Question

Find the general solution of $y''-y=2e^x-10\sin(x)$, using the method of undetermined coefficients.

Answer 1

The general solution is $y(x) = c_1e^x + c_2e^{-x} + xe^x + 5\sin(x)$.

Answer 2

1. Complementary Solution ($y_c$): Solve the homogeneous equation $y'' - y = 0$. The characteristic equation is $m^2 - 1 = 0$, which has roots $m = \pm 1$. Therefore, the complementary solution is $y_c(x) = c_1e^x + c_2e^{-x}$.

2. Particular Solution ($y_p$): We find a particular solution for each term on the right-hand side of the non-homogeneous equation $y'' - y = 2e^x - 10\sin(x)$.

   • For $2e^x$: Since $e^x$ corresponds to a root of the characteristic equation ($m=1$), we use the form $y_{p1}(x) = Axe^x$. Substituting this into the equation yields $A=1$, so $y_{p1}(x) = xe^x$.

   • For $-10\sin(x)$: Since $\pm i$ are not roots of the characteristic equation, we use the form $y_{p2}(x) = B\sin(x) + C\cos(x)$. Substituting this into the equation yields $B=5$ and $C=0$, so $y_{p2}(x) = 5\sin(x)$.

   The particular solution is the sum of these: $y_p(x) = y_{p1}(x) + y_{p2}(x) = xe^x + 5\sin(x)$.

3. General Solution: The general solution is the sum of the complementary and particular solutions: $y(x) = y_c(x) + y_p(x) = c_1e^x + c_2e^{-x} + xe^x + 5\sin(x)$.