Find the general solution of the given second-order differential equation. $$ 2 y^{\prime \prime}-5 y^{\prime}+6 y=0 $$ $$ y(x)=\square $$

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UpStudy AI · Accepted Answer

To find the general solution of the second-order differential equation:

$$
2 y'' - 5 y' + 6 y = 0
$$

we start by finding the characteristic equation associated with the differential equation. Assume a solution of the form $$ y = e^{rx} $$. Plugging this into the differential equation gives:

$$
2r^2 e^{rx} - 5r e^{rx} + 6 e^{rx} = 0
$$

Dividing through by $$ e^{rx} $$ (which is never zero):

$$
2r^2 - 5r + 6 = 0
$$

This is a quadratic equation in $$ r $$. To solve it, we calculate the discriminant:

$$
D = (-5)^2 - 4 \cdot 2 \cdot 6 = 25 - 48 = -23
$$

Since the discriminant is negative ($$ D < 0 $$), the roots are complex:

$$
r = \frac{5 \pm \sqrt{-23}}{4} = \frac{5}{4} \pm \frac{\sqrt{23}}{4}i
$$

Letting $$ \alpha = \frac{5}{4} $$ and $$ \beta = \frac{\sqrt{23}}{4} $$, the general solution for the differential equation with complex roots is:

$$
y(x) = e^{\alpha x} \left( C_1 \cos(\beta x) + C_2 \sin(\beta x) \right)
$$

Substituting the values of $$ \alpha $$ and $$ \beta $$:

$$
y(x) = e^{\frac{5}{4}x} \left( C_1 \cos\left(\frac{\sqrt{23}}{4}x\right) + C_2 \sin\left(\frac{\sqrt{23}}{4}x\right) \right)
$$

where $$ C_1 $$ and $$ C_2 $$ are arbitrary constants determined by initial conditions.

**Final Answer:**

After simplifying, a general solution is e^(5⁄4 x) times (C₁ cos(√23⁄4 x) plus C₂ sin(√23⁄4 x)). Thus,

y(x) = e^(5/4x)[C₁ cos(√23/4 x) + C₂ sin(√23/4 x)]
The general solution is: $$ y(x) = e^{\frac{5}{4}x} \left( C_1 \cos\left(\frac{\sqrt{23}}{4}x\right) + C_2 \sin\left(\frac{\sqrt{23}}{4}x\right) \right) $$

Find the general solution of the given second-order differential equation.

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