Question

In Problems $\mathbf{1}\mathbf{-}\mathbf{16}$ the indicated function${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ is a solution of the given differential equation. Use reduction of order or formula (5), as instructed, to find a second solution${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{16}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{4}\mathit{x}$

Short answer

The second solution of the given differential equationis $\sin 4x$.

Step-by-step solution

Definition of Reduction of order

• Reduction of order is a method for solving second-order linear ordinary differential equations in mathematics.
• When one solution is known and a second linearly independent solution is sought, this method is used.
• The approach can be used to solve n-th order equations as well.
• $y_2=y_1\left(x\right)\int \frac{e^{-\int P\left(x\right)dx}}{y_1^2\left(x\right)}dx$

Find the second solution

The second order differential equation

$y\text{'}\text{'}+16y=0$

as the form

$y\text{'}\text{'}+p\left(x\right)y\text{'}+q\left(x\right)y=0,\text{then we have}p\left(x\right)=0\text{and}q\left(x\right)=16$

with a first solution$y_1\left(x\right)=\cos 4x$, and we have to obtain the second solution using the formula of reduction of order as

$$\begin{gathered} y_2\left(x\right)=y_1\left(x\right)\int \frac{e^{-\int p\left(x\right)dx}}{{\left(y_1\right)}^2}dx \\ =\cos 4x\int \frac{e^{-\int 0dx}}{{(\cos 4x)}^2}dx \\ =\cos 4x\int \frac{k}{{\cos }^{2}4x}dx \\ =\cos 4x\int k{\sec }^{2}4xdx \\ =\cos 4x\times k\times \frac{1}{4}\tan 4x \\ =\frac{k}{4}\cos 4x\times \frac{\sin 4x}{\cos 4x} \\ =\frac{k}{4}\sin 4x \\ =c_2\sin 4x \\ =\sin 4x \end{gathered}$$

Then, the second solution of the given differential equation$y\text{'}\text{'}+16y=0$, where $c_2=\frac{k}{4}$and we let it equal 1 .

Therefore, the secondsolution of given differential equationis$\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{4}\mathit{x}$