Answer to Question #199969 in Differential Equations for sania khan

Question

solve undetermined coefficient y''+y=2x+3e^x

Answer 1

$y'' +y=2x+3e^x$

1)$y''+y=0$

$k^2 +1=0$

$k_1 = i, k_2=-i$

$y=C_1\cos(x)+C_2\sin(x)$

$k_1+ik_2=i+1\not= k_1\not= k_2$

$y=e^{k_1}A_n(x)=Ae^{i}$

2)$y''+y=2x+3e^x$

$y=Ae^x+Bx$

$y'= Ae^x+B$

$y'' = Ae^x$

$y'' + y' = Ae^x+Ae^x+Bx=2x+3e^x$

$\begin{cases} 2A=3=>A=3/2\\ B=2 \end{cases}$

$y=C_1\cos(x) +C_2\sin(x)+\dfrac{3}{2}e^x+2x$

Checking:

$y'' = -C_1\cos(x) - C_2\sin(x)+\dfrac{3}{2}e^x$

$y'' + y = -C_1\cos(x) - C_2\sin(x)+\dfrac{3}{2}e^x+C_1\cos(x) +C_2\sin(x)+\dfrac{3}{2}e^x+2x=3e^x+2x$

Answer:

$y=C_1\cos(x) +C_2\sin(x)+\dfrac{3}{2}e^x+2x$