Given three solutions of differential equation, to find its general solution 
According to me answer is just linear combination of three given solutions but i donot know whether it is adequate , or i am missingsome trick.Thanks
 A: No it is not just any linear combination it should be $c_1y_1+c_2y_2+c_3y_3$ such that $c_1+c_2+c_3=1$. All solutions are of the form $a_1y_1+a_2y_2+y_p$ where $y_1,y_2$ are solutions of the homogeneous part and $y_p$ is the particular solution. So here when you take linear combinations the solutions of the homogenous part do not cause a problem but $(c_1+c_2+c_3)y_p$ should be $y_p$. that is why the second condition is required
Also if it is a second order linear differential equation then the solution space is of dimension $2$. So one of the solutions should be a linear combination of the other two. So linearly combining just two of them is sufficient.
A: A short way :
One observe that :  $y_2-y_1=x$ and $y_3+y_2-2y_1=xe^x$ which already gives two independant functions $x$ and $xe^x$. So the general solution is on the form $y=c_1x+c_2xe^x+f(x)$, where $f(x)$ doesn't contains an arbitrary constant. The three given solutions include the same term $-2x^2$. So $f(x)=-2x^2$ and the general solution is :
$$y=c_1x+c_2xe^x-2x^2$$
A longer way : 
If you are not convinced, a systematic demonstration consists in bringing the three given solutions into the ODE :
$$-4+(1-4x)P(x)+x(1-2x)Q(x)=R(x)$$
$$-4+(2-4x)P(x)+2x(1-x)Q(x)=R(x)$$
$$\big(-4+(2+x)e^x\big)+\big(-4x+(1+x)e^x)\big)P(x)+x(e^x-2x)Q(x)=R(x)$$
Solve for $P$ , $Q$, $R$ the above system of three equations. This leads to :
$$P=-\frac{2+x}{x}$$
$$Q=\frac{2+x}{x^2}$$
$$R=2x$$
Then solve the ODE :
$$y''-\frac{2+x}{x}y'+\frac{2+x}{x^2}y=2x$$
The result is :
$$y=c_1x+c_2xe^x-2x^2$$
