Solutions of the differential equation x2y″−4xy′+6y=0.In one of my test it given to prove that x3 and x2|x| are linear independent solutions of the differential equation x2y″−4xy′+6y=0 on R( here x is independent variable).But according to me it’s Cauchy Euler equation having general solution as y=c1x3+c2x2, where c1 and c2 are arbitrary constants. How can be x2|x| a solution of given ODE as I am unable to find its by giving particular values of constants c1 and c2? Please help me to solve it.

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Solutions of the differential equation x2y″−4xy′+6y=0.In one of my test it given to prove that x3 and x2|x| are linear independent solutions of the differential equation x2y″−4xy′+6y=0 on R( here x is independent variable).But according to me it’s Cauchy Euler equation having general solution as y=c1x3+c2x2, where c1 and c2 are arbitrary constants. How can be x2|x| a solution of given ODE as I am unable to find its by giving particular values of constants c1 and c2? Please help me to solve it.

Isla Preston · Accepted Answer

Explanation:The differential equation has a singularity at x=0, so the Existence and Uniqueness Theorem doesn&#039;t apply there. On each of the intervals (−∞,0) and (0,∞) where the theorem does apply, you have two-parameter families of solutions. But it turns out any solution on (−∞,0) and any solution on (0,∞) with the same c2 can be put together to make a solution on R.

Leia Wiggins · Accepted Answer

Step 1c1x3+c2x2is not the general solution of the Cauchy-Euler equation.Step 2If you set x=eu, you indeed linearize the original equation toy¨−5y˙+6y=0with the obvious solutiony=c3e3u+c2e2u=c3x3+c2x2.But recall that by our change of variable, x was assumed positive. Now we can solve again by setting x=−eu and get a solution of a similar shape, but the constants have no reason to be equal on both sides.

Solutions of the differential equation x^2y’’-4xy’+6y=0. In one of my test it given to pro

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