Solve the given differential equation: $49x^2 y'' + 49x y' + y = 0$

Neetesh Kumar | October 28, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh4Sec07 (Homework) Question - 3
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Step-by-step solution:

This is a Cauchy-Euler differential equation of the form:

$x^2 y'' + bx y' + c y = 0$

In our case, the equation is: $49x^2 y'' + 49x y' + y = 0$

Divide the entire equation by $49$ to simplify: $x^2 y'' + x y' + \frac{1}{49} y = 0$

Step 1: Assume a Solution of the Form $y = x^r$

For Cauchy-Euler equations, we assume a solution of the form: $y = x^r$

Then, the first and second derivatives are: $y' = r x^{r-1}$ $y'' = r(r - 1) x^{r - 2}$

Step 2: Substitute into the Differential Equation

Substitute $y = x^r$, $y' = r x^{r-1}$, and $y'' = r(r - 1) x^{r-2}$ into the equation:

$x^2 \cdot r(r - 1) x^{r - 2} + x \cdot r x^{r-1} + \frac{1}{49} x^r = 0$

Simplify each term:

$r(r - 1) x^r + r x^r + \frac{1}{49} x^r = 0$

Factor out $x^r$:

$x^r (r(r - 1) + r + \frac{1}{49}) = 0$

Since $x^r \neq 0$, we get the characteristic equation:

$r(r - 1) + r + \frac{1}{49} = 0$

Step 3: Solve the Characteristic Equation

Expanding the terms in the characteristic equation:

$r^2 - r + r + \frac{1}{49} = 0$

This simplifies to:

$r^2 + \frac{1}{49} = 0$

Rearrange to solve for $r$:

$r^2 = -\frac{1}{49}$

Taking the square root of both sides, we get:

$r = \pm \frac{i}{7}$

Step 4: Write the General Solution

Since the roots are complex, $r = \pm \frac{i}{7}$, we can write the general solution as:

$y(x) = C_1 \cos\left(\frac{\ln(x)}{7}\right) + C_2 \sin\left(\frac{\ln(x)}{7}\right)$

where $C_1$ and $C_2$ are constants.

Final Answer

$y(x) = C_1 \cos\left(\frac{\ln(x)}{7}\right) + C_2 \sin\left(\frac{\ln(x)}{7}\right)$

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