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.

  • $\begingroup$ b d look like solution to Euler Cauchy inhomogeneous equation . For d we may have $y''-y'=2e^{2t}$ Check wolframalpha.com/input/?i=y%27%27-y%27%3D2e%5E%282t%29 $\endgroup$ Commented Dec 28, 2019 at 17:39
  • $\begingroup$ What makes you think b is not a correct answer ? $\endgroup$ Commented Dec 28, 2019 at 19:02
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    $\begingroup$ 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. $\endgroup$
    – Axion004
    Commented Dec 28, 2019 at 19:19
  • $\begingroup$ 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. $\endgroup$
    – Axion004
    Commented Dec 28, 2019 at 19:44
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    $\begingroup$ @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 $x^2$ $\endgroup$
    – user435638
    Commented Dec 28, 2019 at 20:24

1 Answer 1


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: $$y(x)=c_1x^2+c_2x+1$$


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