# Background

# Method

We want to figure out what the u's are, so we need two conditions. The first condition is $L(y_p) = f(x)$. We still need another, and we'll find a condition that simplifies our arithmetic. $$y_p = u_1y_1 + u_2y_2$$ $$y'_p = (u_1y'_1 + u'_1y_1) + (u_2y'_2 + u'_2y_2)$$ Here is where we get our second condition: $u'_1y_1 + u'_2y_2 = 0$. This will really simplify arithmetic. $$y''_p = (u_1y''_1 + u_2y''_2) + (u'_1y'_1 + u'_2y'_2)$$ Now, we go back to our original equation. $y'' + Py' + Qy = 0$. This works for any y. So, we rewrite $y''_p$. $$y''_p = (u_1y''_1 + u_2y''_2) + (u'_1y'_1 + u'_2y'_2)$$ $$y''_p = u_1(-Py'_1 - Qy_1) + u_2(-Py'_2 - Qy_2) + (u'_1y'_1 + u'_2y'_2)$$ $$y''_p = (u'_1y'_1 + u'_2y'_2) - P(u_1y'_1 + u_2y'_2) - Q(u_1y_1 + u_2y_2)$$ Finally, we can rewrite this one last time, noting that the terms multiplied by P and Q are $y_p$ and $y'_p$. $$y''_p = (u'_1y'_1 + u'_2y'_2) - Py'_p - Qy_p \to y''_p + Py'_p + Qy_p = u'_1y'_1 + u'_2y'_2 = f(x)$$ We have our two conditions: $$\boxed{u'_1y_1 + u'_2y_2 = 0}$$ $$\boxed{u'_1y'_1 + u'_2y'_2 = f(x)}$$

# General Case

# Example

**many**times. The determinant of that matrix is $\dfrac{30}{x}$. In order to solve this system of equations, we must use something called Cramer's rule. The rule (in this case) is this: $$u'_1 = \dfrac{\begin{vmatrix}0 & x^3 & x^{-2} \\ 0 & 3x^2 & -2x^{-3} \\ 30x^{-2} & 6x & 6x^{-4}\end{vmatrix}}{\frac{30}{x}} = \frac{-5}{x}$$ You replace the column you want to solve with the solution column. $$u'_2 = \dfrac{\begin{vmatrix}x & 0 & x^{-2} \\ 1 & 0 & -2x^{-3} \\ 0 & 30x^{-2} & 6x^{-4}\end{vmatrix}}{\frac{30}{x}} = \frac{3}{x^3}$$ $$u'_3 = \dfrac{\begin{vmatrix}x & x^3 & 0 \\ 1 & 3x^2 & 0 \\ 0 & 6x & 30x^{-2}\end{vmatrix}}{\frac{30}{x}} = 2x^2$$ Because all of the terms are derivatives, we must integrate. $$u_1 = -5\ln(x), u_2 = \dfrac{-3}{2x^2}, u_3 = \frac{2}{3}x^3$$ We finally get our particular solution. $$y_p = -5\ln(x)(x) + \dfrac{-3}{2x^2}(x^3) + \dfrac{2}{3}x^3(x^{-2}) = -\dfrac{5}{6}x - 5x\ln(x)$$ You may have noticed that there was an x term in the particular solution

**and**the homogeneous solution. That doesn't mean you did anything wrong. When you get your full solution, that will be swallowed up by the constant. The entire solution for this different equation is $$y = c_1x + c_2x^3 + c_3x^{-2} - 5x\ln(x)$$

David Witten