Question

In Problems 39-44 solve the given differential equation subject to the indicated conditions.

${\mathit{y}}^{\mathbf{\text{'}\text{'}}}\mathbf{+}\mathit{y}\mathbf{=}{\mathbf{sec}}^{\mathbf{3}}\mathit{x}\mathbf{,}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$

Short answer

The solution is$\mathrm{y}\left(\mathrm{x}\right)={\mathrm{y}}_{\mathrm{h}}\left(\mathrm{x}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\frac{1}{2}\text{sin}\mathrm{x}+\frac{3}{2}\text{cos}\mathrm{x}-\frac{1}{2}\text{cos}\left(2\mathrm{x}\right)\text{sec}\left(\mathrm{x}\right)$

Step-by-step solution

Given Data

Differential equation is the equation which relate one or more unknown functions and their derivatives.

Given differential equation,

$y^{\text{'}\text{'}}+y={\text{sec}}^{3}x,y\left(0\right)=1,y^{\text{'}}\left(0\right)=\frac{1}{2}$

Find homogeneous equation

Given differential equation:

$\begin{array}{c}{y}^{\text{'}\text{'}}+y={\text{sec}}^{3}x\\ y\left(0\right)=1\\ y^{\text{'}}\left(0\right)=\frac{1}{2}\end{array}$

First determine the solution of the homogeneous differential equation

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

Subtract$y$from each side:

$y^{\text{'}\text{'}}=-y$

We know that the sine and the cosine both satisfy this differential equation and thus the solution is a linear combination of the sine and cosine:

$y_h=c_1\text{sin}x+c_2\text{cos}x$

Find particular solution and general solution

Next determine a particular solution of$y^{\text{'}\text{'}}+y={\text{sec}}^{3}x$.

$q\left(x\right)={\text{sec}}^{3}x$

Result homogeneous solution:

$\begin{array}{rr}& y_1\left(x\right)=\text{cos}\left(x\right)\\ & y_2\left(x\right)=\text{sin}\left(x\right)\end{array}$

Determine the Wronskian:

$\omega \left(t\right)=\left|\begin{array}{ll}{y}_1\left(x\right)& y_2\left(x\right)\\ y_1^{\text{'}}\left(x\right)& y_2^{\text{'}}\left(x\right)\end{array}\right|=\left|\begin{array}{cc}\text{cos}x& \text{sin}x\\ -\text{sin}x& \text{cos}x\end{array}\right|=1$

Then determine the function $G(x,t)$:

$G(x,t)=\frac{y_1\left(t\right)y_2^{\text{'}}\left(x\right)+y_1^{\text{'}}\left(t\right)y_2\left(x\right)}{\omega \left(t\right)}=\frac{\text{cos}t\text{sin}x-\text{sin}t\text{cos}x}{1}=\text{sin}(x-t)$

Finally, we then know that the particular solution is:

$y_p\left(x\right)={\int }^{\text{}}G(x,t)q\left(t\right)dt={\int }^{\text{}}\text{sin}(x-t){\text{sec}}^{3}tdt=-\frac{1}{2}\text{cos}\left(2x\right)\text{sec}\left(x\right)$

Note: You can also rewrite this particular solution as$y_p=\text{sin}{\left(x\right)}^2\text{sec}x-\frac{1}{2}\text{sec}x=\text{sin}x\text{tan}x-\frac{1}{2}\text{sec}x$

Final Answer

The solution of the differential equation is the sum of the homogeneous and particular solution:

$y\left(x\right)=y_h\left(x\right)+y_p\left(x\right)=c_1\text{sin}x+c_2\text{cos}x-\frac{1}{2}\text{cos}\left(2x\right)\text{sec}\left(x\right)$

Finally determine the constants:

$\begin{array}{c}y\left(0\right)=c_2-\frac{1}{2}\\ y^{\text{'}}\left(0\right)=c_1\end{array}$

We also know that$y\left(0\right)$is equal to 1 and$y^{\text{'}}\left(0\right)$is equal to$1/2$, thus

$\begin{array}{c}{c}_2=1+\frac{1}{2}=\frac{3}{2}\\ c_1=\frac{1}{2}\end{array}$

Thus, the solution then becomes:

$\mathrm{y}\left(\mathrm{x}\right)={\mathrm{y}}_{\mathrm{h}}\left(\mathrm{x}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\frac{1}{2}\text{sin}\mathrm{x}+\frac{3}{2}\text{cos}\mathrm{x}-\frac{1}{2}\text{cos}\left(2\mathrm{x}\right)\text{sec}\left(\mathrm{x}\right)$