Calculus Examples

$y'' - y = 0$

Step 1

Rewrite the differential equation.

$\frac{d^{2} y}{d x^{2}} - y = 0$

Step 2

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

Step 3

Find the characteristic equation for $\frac{d^{2} y}{d x^{2}} - y = 0$ .

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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.

$r^{2} e^{r x} - e^{r x} = 0$

Step 3.4

Factor out $e^{r x}$ .

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Step 3.4.1

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

$e^{r x} r^{2} - e^{r x} = 0$

Step 3.4.2

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

$e^{r x} r^{2} + e^{r x} \cdot - 1 = 0$

Step 3.4.3

Factor $e^{r x}$ out of $e^{r x} r^{2} + e^{r x} \cdot - 1$ .

$e^{r x} (r^{2} - 1) = 0$

Step 3.5

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

$r^{2} - 1 = 0$

Step 4

Solve for $r$ .

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Step 4.1

Add $1$ to both sides of the equation.

$r^{2} = 1$

Step 4.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$r = \pm \sqrt{1}$

Step 4.3

Any root of $1$ is $1$ .

$r = \pm 1$

Step 4.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 4.4.1

First, use the positive value of the $\pm$ to find the first solution.

$r = 1$

Step 4.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$r = - 1$

Step 4.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$r = 1, - 1$

Step 5

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

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

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

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.

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

Step 7

Multiply $x$ by $1$ .

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