Calculus Examples

$\frac{d^{2} y}{d x^{2}} - 4 y = 4 e^{- x}$

Step 1

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

Step 2

Find the characteristic equation for $\frac{d^{2} y}{d x^{2}} - 4 y = 4 e^{- x}$ .

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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.

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

Step 2.4

Factor out $e^{r x}$ .

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Step 2.4.1

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

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

Step 2.4.2

Factor $e^{r x}$ out of $- 4 e^{r x}$ .

$e^{r x} r^{2} + e^{r x} \cdot - 4 = 4 e^{- x}$

Step 2.4.3

Factor $e^{r x}$ out of $e^{r x} r^{2} + e^{r x} \cdot - 4$ .

$e^{r x} (r^{2} - 4) = 4 e^{- x}$

Step 2.5

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

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

Step 3

Solve for $r$ .

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Step 3.1

Add $4$ to both sides of the equation.

$r^{2} = 4 e^{- x} + 4$

Step 3.2

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

$r = \pm \sqrt{4 e^{- x} + 4}$

Step 3.3

Simplify $\pm \sqrt{4 e^{- x} + 4}$ .

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Step 3.3.1

Factor $4$ out of $4 e^{- x} + 4$ .

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Step 3.3.1.1

Factor $4$ out of $4 e^{- x}$ .

$r = \pm \sqrt{4 (e^{- x}) + 4}$

Step 3.3.1.2

Factor $4$ out of $4$ .

$r = \pm \sqrt{4 (e^{- x}) + 4 (1)}$

Step 3.3.1.3

Factor $4$ out of $4 (e^{- x}) + 4 (1)$ .

$r = \pm \sqrt{4 (e^{- x} + 1)}$

Step 3.3.2

Rewrite $4 (e^{- x} + 1)$ as $2^{2} (e^{- x} + 1^{2})$ .

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Step 3.3.2.1

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

$r = \pm \sqrt{2^{2} (e^{- x} + 1)}$

Step 3.3.2.2

Rewrite $1$ as $1^{2}$ .

$r = \pm \sqrt{2^{2} (e^{- x} + 1^{2})}$

Step 3.3.3

Pull terms out from under the radical.

$r = \pm 2 \sqrt{e^{- x} + 1^{2}}$

Step 3.3.4

One to any power is one.

$r = \pm 2 \sqrt{e^{- x} + 1}$

Step 3.4

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

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Step 3.4.1

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

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

Step 3.4.2

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

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

Step 3.4.3

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

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

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

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

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

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

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

Step 4

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

$y_{1} = e^{2 \sqrt{e^{- x} + 1} x}$

$y_{2} = e^{- 2 \sqrt{e^{- x} + 1} x}$

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.

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