solving second order linear ODE a/x y′(x)+1/2y(x)=0

Tara Alvarado Answered2022-01-20

solving second order linear ODE

\frac{a}{x} y^{'} (x) + \frac{1}{2} y^{″} (x) = 0

where a is a constant. Also how can I solve this equation if the right hand side is different than zero?

Answer & Explanation

Papilys3q

Expert

2022-01-20Added 34 answers

Step 1
Consider

z = y^{'}

. Then this is just

\frac{a}{x} z (x) + \frac{1}{2} z^{'} (x) = 0

. That is

z^{'} = - \frac{2 a}{x} z

which is separable.
Step 2

\ln | z | = \int \frac{d z}{z} = \int - \frac{2 a}{x} d x = - 2 a \ln | x | + C_{0}

and so

y^{'} = z = C_{1} x^{- 2 a}

.
Integrating yields

y = \frac{C_{1}}{1 - 2 a} x^{1 - 2 a} + C_{2}

Karen Robbins

Expert

2022-01-21Added 49 answers

Step 1
Assuming a solution in an interval not containing

x = 0

, lets rewrite the equation as:

2 a y^{'} + x y^{″} = 0

Now put

y^{'} = v, y^{″} = v^{'}

to get

2 a v + x v^{'} = 0

Step 2
Separating variables produces

\frac{2 a}{x} = - \frac{v^{'}}{v}

2 a \log C x = - \log v

2 a \log C x = - \log v

K x^{- 2 a} = y^{'}

\frac{K x^{1 - 2 a}}{1 - 2 a} + C = y and a \neq \frac{1}{2}

RizerMix

Expert

2022-01-27Added 437 answers

Substitute: \(y=u

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