Question

Solve the given initial-value problem. y′′+4y=−3,y(π/8)=1/4​,y′(π/8)=2 y(x)=___

Answer 1

Solution of the given initial-value problemy′′+4y=−3,y(π/8)=1/4​,y′(π/8)=2 is
`\[ y(x) = (1)/(4) \cos(2x) + (1)/(2) \sin(2x) \]`

Step-by-step explanation:

To solve the given initial-value problem, we start by finding the characteristic equation of the differential equation (y'' + 4y = -3). The characteristic equation is given by (
`r^2` + 4 = 0), which yields complex roots
`\(r = \pm 2i\)`. The general solution is then
`\(y(x) = A \cos(2x) + B \sin(2x)\)`, where (A) and (B) are constants.

Next, we apply the initial conditions
`\(y(\pi/8) = 1/4\)` and
`\(y'(\pi/8) = 2\)` to determine the values of (A) and (B). Plugging in
`\(\pi/8\)` into the general solution and its derivative, we get two equations:

`\[ A \cos(\pi/4) + B \sin(\pi/4) = (1)/(4) \]`

`\[ -2A \sin(\pi/4) + 2B \cos(\pi/4) = 2 \]`

Solving these equations simultaneously, we find
`\(A = (1)/(4)\)` and
`\(B = (1)/(2)\)`.

Substituting these values back into the general solution, we arrive at the final answer:

`\[ y(x) = (1)/(4) \cos(2x) + (1)/(2) \sin(2x) \]`

In summary, the solution to the initial-value problem is a linear combination of cosine and sine functions with specific coefficients determined by the given initial conditions. The final expression represents the function y(x) that satisfies the differential equation and the specified initial values at
`\(x = \pi/8\)`.

Answer 2

The solution to the initial-value problem y''+4y=-3 with given initial conditions is y(x) = 1/4 cos(2x) + 7/2 sin(2x) - 3/4, derived by finding the general solution to the homogeneous equation, a particular solution to the non-homogenous equation, and applying the given initial conditions to solve for the constants.

Step-by-step explanation:

To solve the initial-value problem y''+4y=-3, with initial conditions y(\frac{\pi}{8})=\frac{1}{4}, and y'(\frac{\pi}{8})=2, we first find the general solution to the homogeneous equation y''+4y=0, and then find a particular solution to the non-homogenous equation.

The general solution to the homogeneous equation is y_h(x) = C_1 \cos(2x) + C_2 \sin(2x), where C_1 and C_2 are constants to be determined. A particular solution to the non-homogenous equation is y_p(x) = -\frac{3}{4}, since the non-homogeneous term is a constant.

The general solution to the initial-value problem is y(x) = y_h(x) + y_p(x). Using the initial conditions, we substitute x = \frac{\pi}{8} into y(x) and y'(x) to find C_1 and C_2. After substituting and solving the system of equations, we obtain the particular solution:
y(x) = \frac{1}{4} \cos(2x) + \frac{7}{2} \sin(2x) - \frac{3}{4}.