Calculus Examples

$\frac{d^{2} x}{d t^{2}} + 4 \frac{d x}{d t} + 5 x = 10$

Step 1

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

Step 2

Find the characteristic equation for $\frac{d^{2} x}{d t^{2}} + 4 \frac{d x}{d t} + 5 x = 10$ .

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Step 2.1

Find the first derivative.

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

Step 2.2

Find the second derivative.

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

Step 2.3

Substitute into the differential equation.

$r^{2} e^{r t} + 4 (r e^{r t}) + 5 e^{r t} = 10$

Step 2.4

Remove parentheses.

$r^{2} e^{r t} + 4 r e^{r t} + 5 e^{r t} = 10$

Step 2.5

Factor out $e^{r t}$ .

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Step 2.5.1

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

$e^{r t} r^{2} + 4 r e^{r t} + 5 e^{r t} = 10$

Step 2.5.2

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

$e^{r t} r^{2} + e^{r t} (4 r) + 5 e^{r t} = 10$

Step 2.5.3

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

$e^{r t} (r^{2} + 4 r) + 5 e^{r t} = 10$

Step 2.5.4

Factor $e^{r t}$ out of $5 e^{r t}$ .

$e^{r t} (r^{2} + 4 r) + e^{r t} \cdot 5 = 10$

Step 2.5.5

Factor $e^{r t}$ out of $e^{r t} (r^{2} + 4 r) + e^{r t} \cdot 5$ .

$e^{r t} (r^{2} + 4 r + 5) = 10$

Step 2.6

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

$r^{2} + 4 r + 5 = 10$

Step 3

Solve for $r$ .

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Step 3.1

Subtract $10$ from both sides of the equation.

$r^{2} + 4 r + 5 - 10 = 0$

Step 3.2

Subtract $10$ from $5$ .

$r^{2} + 4 r - 5 = 0$

Step 3.3

Factor $r^{2} + 4 r - 5$ using the AC method.

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Step 3.3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 5$ and whose sum is $4$ .

$- 1, 5$

Step 3.3.2

Write the factored form using these integers.

$(r - 1) (r + 5) = 0$

Step 3.4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$r - 1 = 0$

$r + 5 = 0$

Step 3.5

Set $r - 1$ equal to $0$ and solve for $r$ .

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Step 3.5.1

Set $r - 1$ equal to $0$ .

$r - 1 = 0$

Step 3.5.2

Add $1$ to both sides of the equation.

$r = 1$

Step 3.6

Set $r + 5$ equal to $0$ and solve for $r$ .

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Step 3.6.1

Set $r + 5$ equal to $0$ .

$r + 5 = 0$

Step 3.6.2

Subtract $5$ from both sides of the equation.

$r = - 5$

Step 3.7

The final solution is all the values that make $(r - 1) (r + 5) = 0$ true.

$r = 1, - 5$

Step 4

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

$x_{1} = e^{1 t}$

$x_{2} = e^{- 5 t}$

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.

$x = c_{1} e^{1 t} + c_{2} e^{- 5 t}$

Step 6

Multiply $t$ by $1$ .

$x = c_{1} e^{t} + c_{2} e^{- 5 t}$