Eliza Shields

2023-03-24

The solution of a differential equation y′′+3y′+2y=0 is of the form

A) ${c}_{1}{e}^{x}+{c}_{2}{e}^{2x}$

B) ${c}_{1}{e}^{-x}+{c}_{2}{e}^{3x}$

C) ${c}_{1}{e}^{-x}+{c}_{2}{e}^{-2x}$

D) ${c}_{1}{e}^{-2x}+{c}_{2}{2}^{-x}$

A) ${c}_{1}{e}^{x}+{c}_{2}{e}^{2x}$

B) ${c}_{1}{e}^{-x}+{c}_{2}{e}^{3x}$

C) ${c}_{1}{e}^{-x}+{c}_{2}{e}^{-2x}$

D) ${c}_{1}{e}^{-2x}+{c}_{2}{2}^{-x}$

Haylee French

Beginner2023-03-25Added 7 answers

Step 1

The correct answer is C) ${c}_{1}{e}^{-x}+{c}_{2}{2}^{-2x}$

Given DE is y′′+3y′+2y=0

$\Rightarrow ({D}^{2}+3D+2)y=\mathrm{0...}(i)$

AE is ${m}^{2}+3m+2=0$

⇒m=−1 and −2

Step 2

General solution is

$\therefore y={c}_{1}{e}^{-x}+{c}_{2}{e}^{-2x}$

The correct answer is C) ${c}_{1}{e}^{-x}+{c}_{2}{2}^{-2x}$

Given DE is y′′+3y′+2y=0

$\Rightarrow ({D}^{2}+3D+2)y=\mathrm{0...}(i)$

AE is ${m}^{2}+3m+2=0$

⇒m=−1 and −2

Step 2

General solution is

$\therefore y={c}_{1}{e}^{-x}+{c}_{2}{e}^{-2x}$

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