Question

Solve Laplace transforms and the convolution integral or by Green functions.

$\mathbf{y}\mathbf{\text{'}\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{sec}}^{\mathbf{2}}\mathbf{}\mathbf{t}$

Short answer

The general solution of the equation $y\text{'}\text{'}+y={\sec }^{2}t$is$y\left(t\right)=A\sin \left(t\right)+B\cos \left(t\right)+\sin \left(t\right)\mathrm{In}\left(sec\right(t)+\tan (t\left)\right)-1$

Step-by-step solution

Given Information

The given expressions are$y\text{'}\text{'}+y=se{c}^{2}t$

Green Function

The impulse response of an inhomogeneous linear differential operator defined on a domain with specified initial conditions or boundary conditions is known as a Green's function.

Green Function

Use the green function, where the given differential equation have the following form

$y\text{'}\text{'}+{\omega }^{2}y=f\left(t\right)...\left(2\right)$

Compare equation (1) and (2)

$$\begin{gathered} \omega =1 \\ f\left(t\right)=\frac{1}{\cos {\left(t\right)}^2} \end{gathered}$$

And, the solution to such differential equation is,

$y{\left(t\right)}_p=\frac{1}{\omega }{\int }_0^t\sin \left(\omega \left(t-t\text{'}\right)\right)f\left(t\text{'}\right)dt$

Hence, the solution is thus,

$y{\left(t\right)}_p=\frac{1}{\omega }{\int }_0^t\sin \left(t-t\text{'}\right)sec{\left(t\text{'}\right)}^{2}dt\text{'}$

The general solution of the equation

The sec function is

$\sec {\left(t\right)}^2=\frac{1}{\cos {\left(t\right)}^2}$

And, use double angle identity that

$$\begin{gathered} \alpha =0 \\ \beta =1 \end{gathered}$$

Thus,

$y\left(t\right)=e^0\left[A\sin \left(t\right)+B\cos \left(t\right)\right]=A\sin \left(t\right)+B\cos \left(t\right)$

And, hence the general solution to the differential equation, is thus

$y\left(t\right)=y{\left(t\right)}_c=y{\left(t\right)}_p$

And, thus

$y\left(t\right)=A\sin \left(t\right)+B\cos \left(t\right)+\sin \left(t\right)In\left(sec\left(t\right)+\tan \left(t\right)\right)+\cos \left(t\right)-1$

$B=B+1$

Thus,

$$\begin{gathered} y\left(t\right)=A\sin \left(t\right)+(B+1)\cos \left(t\right)+\sin \left(t\right)\mathrm{In}\left(sec\right(t)+\tan (t\left)\right)+\cos \left(t\right)-1n\&y\left(t\right) \\ =A\sin \left(t\right)+B\cos \left(t\right)+\sin \left(t\right)In\left(sec\right(t)+\tan (t\left)\right)-1 \end{gathered}$$

Hence, the general solution of the $y\text{'}\text{'}+y=se{c}^{2}t$

$y\left(t\right)=A\sin \left(t\right)+B\cos \left(t\right)+\sin \left(t\right)In\left(sec\right(t)+\tan (t\left)\right)-1$