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Let $1 ,x$ and $x^2$ be the solution of a second order linear Non homogenous differential equation on $-1 < x < 1$, then it's general solution involving arbitrary constants can be written as :

(a) $c_1(1-x) + c_2(x - x^2) +1$

(b) $c_1(x) + c_2 ( x^2) +1$

(c) $c_1(1+x) + c_2(1 + x^2) +1$

(d) $c_1 + c_2 x + x^2$

Now, I know this : The general solution of such a differential equation is written as:

$Y = c_1 f + c_2 g + \text{P.I.}$

where $f$ and $g$ are two Linearly Independent solutions and $P.I.$ denotes the particular integral obtained by solving the non homogeneous part.

So, Using this fact I know that options (b) and (c) are false because the function are linearly Dependent on given interval.

However I am confused between (a) and (d) .The given functions are Linearly Independent but I have no idea how to decide the Particular Integral.

Can anyone tell me how should I tackle options (a) and (d) ?

Thank you.

  • b d look like solution to Euler Cauchy inhomogeneous equation . For d we may have $y''-y'=2e^{2t}$ Check https://www.wolframalpha.com/input/?i=y%27%27-y%27%3D2e%5E%282t%29 – user577215664 Dec 28 '19 at 17:39
  • What makes you think b is not a correct answer ? – user577215664 Dec 28 '19 at 19:02
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    Two functions $y_1$ and $y_2$ are said to be linearly independent if neither function is a constant multiple of the other. Therefore, $f(x)=1,g(x)=x,h(x)=x^2$ are linearly independent and $(b)$ could certainly be the general solution. – Axion004 Dec 28 '19 at 19:19
  • I am able to rewrite $(a),(c)$ as $c_1+c_2(x)+c_3(x^2)$ while $(b)$ is $c_1(x)+c_2(x^2)+1$ (set $c_3=1$) and $(d)$ is $c_1+c_2x+x^2$ (set $c_3=1$). So, it appears that one or two of these should be incorrect. – Axion004 Dec 28 '19 at 19:44
  • @Isham: I agree with your observation for option (b),but I have a confusion if I compute the wronskian of given function then it becomes zero for $x = -1 + \sqrt{2}$ and so I believe functions are Linearly Dependent . –  Dec 28 '19 at 19:59
  • @Isham: Can you tell me what is the flaw in above reasoning ? –  Dec 28 '19 at 20:01
  • it must be zero everywhere and thats not the case. As you can easily show for option b https://en.wikipedia.org/wiki/Wronskian#The_Wronskian_and_linear_independence – user577215664 Dec 28 '19 at 20:01
  • Integrate the equation I gave in my answer for option B and D . You will get the solutions provided in your book. Integrate $y''=2$ for option b for example. B option is a correct answer – user577215664 Dec 28 '19 at 20:10
  • You a re a bit confused about the Wronskian it seems to me. When there is dependance the Wronskian must be at least zero everywhere. Thats not the case for b. You can easily find an x in the given interval for which the Wronskian is not zero. – user577215664 Dec 28 '19 at 20:14
  • @Isham: I know that if a Wronskian vanishes at certain point it does Not necessarily mean that functions are L.d. But please take a look at this question :https://math.stackexchange.com/questions/3354468/which-among-the-following-is-true-for-a-wronskian-of-a-differential-equation, Here it is concluded wronskian for solution of a Differential Equation is either identically zero or Never zero. My question is : Is this is statement only true for homogeneous Differential Equations ? I would be very grateful if you can clear this doubt . –  Dec 28 '19 at 20:16
  • Then try to compute the wronskian of $x$ and $x^2$ its equal to $x^2$. But it vanishes at $x=0$ and zero is in the interval given...So ? According to you $x,x^2$ are linearly dependant ? – user577215664 Dec 28 '19 at 20:21
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    @Isham: No, they are independent ,if we use the definition that $c_1f +c_2g =0$ are Linearly independent iff $c_1 =c_2 =0$ then this is true for $x$ and $x^2$ –  Dec 28 '19 at 20:24
  • Thus, the Wronskian can be used to show that a set of differentiable functions is linearly independent on an interval by showing that it does not vanish identically. It may, however, vanish at isolated points.[1] (wikisource).... – user577215664 Dec 28 '19 at 20:24
  • True because the Wronskian of $x,x^2$ vanishes at certain points not everywhere. – user577215664 Dec 28 '19 at 20:27

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For the d option. $$y(x)=c_1+c_2x+x^2$$ substitute $x=e^t$ $$y(t)=c_1+c_2e^t+e^{2t}$$ $$r=0, r=1 \implies r(r-1)=0 \implies y''-y'=0$$ $$y''-y'=f(x)$$ Particular solution is $e^{2t}$ $$4e^{2t}-2e^{2t}=f(x) \implies f(x)=2e^{2t}$$ The equation is $$y''-y'=2e^{2t}$$ $$\implies x^2y(x)''=2x^2 \implies y''(x)=2$$

And $y''=2$ has option d as solution. Integrate it.

For option $(b)$ I got the equation

$$y''(t)-3y'(t)+2y(t)=2$$ It gives Cauchy-Euler's equation: $$\implies x^2y''(x)-2xy'(x)+2y(x)=2$$ Has solution: $$y(x)=c_1x^2+c_2x+1$$

user577215664
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