Answer:
See explanation
Step-by-step explanation:
1. If [tex]y=ae^x+be^{2x}+ce^{3x},[/tex] then
[tex]\dfrac{dy}{dx}=ae^x+b\cdot 2e^{2x}+c\cdot 3e^{3x}=ae^x+2be^{2x}+3ce^{3x}\\ \\\dfrac{d^2y}{dx^2}=ae^x+2b\cdot 2e^{2x}+3c\cdot 3e^{3x}=ae^x+4be^{2x}+9ce^{3x}\\ \\\dfrac{d^3y}{dx^3}=ae^x+4b\cdot 2e^{2x}+9c\cdot 3e^{3x}=ae^x+8be^{2x}+27ce^{3x}[/tex]
Consider expression
[tex]\dfrac{d^3y}{dx^3}-6\dfrac{d^2y}{dx^2}+11\dfrac{dy}{dx}-6y:\\ \\ \\ae^x+8be^{2x}+27ce^{3x}-6(ae^x+4be^{2x}+9ce^{3x})+11(ae^x+2be^{2x}+3ce^{3x})-6(ae^x+be^{2x}+ce^{3x})\\ \\=e^x(a-6a+11a-6a)+e^{2x}(8b-24b+22b-6b)+e^{3x}(27c-54c+33c-6c)\\ \\=e^x\cdot 0+e^{2x}\cdot 0+e^{3x}\cdot 0\\ \\=0[/tex]
Hence,
[tex]\dfrac{d^3y}{dx^3}-6\dfrac{d^2y}{dx^2}+11\dfrac{dy}{dx}-6y=0[/tex]
2. If [tex]y=(a+bx)e^x,[/tex] then
[tex]\dfrac{dy}{dx}=b\cdot e^x+(a+bx)e^x=e^x(a+b+bx)\\ \\ \dfrac{d^2y}{dx^2}=b\cdot e^x+(a+b+bx)\cdot e^x=e^x(a+2b+bx)[/tex]
Then
[tex]\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+y\\ \\=e^x(a+2b+bx)-2(a+b+bx)e^x+(a+bx)e^x\\ \\=e^x(a+2b+bx-2a-2b-2bx+a+bx)\\ \\=e^x((a-2a+a)+(2b-2b)+(bx-2bx+bx))\\ \\=e^x(0+0+0)\\ \\=e^x\cdot 0\\ \\=0[/tex]
Hence,
[tex]\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+y=0[/tex]
