Calculus Examples

$3 \frac{d^{2} y}{d t^{2}} + 4 \frac{d y}{d t} - y = 0$

Step 1

Assume all solutions are of the form $y = e^{r t}$ .

Step 2

Find the characteristic equation for $3 \frac{d^{2} y}{d t^{2}} + 4 \frac{d y}{d t} - y = 0$ .

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Step 2.1

Find the first derivative.

$\frac{d y}{d t} = r e^{r t}$

Step 2.2

Find the second derivative.

$\frac{d^{2} y}{d t^{2}} = r^{2} e^{r t}$

Step 2.3

Substitute into the differential equation.

$3 (r^{2} e^{r t}) + 4 (r e^{r t}) - e^{r t} = 0$

Step 2.4

Remove parentheses.

$3 r^{2} e^{r t} + 4 r e^{r t} - e^{r t} = 0$

Step 2.5

Factor out $e^{r t}$ .

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Step 2.5.1

Factor $e^{r t}$ out of $3 r^{2} e^{r t}$ .

$e^{r t} (3 r^{2}) + 4 r e^{r t} - e^{r t} = 0$

Step 2.5.2

Factor $e^{r t}$ out of $4 r e^{r t}$ .

$e^{r t} (3 r^{2}) + e^{r t} (4 r) - e^{r t} = 0$

Step 2.5.3

Factor $e^{r t}$ out of $e^{r t} (3 r^{2}) + e^{r t} (4 r)$ .

$e^{r t} (3 r^{2} + 4 r) - e^{r t} = 0$

Step 2.5.4

Factor $e^{r t}$ out of $- e^{r t}$ .

$e^{r t} (3 r^{2} + 4 r) + e^{r t} \cdot - 1 = 0$

Step 2.5.5

Factor $e^{r t}$ out of $e^{r t} (3 r^{2} + 4 r) + e^{r t} \cdot - 1$ .

$e^{r t} (3 r^{2} + 4 r - 1) = 0$

Step 2.6

Since exponentials can never be zero, divide both sides by $e^{r t}$ .

$3 r^{2} + 4 r - 1 = 0$

Step 3

Solve for $r$ .

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Step 3.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.2

Substitute the values $a = 3$ , $b = 4$ , and $c = - 1$ into the quadratic formula and solve for $r$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (3 \cdot - 1)}}{2 \cdot 3}$

Step 3.3

Simplify.

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Step 3.3.1

Simplify the numerator.

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Step 3.3.1.1

Raise $4$ to the power of $2$ .

$r = \frac{- 4 \pm \sqrt{16 - 4 \cdot 3 \cdot - 1}}{2 \cdot 3}$

Step 3.3.1.2

Multiply $- 4 \cdot 3 \cdot - 1$ .

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Step 3.3.1.2.1

Multiply $- 4$ by $3$ .

$r = \frac{- 4 \pm \sqrt{16 - 12 \cdot - 1}}{2 \cdot 3}$

Step 3.3.1.2.2

Multiply $- 12$ by $- 1$ .

$r = \frac{- 4 \pm \sqrt{16 + 12}}{2 \cdot 3}$

Step 3.3.1.3

Add $16$ and $12$ .

$r = \frac{- 4 \pm \sqrt{28}}{2 \cdot 3}$

Step 3.3.1.4

Rewrite $28$ as $2^{2} \cdot 7$ .

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Step 3.3.1.4.1

Factor $4$ out of $28$ .

$r = \frac{- 4 \pm \sqrt{4 (7)}}{2 \cdot 3}$

Step 3.3.1.4.2

Rewrite $4$ as $2^{2}$ .

$r = \frac{- 4 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 3}$

Step 3.3.1.5

Pull terms out from under the radical.

$r = \frac{- 4 \pm 2 \sqrt{7}}{2 \cdot 3}$

Step 3.3.2

Multiply $2$ by $3$ .

$r = \frac{- 4 \pm 2 \sqrt{7}}{6}$

Step 3.3.3

Simplify $\frac{- 4 \pm 2 \sqrt{7}}{6}$ .

$r = \frac{- 2 \pm \sqrt{7}}{3}$

Step 3.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 3.4.1

Simplify the numerator.

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Step 3.4.1.1

Raise $4$ to the power of $2$ .

$r = \frac{- 4 \pm \sqrt{16 - 4 \cdot 3 \cdot - 1}}{2 \cdot 3}$

Step 3.4.1.2

Multiply $- 4 \cdot 3 \cdot - 1$ .

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Step 3.4.1.2.1

Multiply $- 4$ by $3$ .

$r = \frac{- 4 \pm \sqrt{16 - 12 \cdot - 1}}{2 \cdot 3}$

Step 3.4.1.2.2

Multiply $- 12$ by $- 1$ .

$r = \frac{- 4 \pm \sqrt{16 + 12}}{2 \cdot 3}$

Step 3.4.1.3

Add $16$ and $12$ .

$r = \frac{- 4 \pm \sqrt{28}}{2 \cdot 3}$

Step 3.4.1.4

Rewrite $28$ as $2^{2} \cdot 7$ .

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Step 3.4.1.4.1

Factor $4$ out of $28$ .

$r = \frac{- 4 \pm \sqrt{4 (7)}}{2 \cdot 3}$

Step 3.4.1.4.2

Rewrite $4$ as $2^{2}$ .

$r = \frac{- 4 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 3}$

Step 3.4.1.5

Pull terms out from under the radical.

$r = \frac{- 4 \pm 2 \sqrt{7}}{2 \cdot 3}$

Step 3.4.2

Multiply $2$ by $3$ .

$r = \frac{- 4 \pm 2 \sqrt{7}}{6}$

Step 3.4.3

Simplify $\frac{- 4 \pm 2 \sqrt{7}}{6}$ .

$r = \frac{- 2 \pm \sqrt{7}}{3}$

Step 3.4.4

Change the $\pm$ to $+$ .

$r = \frac{- 2 + \sqrt{7}}{3}$

Step 3.4.5

Rewrite $- 2$ as $- 1 (2)$ .

$r = \frac{- 1 \cdot 2 + \sqrt{7}}{3}$

Step 3.4.6

Factor $- 1$ out of $\sqrt{7}$ .

$r = \frac{- 1 \cdot 2 - 1 (- \sqrt{7})}{3}$

Step 3.4.7

Factor $- 1$ out of $- 1 (2) - 1 (- \sqrt{7})$ .

$r = \frac{- 1 (2 - \sqrt{7})}{3}$

Step 3.4.8

Move the negative in front of the fraction.

$r = - \frac{2 - \sqrt{7}}{3}$

Step 3.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 3.5.1

Simplify the numerator.

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Step 3.5.1.1

Raise $4$ to the power of $2$ .

$r = \frac{- 4 \pm \sqrt{16 - 4 \cdot 3 \cdot - 1}}{2 \cdot 3}$

Step 3.5.1.2

Multiply $- 4 \cdot 3 \cdot - 1$ .

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Step 3.5.1.2.1

Multiply $- 4$ by $3$ .

$r = \frac{- 4 \pm \sqrt{16 - 12 \cdot - 1}}{2 \cdot 3}$

Step 3.5.1.2.2

Multiply $- 12$ by $- 1$ .

$r = \frac{- 4 \pm \sqrt{16 + 12}}{2 \cdot 3}$

Step 3.5.1.3

Add $16$ and $12$ .

$r = \frac{- 4 \pm \sqrt{28}}{2 \cdot 3}$

Step 3.5.1.4

Rewrite $28$ as $2^{2} \cdot 7$ .

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Step 3.5.1.4.1

Factor $4$ out of $28$ .

$r = \frac{- 4 \pm \sqrt{4 (7)}}{2 \cdot 3}$

Step 3.5.1.4.2

Rewrite $4$ as $2^{2}$ .

$r = \frac{- 4 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 3}$

Step 3.5.1.5

Pull terms out from under the radical.

$r = \frac{- 4 \pm 2 \sqrt{7}}{2 \cdot 3}$

Step 3.5.2

Multiply $2$ by $3$ .

$r = \frac{- 4 \pm 2 \sqrt{7}}{6}$

Step 3.5.3

Simplify $\frac{- 4 \pm 2 \sqrt{7}}{6}$ .

$r = \frac{- 2 \pm \sqrt{7}}{3}$

Step 3.5.4

Change the $\pm$ to $-$ .

$r = \frac{- 2 - \sqrt{7}}{3}$

Step 3.5.5

Rewrite $- 2$ as $- 1 (2)$ .

$r = \frac{- 1 \cdot 2 - \sqrt{7}}{3}$

Step 3.5.6

Factor $- 1$ out of $- \sqrt{7}$ .

$r = \frac{- 1 \cdot 2 - (\sqrt{7})}{3}$

Step 3.5.7

Factor $- 1$ out of $- 1 (2) - (\sqrt{7})$ .

$r = \frac{- 1 (2 + \sqrt{7})}{3}$

Step 3.5.8

Move the negative in front of the fraction.

$r = - \frac{2 + \sqrt{7}}{3}$

Step 3.6

The final answer is the combination of both solutions.

$r = - \frac{2 - \sqrt{7}}{3}, - \frac{2 + \sqrt{7}}{3}$

Step 4

With the two found values of $r$ , two solutions can be constructed.

$y_{1} = e^{- \frac{2 - \sqrt{7}}{3} t}$

$y_{2} = e^{- \frac{2 + \sqrt{7}}{3} t}$

Step 5

By the principle of superposition, the general solution is a linear combination of the two solutions for a second order homogeneous linear differential equation.

$y = c_{1} e^{- \frac{2 - \sqrt{7}}{3} t} + c_{2} e^{- \frac{2 + \sqrt{7}}{3} t}$

Step 6

Simplify each term.

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Step 6.1

Combine $t$ and $\frac{2 - \sqrt{7}}{3}$ .

$y = c_{1} e^{- \frac{t (2 - \sqrt{7})}{3}} + c_{2} e^{- \frac{2 + \sqrt{7}}{3} t}$

Step 6.2

Combine $t$ and $\frac{2 + \sqrt{7}}{3}$ .

$y = c_{1} e^{- \frac{t (2 - \sqrt{7})}{3}} + c_{2} e^{- \frac{t (2 + \sqrt{7})}{3}}$