Answer to Question #199969 in Differential Equations for sania khan

Question #199969

solve undetermined coefficient y''+y=2x+3ex

1
Expert's answer
2021-06-17T09:48:09-0400

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

1)y+y=0y''+y=0

k2+1=0k^2 +1=0

k1=i,k2=ik_1 = i, k_2=-i

y=C1cos(x)+C2sin(x)y=C_1\cos(x)+C_2\sin(x)

k1+ik2=i+1k1k2k_1+ik_2=i+1\not= k_1\not= k_2

y=ek1An(x)=Aeiy=e^{k_1}A_n(x)=Ae^{i}

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

y=Aex+Bxy=Ae^x+Bx

y=Aex+By'= Ae^x+B

y=Aexy'' = Ae^x

y+y=Aex+Aex+Bx=2x+3exy'' + y' = Ae^x+Ae^x+Bx=2x+3e^x

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

y=C1cos(x)+C2sin(x)+32ex+2xy=C_1\cos(x) +C_2\sin(x)+\dfrac{3}{2}e^x+2x

Checking:

y=C1cos(x)C2sin(x)+32exy'' = -C_1\cos(x) - C_2\sin(x)+\dfrac{3}{2}e^x

y+y=C1cos(x)C2sin(x)+32ex+C1cos(x)+C2sin(x)+32ex+2x=3ex+2xy'' + 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=C1cos(x)+C2sin(x)+32ex+2xy=C_1\cos(x) +C_2\sin(x)+\dfrac{3}{2}e^x+2x


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