Question

Given that y1(x)= x^-1is one solution of the differential equation 2x^2y"+3xy'-y=0,x>0, find a second linearly independent solution of the equation.

Answer 1

ANSWER ON QUESTION #66145 – Math – Differential Equations

QUESTION

Given that $y_{1} = \frac{1}{x}$ is one solution of the differential equation

$$2 x ^ {2} y ^ {\prime \prime} + 3 x y ^ {\prime} - y = 0, x > 0,$$

find a second linearly independent solution of the equation.

SOLUTION

We transform the original equation to the form

$$y ^ {\prime \prime} + p (x) y ^ {\prime} + q (x) y = 0$$ $$2 x ^ {2} y ^ {\prime \prime} + 3 x y ^ {\prime} - y = 0 \left|: \frac {1}{2 x ^ {2}} \rightarrow y ^ {\prime \prime} + \frac {3 x}{2 x ^ {2}} y ^ {\prime} - \frac {y}{2 x ^ {2}} = 0 \right.$$ $$y ^ {\prime \prime} + \frac {3}{\frac {2 x}{p (x)}} y ^ {\prime} - \frac {1}{\frac {2 x ^ {2}}{q (x)}} y = 0$$

$y_{1}(x) = \frac{1}{x}$ is the first solution of the differential equation (1)

Using the Liouville formula ( https://en.wikipedia.org/wiki/Liouville%27s_formula ) we can find a second solution:

$$\begin{array}{l} y _ {2} (x) = y _ {1} \int \frac {e ^ {- \int p (x) d x}}{y _ {1} ^ {2}} d x = \frac {1}{x} \int \frac {e ^ {- \int \frac {3}{2 x} d x}}{\frac {1}{x ^ {2}}} d x = \frac {1}{x} \int x ^ {2} e ^ {- \frac {3}{2} \ln (x)} d x \\ = \frac {1}{x} \int x ^ {2} e ^ {\ln \left(x ^ {- \frac {3}{2}}\right)} d x = \frac {1}{x} \int x ^ {2} x ^ {- \frac {3}{2}} d x = \frac {1}{x} \int x ^ {2 - \frac {3}{2}} d x = \frac {1}{x} \int x ^ {\frac {1}{2}} d x \\ \end{array}$$ $$= \frac {1}{x} \cdot \frac {x ^ {\frac {1}{2} + 1}}{\frac {1}{2} + 1} = \frac {1}{x} \cdot \frac {x ^ {\frac {3}{2}}}{\frac {3}{2}} = \frac {2}{3} \cdot \frac {x ^ {\frac {3}{2}}}{x} = \frac {2}{3} \cdot x ^ {\frac {1}{2}} \equiv \frac {2 \sqrt {x}}{3}.$$

**ANSWER:**

$y_{2}(x) = \frac{2\sqrt{x}}{3}$ is a second linearly independent solution of the equation

Answer provided by https://www.AssignmentExpert.com