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Solve the initial-value problem $\frac{d^2y}{dt^2}+14\frac{dy}{dt}+49y=0,\:y(1)=0,\:y′(1)=1$

I tried to solve this problem however the answer that I get is wrong. Since I don't have access to any answer key I don't understand where I am making the mistake.

I started with finding the roots:

$$r^2 +14r +49=0$$ $$(r+7)^2$$

so the equation has a double root at $r=-7$

$$y(t)=c_1e^{-7t} + c_2te^{-7t}$$ $$y'(t)=-7c_1e^{-7t} + c_2(t*-7e^{-7t}+ e^{-7t})$$

using the inital values given I got:

$$y(1)=c_1e^{-7} + c_2e^{-7}=0$$ $$y'(1)=-7c_1e^{-7} + c_2(-7e^{-7}+ e^{-7})=1$$

When I then solve for $c_1$ in the second equation and substitute it in in the first equation I get that:

$$c_1 =\frac{1}{13e^{-7}}e^{-7t}$$ $$c_2 =-\frac{1}{13e^{-7}}e^{-7t}$$

so $$y(t)= \frac{1}{13e^{-7}}e^{-7t} -\frac{1}{13e^{-7}}te^{-7t}$$

I would appreciate it if someone could tell me what I am doing wrong.

EDIT

I noticed that I substituted in $c_1=c_2$ instead of $c_1=-c_2$. Thank you all for your help!

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    $\begingroup$ Just a typo notice, it is not $x$ but $t$. i'm continuing reading the rest... $\endgroup$
    – zwim
    Sep 29, 2019 at 11:38
  • $\begingroup$ You probably made a sign error where you should have computed $7+(-6)=1$. $\endgroup$ Sep 29, 2019 at 11:45

2 Answers 2

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Your working is right until near the end, where you have solved for $c_1$ and $c_2$ incorrectly.

$y(1)=0$ gives $c_1+c_2=0$.

$y'(1)=0$ gives $-7c_1-6c_2=e^7$.

Hence $c_2=e^7$ and $c_1=-e^7$.

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When you have a condition at $1$ like this you can use $(t-1)$ as a variable instead, since $d(t-1)=dt$ the ODE is not changed.

$y(t)=(c_1+c_2(t-1))e^{-7(t-1)}$

This gives you $\begin{cases}y(1)=c_1=0\\y'(1)=c_2=1\end{cases}\quad$ and $y(t)=(t-1)e^{-7(t-1)}$

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