Solve x^2y'' - 3xy' - 9y = 0 Cauchy - Euler Differential Equation

[@enashameed471]

Thanks ❤❤❤❤❤❤❤❤❤جزاك الله خيرا

[@TheMathSorcerer]

You are welcome👍

[@olivioarmandocordeirojunio8280]

Thanks for sharing your knowledge.

[@saumyasingh7670]

Thank you ☺😊

[@Speak4Yourself2]

Thanks a lot!

[@ingamado]

Gracias.

[@TheMathSorcerer]

👍

[@VndNvwYvvSvv]

Dividing by x^m almost seems like cheating.

[@SandeepSingh-eb5ks]

And what happen when m give two equal values

[@TheMathSorcerer]

You multiply by lnx, check this at 4min 50 sec or so th-cam.com/video/oqpc89Lfx-E/w-d-xo.html

[@johndavecusi6990]

Ohhh you did something wrong on your last solution. The final answer should be in the form
Y= x^a( C1cosB lnx +c2 sinB lnx)

[@TheMathSorcerer]

that's when it's complex conjugates, here we have distinct real roots so it's different

[@jenniferbeach8457]

There isn't a negative under the square root. There fore no "i", so not Beta, not complex

[@JMCK01]

what form is the general solution if the quadratic only has one real root? (b^2-4ac=0)

[@mariomuysensual]

Use variation of parameters

[@ibrahimcivanoglu2758]

It should be y=x^m1+x^m2*lnx. You can follow the whole playlist for Cauchy - Euler differential equation. It takes a relatively small amount of time to cover.

[@radicalrals9973]

You only did the Yh form, don't you need the Yp form as well, since y=yp+yh?

[@hafizmuhamad5966]

you dont need to do yp since it is homogeneous eqn

[@chaseschnell1026]

If it = 0 on the right hand side you do not

[@ildersonfelipe5795]

muito bom, me salvou kskks

[@modern_genghis_khan0393]

Suppose those roots are imaginary then what can I do ?

[@robertbeurre1825]

cry

[@modern_genghis_khan0393]

@@robertbeurre1825 alright come let us cry 😂

[@justinernest6881]

if the roots are in imaginary form, ie a+ib, then the characteristic equation would be given by e^(ax)(c1cos(bx)+c2sin(bx)) which is in real form

[@robertbeurre1825]

Shouldn't it be e and not x in your final solution?

[@modern_genghis_khan0393]

X be an final variable in solution

[@alyssameylsaechao]

What if the problem didn't = 0. What if the problem = to 4? How do we get rid of X^m?

[@abramar123]

I'm assuming you would apply log on both sides.

[@VndNvwYvvSvv]

First solve with the RHS = 0. Thats the complementary solution Yc. Then apply other techniques to solve for the particular solution Yp, and the full solution y = Yc +yp. Those techniques are usually covered before this.

[@Mangan1025]

I did quadratic formula, the -b+/- …. and I got 2+/-squ-root of 5 i

[@VndNvwYvvSvv]

Then you must have done something wrong. If you get that as the correct roots of a different problem, the first number (2) is alpha, the 2nd is beta (root 5) and you solve y= e^(-alpha x)(C1 cos (beta x) + C2 sin(beta x))

[@ahmadarif6893]

@@VndNvwYvvSvv same answer with me so the general solution will be , Y= (x^alpha x) ( c1 cos ( beta lnx ) + (c^alpha x) ( c2 sin ( beta lnx ) )

[@_vblax]

if 4 years later anyone else also found this... you're doing it wrong because b²-4ac is (-4)²-4×1×(-9) = 16+36 which is 52, you're probably doing 16-36 because you didn't see the negative on c.