$\dfrac{\mathrm dy}{\mathrm dt}=-4x\implies x=-\dfrac14\dfrac{\mathrm dy}{\mathrm dt}\implies\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac14\dfrac{\mathrm d^2y}{\mathrm dt^2}$

Substituting this into the other ODE gives

$-\dfrac14\dfrac{\mathrm d^2y}{\mathrm dt^2}=-4y\implies y''-16y=0$

Since $x(t)=-\dfrac14y'(t)$ , it follows that $x(0)=-\dfrac14y'(0)=4\implies y'(0)=-16$ . The ODE in $y$ has characteristic equation

$r^2-16=0$

with roots $r=\pm4$ , admitting the characteristic solution

$y_c=C_1e^{4t}+C_2e^{-4t}$

From the initial conditions we get

$y(0)=5\implies 5=C_1+C_2$

$y'(0)=16\implies-16=4C_1-4C_2$

$\implies C_1=\dfrac12,C_2=\dfrac92$

So we have

$\boxed{y(t)=\dfrac12e^{4t}+\dfrac92e^{-4t}}$

Take the derivative and multiply it by -1/4 to get the solution for $x(t)$ :

$-\dfrac14y'(t)=\boxed{x(t)=-\dfrac12e^{4t}+\dfrac92e^{-4t}}$

7 0

3 years ago

What is <br> 3.149<br> 3.149 rounded to the nearest hundredth?

grin007 [14]

Answer:

3.15

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3 0

3 years ago

Need help for this question

Bond [772]

Answer:

Step-by-step explanation:

608 square inches

5 0

3 years ago