Help in solving Differential equation

[divinejun]

dy/dx = x^2 + y ^ 2

find general solution for this differential equation...

thx for ur help and pls show the step taken and method use ya..
thank you very much..

 

[matematikawan]

Try checking this page
https://www.physicsforums.com/showthread.php?t=221644

 

[divinejun]

thx for the help.. however, for the website there, i try to reduce the equation into
d^2u/dx^2 + x^2u = 0
but i fail.. mind if u show all the step to get the equation?
and how to solve that using power series? i haven learn power series yet... thanks for the help

 

[divinejun]

i try to reduce it using substitution y = 1/u(du/dx)
dy/dx = -1/u^2(dy/dx)^2 + 1/u(d^2u/dx^2)
substitute into the equation i get
d^2u/dx^2 = ux^2 + 2/u(du/dx)^2

 

[divinejun]

after some tries.. i get d^2u/dx^2 + ux^2 = 0.
how to get the general solution for tis 1?
i check the website.. by using power series.. can any1 show me how? thanks for the help

 

[HallsofIvy]

Let [itex]u(x)= \sum_{n=0}^\infty a_nx^n[/tex]. Then $u'= \sum_{n=1}^\infty na_nx^{n-1}$ and $u''= \sum_{n=2}^\infty n(n-1)a_nx^{n-2}$.<br /> <br /> (I have changed the beginning index from 0 to 1 to 2 because each term, in the sum for u', is multiplied by n and so is 0 when n= 0 and, in the sum for u'' is multiplied by n(n- 1) and so is 0 when n= 0 and when n= 1.)<br /> <br /> Replacing u'' and u in your equation with those, <br /> $$\sum_{n= 2}^\infty n(n-1)a_nx^{n-2}+ x^2\sum_{n=0}^\infty a_nx^n= 0$$<br /> $$\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+ \sum_{n=0}^\infty a_nx^{n+2}= 0$$<br /> <br /> In order to combine "like powers", let i= n-2 in the first sum. Then n= i+ 2, when n=2, i= 0 so the sum is<br /> $$\sum_{i=0}^\infty (i+2)(i+1)a_{i+2}x^i$$.<br /> <br /> Let i= n+ 2 in the second sum (since the indices have meaning only in the individual sums, we can do this.). Now n= i- 2. When n= 0, i= 2 so the sum becomes<br /> $$\sum_{i= 2}^\infty a_{i-2}x^i$$<br /> <br /> The differential equation becomes <br /> $$\sum_{i=0}^\infty (i+2)(i+1)a_{i+2}x^i+ \sum_{i=2}^\infty a_{i-2}x^i$$<br /> and can compare "like powers" by looking at specific values of i. <br /> <br /> The second sum does not start until i= 2 so for i= 0 and i= 1 we have<br /> $(0+2)(0+ 1)a_2= 2a_2= 0$ which tells us that $a_2= 0$ and $(1+ 2)(1+ 1)a_3= 6a_3= 0$ which tells us that $a_3= 0$.<br /> <br /> For i greater than 1, we have <br /> $$(i+ 2)(i+ 1)a_{i+2}+ a_{i-2}= 0$$<br /> or<br /> $$a_{i+2}= -\frac{a_{i-2}}{(i+ 2)(i+1)}$$<br /> <br /> That gives a recursive equation for the coefficients. Since "n-2" will not be non-negative until n= 2, where a+ 2= 4, we cannot, of course, determine $a_0$, $a_1$, $a_2$, or $a_3$ from this, but that's okay- we already know that $a_2= a_3= 0$ and we expect a second order d.e. to involve two "constants". We can take those constants to be $a_0$ and $a_1$. (In fact, it is easy to see that $y(0)= a_0$ and $y'(0)= a_1$, "initial conditions" for this problem. If you were given other initial conditions, say $y(x_0)$ and $y'(x_0)$, write the power series in terms of $(x- x_0)^n$.)<br /> <br /> Now, we have, with i= 2, <br /> $$a_4= \frac{a_0}{(2+ 2)(2+1)}= \frac{a_0}{8}$$<br /> with i= 3,<br /> $$a_5= \frac{a_1}{(3+2)(3+ 1)}= \frac{a_1}{18}$$<br /> with i= 4,<br /> $$a_6= \frac{a_2}{(4+ 2)(4+ 1)}= 0$$<br /> because $a_2= 0$.<br /> Similarly, with i= 5, <br /> $$a_7= \frac{a_3}{(5+2)(5+1)}= 0$$<br /> because $a_3= 0$<br /> With i= 6,<br /> $$a_8= \frac{a_4}{(6+2)(6+1)}= \frac{a_4}{56}$$<br /> $$= \frac{a_0}{8(56)}= \frac{a_0}{448}$$<br /> <br /> etc.<br /> <br /> It might be best to look at the indices "modulo 4". If i is 3 mod 4 (3, 7, 11, 15,...) $a_i= 0$. If i is 2 mod 4 (2, 6, 10, 14, ...), $a_i= 0$.<br /> <br /> If i is 0 mod 4 (0, 4, 8, 12, ...) $a_i$ will be $a_0$ divided by an integer. We could factor $a_0$ out of that sum and have $a_0$ times some function. If i is 1 mod 4 (1, 5, 9, 13, ...) [itex]a_i[/tex] will be $a_1$ divided by some integer. We could factor $a_1$ out of that sum and have $a_1$ times some function. The result would be <br /> $$u(x)= a_0f(x)+ a_1g(x)$$<br /> where f and g are solutions to the differential equation. Of course f and g may be extremely complicated sums. You cannot, in general, expect to be able to determine a formula for the sums or even for the general coefficients.[/itex][/itex]

 

[divinejun]

thx a lot hallsofivy... i try go search the web again.. i found out a similar situation with this. the special ricatti equation and i totally gone blur after seeing it.. comparing with your answer i get more confused.
http://eqworld.ipmnet.ru/en/solutions/ode/ode0106.pdf
the website show the special ricatti equation and final solution only.. it uses the same substitution but i can't get the answer.. mind helping me?

 

[Redbelly98]

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