r1fa8dy5
2022-03-22
Answered

In this problem, $y={c}_{1}{e}^{x}+{c}_{2}{e}^{-x}$ is a two-parameter family of solutions of the second-order DE $y{}^{\u2033}-y=0$ . Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions.

$y\left(0\right)=1,\text{}{y}^{\prime}\left(0\right)=8$

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Kingston Lowery

Answered 2022-03-23
Author has **10** answers

Solution:

$y{}^{\u2033}-y=0$ has solution $y={c}_{1}\cdot {e}^{x}+{c}_{2}\cdot {e}^{-x}$

where $c}_{1$ and $c}_{2$ are constants

so, $y\left(0\right)={c}_{1}+{c}_{2}=1$ (i)

and; $y}^{\prime}\left(x\right)={c}_{1}{e}^{x}-{c}_{2}{e}^{-x$

${y}^{\prime}\left(0\right)={c}_{1}-{c}_{2}=8$ (ii)

so, (i)+(ii) $\Rightarrow 2{c}_{1}=9,\text{}{c}_{1}=\frac{9}{2}$

(i)-(ii)$\Rightarrow 2{c}_{2}=-7,{c}_{2}=-\frac{7}{2}$

so the solution is:

$y=\frac{9}{2}\cdot {e}^{x}-\frac{7}{2}\cdot {e}^{-x}$

Jeffrey Jordon

Answered 2022-03-31
Author has **2262** answers

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At first I thought it was a Bernoulli differential equation and I tried:

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${h}^{\prime}(t)-\frac{\theta h(t)}{2}+1=0$

with $h(T)=C$ where $T$ and $C$ are constants and $T\ge 0$

At first I thought it was a Bernoulli differential equation and I tried:

$V(t)=\frac{1}{h(t)}\Rightarrow {V}^{\prime}(t)=\frac{{h}^{\prime}(t)}{h(t{)}^{2}}$

My equation becomes worse:

${V}^{\prime}(t)-\frac{\theta}{2}+\frac{1}{V(t{)}^{2}}=0$

Then I tried with the integrating factor:

$({e}^{t}h(t){)}^{\prime}=2{e}^{t}h(t)-{e}^{t}$

${e}^{t}h(t)=\int 2{e}^{t}h(t)-{e}^{t}$

I'm stuck here because $h(t)$ is not a variable, is a function and I have to find how it looks like.

Edit: I forgot: $\theta $ is a constant.

How can I solve this problem?

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