 1) y = \ln (t) t - t + C$ by $t + 1$ .

$\frac{(t + 1) y}{t + 1} = \frac{\ln (t) t}{t + 1} + \frac{- t}{t + 1} + \frac{C}{t + 1}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $t + 1$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{(t + 1) y}{t + 1} = \frac{\ln (t) t}{t + 1} + \frac{- t}{t + 1} + \frac{C}{t + 1}$

Step 6.2.1.2

Divide $y$ by $1$ .

$y = \frac{\ln (t) t}{t + 1} + \frac{- t}{t + 1} + \frac{C}{t + 1}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Move the negative in front of the fraction.

$y = \frac{\ln (t) t}{t + 1} - \frac{t}{t + 1} + \frac{C}{t + 1}$

Step 6.3.2

Combine the numerators over the common denominator.

$y = \frac{\ln (t) t - t}{t + 1} + \frac{C}{t + 1}$

Step 6.3.3

Combine the numerators over the common denominator.

$y = \frac{\ln (t) t - t + C}{t + 1}$

Step 6.3.4

Reorder factors in $\ln (t) t - t + C$ .

$y = \frac{t \ln (t) - t + C}{t + 1}$