Find the solution of the following Second Order Differential Equationy″−9y=0,y(0)=2,y′(0)=0

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Answered question

2021-12-30

Find the solution of the following Second Order Differential Equation ${y}^{\u2033}-9y=0,y(0)=2,{y}^{\prime}(0)=0$

Answer & Explanation

Mason Hall

Beginner2021-12-31Added 36 answers

We have given $y9y=0$ Auxiliary equation of given differential equation ${m}^{2}-9=0$ $(m-3)(m+3)=0$ $m=3,-3$ Therefore, general solution of differential equation $y={c}_{1}{e}^{3x}+{c}_{2}{e}^{-3x}$ We have $y\left(0\right)=2$ $\Rightarrow y\left(0\right)={c}_{1}{e}^{3\left(0\right)}+{c}_{2}{e}^{-3\left(0\right)}$ $2={c}_{1}+{c}_{2}$ (i) $y}^{\prime}\left(x\right)=3{c}_{1}{e}^{3x}-3{c}_{2}{e}^{-3x$ We have ${y}^{\prime}\left(0\right)=0$ $\Rightarrow {y}^{\prime}\left(x\right)=3{c}_{1}{e}^{3\left(0\right)}-3{c}_{2}{e}^{-3\left(0\right)}$ $0=3{c}_{1}-3{c}_{2}$ $0={c}_{1}-{c}_{2}$ (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". ${m}^{2}-9=0$ ${m}^{2}=9$ $m=\pm \sqrt{9}$ $m=\pm 3$ $m=-3,3$ 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: $y={c}_{1}{e}^{-3x}+{c}_{2}{e}^{3x}$ $y=1{e}^{-3x}+1{e}^{3x}$ $y={e}^{-3x}+{e}^{3x}$