# Solutions of a Non homogeneous Linear Differential Equation .

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 wolframalpha.com/input/?i=y%27%27-y%27%3D2e%5E%282t%29 Commented Dec 28, 2019 at 17:39
• What makes you think b is not a correct answer ? Commented Dec 28, 2019 at 19:02
• 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. Commented Dec 28, 2019 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. Commented Dec 28, 2019 at 19:44
• @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 $x^2$
– user435638
Commented Dec 28, 2019 at 20:24

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