Calculus Examples

$\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} - 3 y = 5$

Step 1

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

Step 2

Find the characteristic equation for $\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} - 3 y = 5$ .

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Step 2.1

Find the first derivative.

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

Step 2.2

Find the second derivative.

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

Step 2.3

Substitute into the differential equation.

$r^{2} e^{r x} - 2 (r e^{r x}) - 3 e^{r x} = 5$

Step 2.4

Remove parentheses.

$r^{2} e^{r x} - 2 r e^{r x} - 3 e^{r x} = 5$

Step 2.5

Factor out $e^{r x}$ .

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Step 2.5.1

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

$e^{r x} r^{2} - 2 r e^{r x} - 3 e^{r x} = 5$

Step 2.5.2

Factor $e^{r x}$ out of $- 2 r e^{r x}$ .

$e^{r x} r^{2} + e^{r x} (- 2 r) - 3 e^{r x} = 5$

Step 2.5.3

Factor $e^{r x}$ out of $e^{r x} r^{2} + e^{r x} (- 2 r)$ .

$e^{r x} (r^{2} - 2 r) - 3 e^{r x} = 5$

Step 2.5.4

Factor $e^{r x}$ out of $- 3 e^{r x}$ .

$e^{r x} (r^{2} - 2 r) + e^{r x} \cdot - 3 = 5$

Step 2.5.5

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

$e^{r x} (r^{2} - 2 r - 3) = 5$

Step 2.6

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

$r^{2} - 2 r - 3 = 5$

Step 3

Solve for $r$ .

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Step 3.1

Subtract $5$ from both sides of the equation.

$r^{2} - 2 r - 3 - 5 = 0$

Step 3.2

Subtract $5$ from $- 3$ .

$r^{2} - 2 r - 8 = 0$

Step 3.3

Factor $r^{2} - 2 r - 8$ 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 $- 8$ and whose sum is $- 2$ .

$- 4, 2$

Step 3.3.2

Write the factored form using these integers.

$(r - 4) (r + 2) = 0$

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 $0$ .

$r - 4 = 0$

$r + 2 = 0$

Step 3.5

Set $r - 4$ equal to $0$ and solve for $r$ .

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Step 3.5.1

Set $r - 4$ equal to $0$ .

$r - 4 = 0$

Step 3.5.2

Add $4$ to both sides of the equation.

$r = 4$

Step 3.6

Set $r + 2$ equal to $0$ and solve for $r$ .

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Step 3.6.1

Set $r + 2$ equal to $0$ .

$r + 2 = 0$

Step 3.6.2

Subtract $2$ from both sides of the equation.

$r = - 2$

Step 3.7

The final solution is all the values that make $(r - 4) (r + 2) = 0$ true.

$r = 4, - 2$

Step 4

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

$y_{1} = e^{4 x}$

$y_{2} = e^{- 2 x}$

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.

$y = c_{1} e^{4 x} + c_{2} e^{- 2 x}$