Calculus Examples

$y'' + 3 y' + 2 y = 6$

Step 1

Rewrite the differential equation.

$\frac{d^{2} y}{d x^{2}} + 3 \frac{d y}{d x} + 2 y = 6$

Step 2

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

Step 3

Find the characteristic equation for $\frac{d^{2} y}{d x^{2}} + 3 \frac{d y}{d x} + 2 y = 6$ .

Tap for more steps...

Step 3.1

Find the first derivative.

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

Step 3.2

Find the second derivative.

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

Step 3.3

Substitute into the differential equation.

$r^{2} e^{r x} + 3 (r e^{r x}) + 2 e^{r x} = 6$

Step 3.4

Remove parentheses.

$r^{2} e^{r x} + 3 r e^{r x} + 2 e^{r x} = 6$

Step 3.5

Factor out $e^{r x}$ .

Tap for more steps...

Step 3.5.1

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

$e^{r x} r^{2} + 3 r e^{r x} + 2 e^{r x} = 6$

Step 3.5.2

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

$e^{r x} r^{2} + e^{r x} (3 r) + 2 e^{r x} = 6$

Step 3.5.3

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

$e^{r x} (r^{2} + 3 r) + 2 e^{r x} = 6$

Step 3.5.4

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

$e^{r x} (r^{2} + 3 r) + e^{r x} \cdot 2 = 6$

Step 3.5.5

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

$e^{r x} (r^{2} + 3 r + 2) = 6$

Step 3.6

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

$r^{2} + 3 r + 2 = 6$

Step 4

Solve for $r$ .

Tap for more steps...

Step 4.1

Subtract $6$ from both sides of the equation.

$r^{2} + 3 r + 2 - 6 = 0$

Step 4.2

Subtract $6$ from $2$ .

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

Step 4.3

Factor $r^{2} + 3 r - 4$ using the AC method.

Tap for more steps...

Step 4.3.1

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 $- 4$ and whose sum is $3$ .

$- 1, 4$

Step 4.3.2

Write the factored form using these integers.

$(r - 1) (r + 4) = 0$

Step 4.4

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

$r - 1 = 0$

$r + 4 = 0$

Step 4.5

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

Tap for more steps...

Step 4.5.1

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

$r - 1 = 0$

Step 4.5.2

Add $1$ to both sides of the equation.

$r = 1$

Step 4.6

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

Tap for more steps...

Step 4.6.1

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

$r + 4 = 0$

Step 4.6.2

Subtract $4$ from both sides of the equation.

$r = - 4$

Step 4.7

The final solution is all the values that make $(r - 1) (r + 4) = 0$ true.

$r = 1, - 4$

Step 5

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

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

$y_{2} = e^{- 4 x}$

Step 6

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

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

Step 7

Multiply $x$ by $1$ .

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