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The equation is of the type complete and its solution is the sum of two terms, the general solution of the incomplete equation and any particular solution of the complete equation, so that first we have to solve the equation...$(1-x^2)y'' - xy' + 4y = x \sqrt{1-x^2} $
Hint use the substitution $x =\sin t$
I used it and end with
$\cos t y'' + \sin t y' - \frac{\sin t}{\cos t} y' + 4y = 2\sin t |\cos t| $
how to solve this i just want the name of the method
What i did My question isI do not think you have gone to the $t$ domain correctly. You have
$$\frac{dy}{dt}=\frac{dy}{dx}\,\frac{dx}{dt}=\cos(t)\,\frac{dy}{dx}.$$
Then you have
$$\frac{d^{2}y}{dt^{2}}=\frac{d}{dt}\left[\cos(t)\,\frac{dy}{dx}\right]
=\cos(t)\,\frac{d}{dt}\,\frac{dy}{dx}+\frac{dy}{dx}\,(-\sin(t))$$
$$=\cos^{2}(t)\,\frac{d^{2}y}{dx^{2}}-\sin(t)\,\frac{dy}{dx}.$$
See if that doesn't do some nice things for you.
I am a bit new to differential equation I do not know what complete and incomplete equations meanThe equation is of the type complete and its solution is the sum of two terms, the general solution of the incomplete equation and any particular solution of the complete equation, so that first we have to solve the equation...
$\displaystyle (1-x^{2})\ y^{\ ''} -x\ y^{\ '} + 4\ y=0$ (1)
The procedure that I will use is a little 'non conventional' and requires a preliminary. The solution of a second order incomplete equation is of the type...
$\displaystyle y(x)= c_{1}\ u(x) + c_{2}\ v(x)$ (2)
... where $u(x)$ and $v(x)$ are two independent solutions of (1). Since u and v both satisfy (1) is...
$\displaystyle (1-x^{2})\ u^{\ ''} -x\ u^{\ '} +4\ u=0$
$\displaystyle (1-x^{2})\ v^{\ ''} -x\ v^{\ '} +4\ v=0$ (3)
Multiplying the first of (3) by v and the second by u and do the difference we obtain...
$\displaystyle (1-x^{2})\ (v\ u^{\ ''} - u\ v^{\ ''}) - x\ (v\ u^{\ '} - u\ v^{\ '}) =0$ (4)
Now we set $z= v\ u^{\ '}- u\ v^{\ '}$ so that (4) becomes...
$\displaystyle z^{\ '}= \frac {x}{1-x^{2}}\ z$ (5)
The (5) is a linear ODE the solution of which is...
$\displaystyle z=\frac{c_{2}}{\sqrt{1-x^{2}}}$ (6)
... so that is...
$\displaystyle \frac{z}{v^{2}}= \frac{d}{dx} (\frac{u}{v})= \frac{c_{2}}{v^{2}\ \sqrt{1-x^{2}}} \implies\ u= c_{1}\ v + c_{2}\ v\ \int \frac{dx}{v^{2}\ \sqrt{1-x^{2}}}$ (7)
Now it is easy enough to see that $u=x$ is solution of (1) so that from (7) we derive that...
$\displaystyle v= x\ \int \frac{dx}{x^{2}\ \sqrt{1-x^{2}}} = - \sqrt{1-x^{2}}$ (7)
... is also solution of (1) so that the general solution of (1) is...
$\displaystyle y(x)= c_{1}\ x + c_{2}\ \sqrt{1-x^{2}}$ (8)
... and half of the work is done. The second half will be done [if possible...] in next post...
Kind regards
$\chi$ $\sigma$
An incomplete second order linear ODE is written as...toi
I am a bit new to differential equation I do not know what complete and incomplete equations mean
can you give me a link about it or a little explanation ?
Thanks
For translation effectiveness: chi sigma's "incomplete" is often termed "homogeneous", and chi sigma's "complete" is often termed "inhomogeneous".An incomplete second order linear ODE is written as...
$\displaystyle a(x)\ y^{\ ''} + b(x)\ y^{\ '} + c(x)\ y=0$ (1)
... and a complete second order linear ODE as...
$\displaystyle a(x)\ y^{\ ''} + b(x)\ y^{\ '} + c(x)\ y= d(x)$ (2)
If u(x) and v(x) are independent solutions of (1) and w(x) is any particular solution of (2), the the general solution of (2) is...
$\displaystyle y(x)= c_{1}\ u(x) + c_{2}\ v(x) + w(x)$ (3)
... where $c_{1}$ and $c_{2}$ are arbitrary constants...
Kind regards
$\chi$ $\sigma$
The reason why I prefer the terms 'complete' and 'incomplete' is that these terms are related to the presence or not of the 'known term' d(x). It is obvious that the correspondence 'homogeneous -> incomplete' and 'inhomogeneous -> complete' is a perfect source of confusion...For translation effectiveness: chi sigma's "incomplete" is often termed "homogeneous", and chi sigma's "complete" is often termed "inhomogeneous".
no i write the question correctly it is -xy'My attempts to do the second part of the work searching a particular solution of the complete equation...
$\displaystyle (1-x^{2})\ y^{\ ''} -x\ y^{\ '} + 4\ y= 2\ x\ \sqrt{1-x^{2}}$ (1)
... didn't produce results but the problem would be easily overcome changing in (1) the sigh of the term in y' so that the ODE becomes...
$\displaystyle (1-x^{2})\ y^{\ ''} + x\ y^{\ '} + 4\ y= 2\ x\ \sqrt{1-x^{2}}$ (2)
I would ask Amer if May be that the equation is (2) and not (1)...
Kind regards
$\chi$ $\sigma$