## Related questions with answers

Solve the IVP. Check that your answer satisfies the ODE as well as the initial conditions. Show the details of your work. y''+25y=0, y(0)=4.6, y'(0)=-1.2

Solution

VerifiedWe have

$y''+25y=0$

The characteristic equation is,

$\begin{align*} \lambda^{2} + 25 &= 0 \\ \lambda^{2} &= -25 \\ \lambda &= \sqrt{-25}\\ \text{ its roots:} \quad \lambda &= \pm 5i \end{align*}$

where we can see that we have a case of a complex conjugate, where

$\begin{gather*} a + bi = 0 \\ 0 \pm 5i = 0 \\ a = 0 \quad \text{and} \quad b= 5 \end{gather*}$

which means that the general solution of this Second-Order Linear ODE, would be of the form

$\mathbf{y=e^{-ax}\left(C_{1}\cos b x + C_{2}\sin b x \right)}$

$\text{where} \quad \mathbf{a} = 0 \quad \text{and} \quad \mathbf{b}= 5$

Therefore,

$\begin{align*} y &= e^{0} \left( C_{1}\cos 5x + C_{2}\sin 5x\right)\\\\ &= C_{1}\cos 5x + C_{2}\sin 5x \end{align*}$

and

$\begin{align*} y' &= -5 C_{1} \sin 5x + 5C_{2}\cos 5x \end{align*}$

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