Solve the given differential equation by undetermined coefficients: $y''' - 6y'' = 2 - \cos(x)$

Neetesh Kumar | November 06, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh4Sec04 (Homework) Question - 6
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Step-by-Step-Solution:

Step 1: Solve the Homogeneous Equation

We start by solving the corresponding homogeneous equation: $y_h''' - 6y_h'' = 0$

To simplify, let’s treat $y_h''$ as a factor: $D^2 (D - 6) = 0$

This implies that either $D = 0 (Repeated)$ or $D = 6$.

Then, $y_h = C_1 + C_2x + C_3e^{6x}$

Step 2: Solve the Particular Solution

The non-homogeneous term is $2 - \cos(x)$, which suggests a particular solution like:

$y_p(x) = Ax^2 + B\cos(x) + C\sin(x)$.

To substitute $y_p(x)$ into the differential equation, we need its second and third derivatives:

$y_p'(x) = 2Ax + C\cos(x) - B\sin(x)$
$y_p''(x) = 2A -B\cos(x) - C\sin(x)$
$y_p'''(x) = -C\cos(x) + B\sin(x)$

Step 4: Substitute $y_p''(x)$ and $y_p'''(x)$ into the Differential Equation

$(-C\cos(x) + B\sin(x)) - 6(2A - B\cos(x) - C\sin(x)) = 2 - \cos(x)$

Combining like terms:

$-12A + (6B-C)\cos(x) + (6C+B)\sin(x) = 2 - \cos(x)$

From this equation, we equate the coefficients of $\sin(x)$ and $\cos(x)$: $6B-C = -1$ and $6C+B = 0$

After solving we will get: $A = -\frac{1}{6}, B = -\frac{6}{37}, C = \frac{1}{37}$

Now, we can write $y_p(x) = -\frac{1}{6}x^2 - \frac{6}{37}\cos(x) + \frac{1}{37}\sin(x)$

So, the General Solution is: $y(x) = y_h(x) + y_p(x)$

Final Answer:

$y(x) = C_1 + C_2x + C_3e^{6x} - \frac{1}{6}x^2 - \frac{6}{37}\cos(x) + \frac{1}{37}\sin(x)$

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