I have three solutions for a non homogeneous, linear, 2nd order D.E
$$y_1(x)=3x−2x^2$$$$y_2(x)=−2x^2+5xe^x$$$$y_3(x)=2x−2x^2+3xe^x$$
I want to find the general solution.
I know that the general solution is a combination of the three but with some coefficients, $c_1y_1+c_2y_2+c_3y_3$ such that $c_1+c_2+c_3=1$.
How do i find the $c_1$, $c_2$, and $c_3$ coefficients, in order to calculate the general solution?
$\endgroup$ 81 Answer
$\begingroup$You are looking for an ODE on the form :$$y''+f(x)y'+g(x)y+h(x)=0$$The first known solution is $\quad y_1=3x-2x^2\quad,\quad y'_1=3-4x \quad,\quad y''=-4. \quad$ Putting them into the above ODE leads to a first equation :$$-4+(3-4x)f+(3x-2x^2)g+h=0 \tag 1$$ On the same manner we put the two other known solutions into the PDE.$$(-4+10e^x+5xe^x)+(-4x+5e^x+5xe^x)f+(-2x^2+5xe^x)g+h=0 \tag 2$$$$(-4+6e^x+3xe^x)+(2-4x+3e^x+3xe^x)f+(2x-2x^2+3xe^x)g+h=0\tag 3$$The system of three equations $(1)$ , $(2)$ , $(3)$ can be solved for the unknowns $f,g,h$. $$\begin{bmatrix} f &\\ g \\ h \end{bmatrix} = \begin{bmatrix} (3-4x) & (3x-2x^2) & 1 &\\ (-4x+5e^x+5xe^x) & (-2x^2+5xe^x) & 1\\ (2-4x+3e^x+3xe^x) & (2x-2x^2+3xe^x) & 1 \end{bmatrix}^{-1} \begin{bmatrix} 4 \\ 4-10e^x-5xe^x \\ 4-6e^x-3xe^x \end{bmatrix} $$The result is :$$\begin{bmatrix} f &\\ g \\ h \end{bmatrix} = \begin{bmatrix} -\frac{x+2}{x}\\ \frac{x+2}{x^2}\\ -2x \end{bmatrix}$$The ODE is :$$ \boxed{y''-\frac{x+2}{x}y'+\frac{x+2}{x^2}y=2x}$$
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