Calculus Examples

$\frac{d^{2} y}{d x^{2}} + y = \cos (x)$

Step 1

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

Step 2

Find the characteristic equation for $\frac{d^{2} y}{d x^{2}} + y = \cos (x)$ .

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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.

$r^{2} e^{r x} + e^{r x} = \cos (x)$

Step 2.4

Factor out $e^{r x}$ .

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Step 2.4.1

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

$e^{r x} r^{2} + e^{r x} = \cos (x)$

Step 2.4.2

Multiply by $1$ .

$e^{r x} r^{2} + e^{r x} \cdot 1 = \cos (x)$

Step 2.4.3

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

$e^{r x} (r^{2} + 1) = \cos (x)$

Step 2.5

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

$r^{2} + 1 = \cos (x)$

Step 3

Solve for $r$ .

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Step 3.1

Subtract $1$ from both sides of the equation.

$r^{2} = \cos (x) - 1$

Step 3.2

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

$r = \pm \sqrt{\cos (x) - 1}$

Step 3.3

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

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Step 3.3.1

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

$r = \sqrt{\cos (x) - 1}$

Step 3.3.2

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

$r = - \sqrt{\cos (x) - 1}$

Step 3.3.3

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

$r = \sqrt{\cos (x) - 1}$

$r = - \sqrt{\cos (x) - 1}$

$r = \sqrt{\cos (x) - 1}$

$r = - \sqrt{\cos (x) - 1}$

$r = \sqrt{\cos (x) - 1}$

$r = - \sqrt{\cos (x) - 1}$

Step 4

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

$y_{1} = e^{\sqrt{\cos (x) - 1} x}$

$y_{2} = e^{- \sqrt{\cos (x) - 1} x}$

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.

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