Solving the IVP $8y''+26y=0$, $y(0)=2$, and $y'(0)=7$

Question

Im supposed to find $y$ as a function of $t$ given the equation:

$$8y''+26y=0$$

Initial values are $y(0)=2$ and $y'(0)=7$.

I found the function of $y$ is $y=c_1\cos (\frac{\sqrt{13}}{2}t)+c_2\sin(\frac{\sqrt{13}}{2}t)$. Then I take the derivative, $y'=-\frac{\sqrt{13}}{2}c_1 \sin (\frac{\sqrt{13}}{2}t)+\frac{\sqrt{13}}{2}c_2\cos (\frac{\sqrt{13}}{2}t)$. Then using the initial values i'd solve for both arbitrary constants $c_1$ and $c_2$ and I ended up getting $2$ and $\frac{14}{\sqrt{13}}$. However this answer isn't correct. There's gotta be a mistake somewhere but I've been struggling to find it.

Answer 1

The DE's characteristic equation has two distinct complex roots: $+\frac{\sqrt{13}}{2}i$ and $-\frac{\sqrt{13}}{2}i$, so the general solution (indefinite integral) is:

$$y=c_1\cos (\frac{\sqrt{13}}{2}t)+c_2\sin(\frac{\sqrt{13}}{2}t)$$

First initial condition, $y(0)=2$:

$$2=c_1\cos (\frac{\sqrt{13}}{2}0)+c_2\sin(\frac{\sqrt{13}}{2}0) \implies c_1=2$$

Second initial condition, $y'(0)=7$:

$$y'=-\frac{\sqrt{13}}{2}c_1 \sin (\frac{\sqrt{13}}{2}t)+\frac{\sqrt{13}}{2}c_2\cos (\frac{\sqrt{13}}{2}t)$$

$$7=-\frac{\sqrt{13}}{2}c_1 \sin (\frac{\sqrt{13}}{2}0)+\frac{\sqrt{13}}{2}c_2\cos (\frac{\sqrt{13}}{2}0) \implies c_2=\frac{14}{\sqrt{13}}$$

Your solution is correct.

Answer 2

What you say you did isnt correct. But you don't show exactly what you did or how you know it was not correct! Yes, the general solution to this equation is $y(t)= c_1\cos\left(\frac{\sqrt{13}}{2}t\right)+ c_2 sin\left(\frac{\sqrt{13}}{2}t\right)$. Yes, $y'(t)= -c_1\frac{\sqrt{13}}{2}\sin\left(\frac{\sqrt{13}}{2}t\right)+ c_2\frac{\sqrt{13}}{2} cos\left(\frac{\sqrt{13}}{2}t\right)$.

Setting t= 0, $y(0)= c_1= 2$ and $y'(0)= c_2\frac{\sqrt{13}}{2}= 7$ so $c_2= \frac{14}{\sqrt{13}}= \frac{14\sqrt{13}}{13}$. You say that is not correct. How do you know that?

Answer 3

Use Laplace transform:

$$8y''(t)+26y(t)=0\Longleftrightarrow$$ $$\mathcal{L}_t\left[8y''(t)+26y(t) \right]_{(s)}=\mathcal{L}_t\left[0\right]_{(s)}\Longleftrightarrow$$ $$\mathcal{L}_t\left[8y''(t)\right]_{(s)}+\mathcal{L}_t\left[26y(t) \right]_{(s)}=\mathcal{L}_t\left[0\right]_{(s)}\Longleftrightarrow$$ $$8\cdot\mathcal{L}_t\left[y''(t)\right]_{(s)}+26\cdot\mathcal{L}_t\left[y(t) \right]_{(s)}=\mathcal{L}_t\left[0\right]_{(s)}\Longleftrightarrow$$

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Now, use:

• $$\mathcal{L}_t\left[y(t) \right]_{(s)}=\text{Y}(s)$$
• $$\mathcal{L}_t\left[y''(t) \right]_{(s)}=s^2\text{Y}(s)-sy(0)-y'(0)$$
• $$\mathcal{L}_t\left[0 \right]_{(s)}=0$$

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$$8\cdot\left(s^2\text{Y}(s)-sy(0)-y'(0)\right)+26\text{Y}(s)=0\Longleftrightarrow$$

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Use the intitial conditions $y(0)=2$ and $y'(0)=7$:

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$$8\cdot\left(s^2\text{Y}(s)-2s-7\right)+26\text{Y}(s)=0\Longleftrightarrow$$ $$8s^2\text{Y}(s)-16s-56+26\text{Y}(s)=0\Longleftrightarrow$$ $$\text{Y}(s)\left[8s^2+26\right]=16s+56\Longleftrightarrow$$ $$\text{Y}(s)=\frac{16s+56}{8s^2+26}\Longleftrightarrow$$ $$\mathcal{L}_s^{-1}\left[\text{Y}(s)\right]_{(t)}=\mathcal{L}_s^{-1}\left[\frac{16s+56}{8s^2+26}\right]_{(t)}\Longleftrightarrow$$ $$y(t)=2\cos\left(\frac{t\sqrt{13}}{2}\right)+\frac{14\sin\left(\frac{t\sqrt{13}}{2}\right)}{\sqrt{13}}$$