fanyattehedzg

Answered question

2021-12-30

Find the solution of the following Second Order Differential Equation
${y}^{″}-9y=0,y\left(0\right)=2,{y}^{\prime }\left(0\right)=0$

Answer & Explanation

Mason Hall

Beginner2021-12-31Added 36 answers

We have given
$y9y=0$
Auxiliary equation of given differential equation
${m}^{2}-9=0$
$\left(m-3\right)\left(m+3\right)=0$
$m=3,-3$
Therefore, general solution of differential equation
$y={c}_{1}{e}^{3x}+{c}_{2}{e}^{-3x}$
We have $y\left(0\right)=2$
$⇒y\left(0\right)={c}_{1}{e}^{3\left(0\right)}+{c}_{2}{e}^{-3\left(0\right)}$
$2={c}_{1}+{c}_{2}$ (i)
${y}^{\prime }\left(x\right)=3{c}_{1}{e}^{3x}-3{c}_{2}{e}^{-3x}$
We have ${y}^{\prime }\left(0\right)=0$
$⇒{y}^{\prime }\left(x\right)=3{c}_{1}{e}^{3\left(0\right)}-3{c}_{2}{e}^{-3\left(0\right)}$
$0=3{c}_{1}-3{c}_{2}$
$0={c}_{1}-{c}_{2}$ (2)
Now, solve equation (1) and (2) we get
${c}_{1}=1,{c}_{2}=1$
Therefore, particular solution is
$y={e}^{3x}+{e}^{-3x}$

Linda Birchfield

Beginner2022-01-01Added 39 answers

$y9y=0,y\left(0\right)=2,{y}^{\prime }\left(0\right)=0$
The auxiliary equation for this equation is:
${m}^{2}-9=0$
On solving this equation for "m".
${m}^{2}-9=0$
${m}^{2}=9$
$m=±\sqrt{9}$
$m=±3$
$m=-3,3$
Then, the solution of this given differential equation is:
$y={c}_{1}{e}^{-3x}+{c}_{2}{e}^{3x}$
Given that,
$y\left(0\right)=2$
So, replace x with 0 and y with 2 in it.
$y={c}_{1}{e}^{-3x}+{c}_{2}{e}^{3x}$
$2={c}_{1}{e}^{-3\left(0\right)}+{c}_{2}{e}^{3\left(0\right)}$
$2={c}_{1}+{c}_{2}$ (1)
Now, differentiate the obtained solution with respect to x.
${y}^{\prime }=-3{c}_{1}{e}^{-3x}+3{c}_{2}{e}^{3x}$
And, also given that,
${y}^{\prime }\left(0\right)=0$
So, replace x with 0 and y with 0 in it.
${y}^{\prime }=-3{c}_{1}{e}^{-3x}+3{c}_{2}{e}^{3x}$
$0=-3{c}_{1}{e}^{-3\left(0\right)}+3{c}_{2}{e}^{3\left(0\right)}$
$0=-3{c}_{1}+3{c}_{2}$ (2)
Now, solve equations (1) and (2) by the elimination or substitution method.
From equation (2), ${c}_{1}={c}_{2}$, put this value in equation (1), which gives ${c}_{1}={c}_{2}=1$
Then, ${c}_{1}=1,{c}_{2}=1$
Thus, the solution of the given differential equation is:
$y={c}_{1}{e}^{-3x}+{c}_{2}{e}^{3x}$
$y=1{e}^{-3x}+1{e}^{3x}$
$y={e}^{-3x}+{e}^{3x}$

karton

Skilled2022-01-09Added 439 answers

$\begin{array}{}{y}^{″}-9y=0\\ {m}^{2}-9=0\\ \left(m-3\right)\left(m+3\right)=0\\ m=3,-3\\ y={c}_{1}{e}^{3x}+{c}_{2}{e}^{-3x}\\ y\left(0\right)=2\\ ⇒y\left(0\right)={c}_{1}{e}^{3\left(0\right)}+{c}_{2}{e}^{-3\left(0\right)}\\ 2={c}_{1}+{c}_{2}\left(i\right)\\ {y}^{\prime }\left(x\right)=3{c}_{1}{e}^{3x}-3{c}_{2}{e}^{-3x}\\ {y}^{\prime }\left(0\right)=0\\ ⇒{y}^{\prime }\left(x\right)=3{c}_{1}{e}^{3\left(0\right)}-3{c}_{2}{e}^{-3\left(0\right)}\\ 0=3{c}_{1}-3{c}_{2}\\ 0={c}_{1}-{c}_{2}\left(2\right)\\ {c}_{1}=1,{c}_{2}=1\\ y={e}^{3x}+{e}^{-3x}\end{array}$

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