Calculus Examples

$\frac{d^{2} y}{d x^{2}} - 3 \frac{d y}{d x} + 2 y = 4 x^{2}$

Step 1

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

Step 2

Find the characteristic equation for $\frac{d^{2} y}{d x^{2}} - 3 \frac{d y}{d x} + 2 y = 4 x^{2}$ .

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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.

$r^{2} e^{r x} - 3 (r e^{r x}) + 2 e^{r x} = 4 x^{2}$

Step 2.4

Remove parentheses.

$r^{2} e^{r x} - 3 r e^{r x} + 2 e^{r x} = 4 x^{2}$

Step 2.5

Factor out $e^{r x}$ .

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Step 2.5.1

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

$e^{r x} r^{2} - 3 r e^{r x} + 2 e^{r x} = 4 x^{2}$

Step 2.5.2

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

$e^{r x} r^{2} + e^{r x} (- 3 r) + 2 e^{r x} = 4 x^{2}$

Step 2.5.3

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

$e^{r x} (r^{2} - 3 r) + 2 e^{r x} = 4 x^{2}$

Step 2.5.4

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

$e^{r x} (r^{2} - 3 r) + e^{r x} \cdot 2 = 4 x^{2}$

Step 2.5.5

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

$e^{r x} (r^{2} - 3 r + 2) = 4 x^{2}$

Step 2.6

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

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

Step 3

Solve for $r$ .

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Step 3.1

Subtract $4 x^{2}$ from both sides of the equation.

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

Step 3.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.3

Substitute the values $a = 1$ , $b = - 3$ , and $c = 2 - 4 x^{2}$ into the quadratic formula and solve for $r$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (1 \cdot (2 - 4 x^{2}))}}{2 \cdot 1}$

Step 3.4

Simplify.

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Step 3.4.1

Simplify the numerator.

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Step 3.4.1.1

Raise $- 3$ to the power of $2$ .

$r = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot (2 - 4 x^{2})}}{2 \cdot 1}$

Step 3.4.1.2

Multiply $- 4$ by $1$ .

$r = \frac{3 \pm \sqrt{9 - 4 \cdot (2 - 4 x^{2})}}{2 \cdot 1}$

Step 3.4.1.3

Apply the distributive property.

$r = \frac{3 \pm \sqrt{9 - 4 \cdot 2 - 4 (- 4 x^{2})}}{2 \cdot 1}$

Step 3.4.1.4

Multiply $- 4$ by $2$ .

$r = \frac{3 \pm \sqrt{9 - 8 - 4 (- 4 x^{2})}}{2 \cdot 1}$

Step 3.4.1.5

Multiply $- 4$ by $- 4$ .

$r = \frac{3 \pm \sqrt{9 - 8 + 16 x^{2}}}{2 \cdot 1}$

Step 3.4.1.6

Subtract $8$ from $9$ .

$r = \frac{3 \pm \sqrt{1 + 16 x^{2}}}{2 \cdot 1}$

Step 3.4.2

Multiply $2$ by $1$ .

$r = \frac{3 \pm \sqrt{1 + 16 x^{2}}}{2}$

Step 3.5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 3.5.1

Simplify the numerator.

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Step 3.5.1.1

Raise $- 3$ to the power of $2$ .

$r = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot (2 - 4 x^{2})}}{2 \cdot 1}$

Step 3.5.1.2

Multiply $- 4$ by $1$ .

$r = \frac{3 \pm \sqrt{9 - 4 \cdot (2 - 4 x^{2})}}{2 \cdot 1}$

Step 3.5.1.3

Apply the distributive property.

$r = \frac{3 \pm \sqrt{9 - 4 \cdot 2 - 4 (- 4 x^{2})}}{2 \cdot 1}$

Step 3.5.1.4

Multiply $- 4$ by $2$ .

$r = \frac{3 \pm \sqrt{9 - 8 - 4 (- 4 x^{2})}}{2 \cdot 1}$

Step 3.5.1.5

Multiply $- 4$ by $- 4$ .

$r = \frac{3 \pm \sqrt{9 - 8 + 16 x^{2}}}{2 \cdot 1}$

Step 3.5.1.6

Subtract $8$ from $9$ .

$r = \frac{3 \pm \sqrt{1 + 16 x^{2}}}{2 \cdot 1}$

Step 3.5.2

Multiply $2$ by $1$ .

$r = \frac{3 \pm \sqrt{1 + 16 x^{2}}}{2}$

Step 3.5.3

Change the $\pm$ to $+$ .

$r = \frac{3 + \sqrt{1 + 16 x^{2}}}{2}$

Step 3.6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 3.6.1

Simplify the numerator.

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Step 3.6.1.1

Raise $- 3$ to the power of $2$ .

$r = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot (2 - 4 x^{2})}}{2 \cdot 1}$

Step 3.6.1.2

Multiply $- 4$ by $1$ .

$r = \frac{3 \pm \sqrt{9 - 4 \cdot (2 - 4 x^{2})}}{2 \cdot 1}$

Step 3.6.1.3

Apply the distributive property.

$r = \frac{3 \pm \sqrt{9 - 4 \cdot 2 - 4 (- 4 x^{2})}}{2 \cdot 1}$

Step 3.6.1.4

Multiply $- 4$ by $2$ .

$r = \frac{3 \pm \sqrt{9 - 8 - 4 (- 4 x^{2})}}{2 \cdot 1}$

Step 3.6.1.5

Multiply $- 4$ by $- 4$ .

$r = \frac{3 \pm \sqrt{9 - 8 + 16 x^{2}}}{2 \cdot 1}$

Step 3.6.1.6

Subtract $8$ from $9$ .

$r = \frac{3 \pm \sqrt{1 + 16 x^{2}}}{2 \cdot 1}$

Step 3.6.2

Multiply $2$ by $1$ .

$r = \frac{3 \pm \sqrt{1 + 16 x^{2}}}{2}$

Step 3.6.3

Change the $\pm$ to $-$ .

$r = \frac{3 - \sqrt{1 + 16 x^{2}}}{2}$

Step 3.7

The final answer is the combination of both solutions.

$r = \frac{3 + \sqrt{1 + 16 x^{2}}}{2}$

$r = \frac{3 - \sqrt{1 + 16 x^{2}}}{2}$

$r = \frac{3 + \sqrt{1 + 16 x^{2}}}{2}$

$r = \frac{3 - \sqrt{1 + 16 x^{2}}}{2}$

Step 4

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

$y_{1} = e^{\frac{3 + \sqrt{1 + 16 x^{2}}}{2} x}$

$y_{2} = e^{\frac{3 - \sqrt{1 + 16 x^{2}}}{2} x}$

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.

$y = c_{1} e^{\frac{3 + \sqrt{1 + 16 x^{2}}}{2} x} + c_{2} e^{\frac{3 - \sqrt{1 + 16 x^{2}}}{2} x}$

Step 6

Simplify each term.

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Step 6.1

Combine $\frac{3 + \sqrt{1 + 16 x^{2}}}{2}$ and $x$ .

$y = c_{1} e^{\frac{(3 + \sqrt{1 + 16 x^{2}}) x}{2}} + c_{2} e^{\frac{3 - \sqrt{1 + 16 x^{2}}}{2} x}$

Step 6.2

Combine $\frac{3 - \sqrt{1 + 16 x^{2}}}{2}$ and $x$ .

$y = c_{1} e^{\frac{(3 + \sqrt{1 + 16 x^{2}}) x}{2}} + c_{2} e^{\frac{(3 - \sqrt{1 + 16 x^{2}}) x}{2}}$