d² xdy3Ddtdt2dy8x = 0%3Ddt2¤(t) = e - e' v3t + (c3 - 3c4) sin /3t]1-2tC2e[(3 c3 + C4) cos43 c4) sin 3t2ty(t) = c1 + c2e+e' (c3 cos 3t + c4 sin v3 t)¤(t) = cze-24 - e' [(c3 - V3ca) cos v3t + (v3cs + c4) sin /3t]1C2e42ty(t) = c1 + c2e + e'(c3 cos V3t + C4 sin 3t)%3D-2tO x(t) = c1 + c2e¯t + e' (c3 cos 3t + c4 sin v3t)1y(t) = ,c2e-4 -e' [(v3c3 + c4) cos 3t + (c3 - V3 c4) sin 3t1÷e [(V3 c3 + c4) cos 3t + (c3 – V3 c4) sin v3t-2t- V3 c4) sin /3t4-2tO x(t) = c1 + c2e+ e' (c3 cos 3t + c4 sin /3t)11y(t) = cze-24 – e' [(cs - V3 ca) cos 3t + (v3 c3 + c4) sin v3t]C2e24.

Name: d² xdy3Ddtdt2dy8x = 0%3Ddt2¤(t) = e - e' v3t + (c3 - 3c4) sin /3t]1-2
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Description: d² xdy3Ddtdt2dy8x = 0%3Ddt2¤(t) = e - e' v3t + (c3 - 3c4) sin /3t]1-2tC2e[(3 c3 + C4) cos43 c4) sin 3t2ty(t) = c1 + c2e+e' (c3 cos 3t + c4 sin v3 t)¤(t) = cze-24 - e' [(c3 - V3ca) cos v3t + (v3cs + c4) sin /3t]1C2e42ty(t) = c1 + c2e + e'(c3 cos V3t

Answered: dy dt2 dt d'y 8x = 0 - dt?

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

d² x
dy
3D
dt
dt2
dy
8x = 0
%3D
dt2
¤(t) = e - e' v3t + (c3 - 3c4) sin /3t]
1
-2t
C2e
[(3 c3 + C4) cos
4
3 c4) sin 3t
2t
y(t) = c1 + c2e
+e' (c3 cos 3t + c4 sin v3 t)
¤(t) = cze-24 - e' [(c3 - V3ca) cos v3t + (v3cs + c4) sin /3t]
1
C2e
4
2t
y(t) = c1 + c2e + e'(c3 cos V3t + C4 sin 3t)
%3D
-2t
O x(t) = c1 + c2e¯t + e' (c3 cos 3t + c4 sin v3t)
1
y(t) = ,c2e-4 -e' [(v3c3 + c4) cos 3t + (c3 - V3 c4) sin 3t
1
÷e [(V3 c3 + c4) cos 3t + (c3 – V3 c4) sin v3t
-2t
- V3 c4) sin /3t
4
-2t
O x(t) = c1 + c2e+ e' (c3 cos 3t + c4 sin /3t)
1
1
y(t) = cze-24 – e' [(cs - V3 ca) cos 3t + (v3 c3 + c4) sin v3t]
C2e
2
4.

Expert Solution

Step 1

Given that,

$\frac{d^{2} x}{d t^{2}} + \frac{d y}{d t} = 0 \frac{d^{2} y}{d t^{2}} - 8 x = 0$

Now take the derivative for the first equation with respect to $t$ ,

$\frac{d^{3}}{d t^{3}} x + \frac{d^{2} y}{d t^{2}} = 0 x''' + 8 x = 0$

Since $\frac{d^{2} y}{d t^{2}} = 8 x$ .

So for $x''' + 8 x = 0$ take $m = \frac{d}{d t}$ .

Then we get,

$m^{3} + 8 = 0 (m + 2) (m^{2} - 2 m + 4) = 0$

Where $m = - 2$ is one of the solution for the cubic equation now to find the roots for the quadratic equation,

$m^{2} - 2 m + 4 = 0$

If the quadratic is of the form $a x^{2} + b x + c = 0$ then the roots are, $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ .

So, the roots for $m^{2} - 2 m + 4 = 0$ is

$\begin{array}{rcl} m & = & \frac{2 \pm \sqrt{4 - 16}}{2} \\ = & 1 \pm \frac{i 2 \sqrt{3}}{2} \\ = & 1 \pm i \sqrt{3} \end{array}$

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