- #1

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## Main Question or Discussion Point

Here are two ways:

$$(x-{ x }_{ 1 })({ x }-{ x }_{ 2 })=0\\ x-{ x }_{ 1 }=0\quad \quad \quad \quad x-{ x }_{ 2 }=0\\ \\$$$$ \\ { e }^{ x }({ c }_{ 1 }-3{ c }_{ 2 })+{ e }^{ -32x }({ c }_{ 5 }-{ c }_{ 4 })=0\\ { c }_{ 1 }-3{ c }_{ 2 }=0\quad \quad \quad \quad \quad { c }_{ 5 }-{ c }_{ 4 }=0$$

Any other ways?

$$(x-{ x }_{ 1 })({ x }-{ x }_{ 2 })=0\\ x-{ x }_{ 1 }=0\quad \quad \quad \quad x-{ x }_{ 2 }=0\\ \\$$$$ \\ { e }^{ x }({ c }_{ 1 }-3{ c }_{ 2 })+{ e }^{ -32x }({ c }_{ 5 }-{ c }_{ 4 })=0\\ { c }_{ 1 }-3{ c }_{ 2 }=0\quad \quad \quad \quad \quad { c }_{ 5 }-{ c }_{ 4 }=0$$

Any other ways?