Question

In Problems 29-36 solve the given initial-value problem.

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{\theta }}^{\mathbf{2}}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{(}\mathit{\pi }\mathbf{/}\mathbf{3}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{(}\mathit{\pi }\mathbf{/}\mathbf{3}\mathbf{)}\mathbf{=}\mathbf{2}$

Short answer

$y=-\sqrt{3}\cos x+\sin x$

Step-by-step solution

Homogeneous differential equation

• Linear Homogeneous Differential Equations of Higher Order with Constant Coefficients where are constants that can be either real or complex is known as the differential equation's characteristic equation.
• A polynomial of degree has exactly roots, counting multiplicity, according to the fundamental theorem of algebra.

Determine the initial value of differential equation

We have the second order differential equation

$\frac{d^{2}y}{d{\theta }^2}+y=0······\left(1\right)$

With the initial conditions

$y\left(\frac{\pi }{3}\right)=0,\text{and}y\text{'}\left(\frac{\pi }{3}\right)=2$

And we have to get its general solution by the following technique:

First, assume that that $y=e^{\lambda \theta }$ is a solution for the homogeneous differential equation.

Second, differentiate the assumption with respect to theta, then we have

$y\text{'}=\lambda e^{\lambda \theta }$

And

$y\text{'}\text{'}={\lambda }^2{e}^{\lambda \theta }······\left(2\right)$

Third, substitute with $y=e^{\lambda \theta }$, equation (2) into the differential equation (1), then we obtain

$$\begin{gathered} {\lambda }^2{e}^{\lambda \theta }+e^{\lambda \theta }=0 \\ {e}^{\lambda \theta }\left({\lambda }^2+1\right)=0 \end{gathered}$$

Since $e^{\lambda \theta }$ cannot be equal to 0, then we have

$$\begin{gathered} {\lambda }^2+1=0 \\ {\lambda }^2=-1 \end{gathered}$$

Then the roots are ${\lambda }_{1,2}=\sqrt{-1}=\pm i$which are conjugate complex

Then the general solution of this homogeneous equation $y^{\text{'}\text{'}}+y=0$ can be given as

Where $k_1=c_1+c_{2}and{k}_2=c_1-c_2$and are an arbitrary constants.

We have to apply with the point $(\theta ,y)=\left(\frac{\pi }{3},0\right)$ in the general solution (3) as

$$\begin{gathered} 0=k_1\cos \left(\frac{\pi }{3}\right)+k_2\sin \left(\frac{\pi }{3}\right) \\ \text{=}\frac{1}{2}{k}_1+\frac{\sqrt{3}}{2}{k}_2 \end{gathered}$$

Then we have

$k_1+\sqrt{3}{k}_2=0······\left(a\right)$

After that, we have to take the first derivative for the general solution of the given differential equation as

$y\text{'}=-k_1\sin x+k_2\cos x$

Then apply with the point $\left(\theta ,y\text{'}\right)=\left(\frac{\pi }{3},2\right)$as

$$\begin{gathered} 2=-k_1\sin \left(\frac{\pi }{3}\right)+k_2\cos \left(\frac{\pi }{3}\right) \\ -\frac{\sqrt{3}}{2}{k}_1+\frac{1}{2}{k}_2=2 \end{gathered}$$

Then we have

$-\sqrt{3}{k}_1+k_2=4······\left(b\right)$

Then by solving the two equations (a) and (b), we find that

$k_1=-\sqrt{3}\text{and}{k}_2=1$

Finally, substitute with the values of $k_{1}and{k}_2$into the general solution (3) of our given differential equation, then we have $y=-\sqrt{3}\cos x+\sin x$is a solution of the differential equation$y^{\text{'}\text{'}}+y=0$ .

Hence, the general solution is$\mathit{y}\mathbf{=}\mathbf{-}\sqrt{\mathbf{3}}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{x}\mathbf{+}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{x}$