Solve a h dif eq by variation of paraders subject to the initial conditions y(0)=1, y’(0)=0. 4y’’-y=xe^(x/2). The solution should be y=c1+c2cosx+c3sinx-ln|cos|-sinxln|sec+tan|


The problem you've provided involves solving the non-homogeneous second-order differential equation:

\[
4y'' - y = xe^{x/2},
\]

subject to the initial conditions:

\[
y(0) = 1, \quad y'(0) = 0.
\]

Additionally, the solution is expected to take the form:

\[
y = c_1 + c_2 \cos x + c_3 \sin x - \ln|\cos x| - \sin x \ln|\sec x + \tan x|.
\]

### Step 1: Homogeneous Solution

First, solve the corresponding homogeneous equation:

\[
4y'' - y = 0.
\]

Rewriting this equation:

\[
y'' - \frac{1}{4}y = 0.
\]

The characteristic equation is:

\[
r^2 - \frac{1}{4} = 0,
\]

which gives:

\[
r = \pm \frac{1}{2}.
\]

Therefore, the general solution to the homogeneous equation is:

\[
y_h(x) = c_1 e^{x/2} + c_2 e^{-x/2}.
\]

### Step 2: Particular Solution via Variation of Parameters

Now, we solve for a particular solution to the non-homogeneous equation using the method of **variation of parameters**. The right-hand side of the equation is \( xe^{x/2} \), which suggests we will vary the constants in the homogeneous solution. 

Let's assume the particular solution takes the form:

\[
y_p(x) = u_1(x) e^{x/2} + u_2(x) e^{-x/2},
\]

where \( u_1(x) \) and \( u_2(x) \) are functions to be determined. The

Solve a h dif eq by variation of paraders subject to the initial conditions y(0)=1, y’(0)=0. 4y’’-y=xe^(x/2). The solution should be y=c1+c2cosx+c3sinx-ln|cos|-sinxln|sec+tan|

Learn how to solve the second-order non-homogeneous differential equation 4y'' - y = xe^(x/2) using the variation of parameters method. The solution also satisfies initial conditions y(0) = 1 and y'(0) = 0. A detailed step-by-step approach covering homogeneous and particular solutions is provided.

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