# Find Laplace transform of $t-\pi$

I am dealing with an Initial Value Problem of a step function:

$$y'' + y = \begin{cases} \cos t, &\text{ if }0\le t \lt \pi\\ t-\pi,&\text{ if }\pi \le t \lt \infty \end{cases}$$

I am trying to solve this using Laplace transforms.

The Laplace transform of $$\cos(t)$$ is $$\mathscr{L}\{\cos t\}=\frac{s}{s^2 + 1}$$, using the Laplace elementary transforms.

But I cannot find the Laplace transform of $$t - \pi$$.

There is no elementary transform I can use.

• Laplace transform is linear, and for a monomial you have $L(x^n)=n!/s^{n+1}$ May 16, 2019 at 23:34
• So what is the transform when i have t-π? May 16, 2019 at 23:36
• @user1584421 For equations, please refer to this MathJax tutorial. May 16, 2019 at 23:59
• Use the shift theorem May 17, 2019 at 8:32

$$\newcommand{\lap}{\mathscr{L}}\newcommand{\C}{\mathbb{C}}$$The Heaviside function $$H_a(t)$$ is defined as

$$$$H_a(t) = \begin{cases} 0, &\text{ if } t\leq a \\ 1, &\text{ otherwise} \end{cases}$$$$

The Laplace transform of $$H_a$$ is $$$$(\lap{} H_a)(s) = \frac{e^{-as}}{s}.$$$$

This function is defined for $$s\in\C$$ with $$\Re(s)>0$$.

Let us denote the right hand side of your ODE by $$u(t)$$. Then, you can verify that

$$u(t) = \cos t \cdot H_0(t) + (t-\pi-\cos t)H_\pi(t)$$

Define $$u_1(t)=\cos t$$ and $$u_2(t)=t-\pi-\cos t$$.

The Laplace transform of $$u_1(t)$$ is $$\frac{s}{s^2+1}$$. For the second term, we need to use the fact that $$\cos(t-\pi)=-\cos t$$, therefore

$$u_2(t) = (t-\pi +\cos(t-\pi))H_{\pi}(t)$$

We can now use the property

$$\lap\{f(t-a)H_a(t)\}(s) = e^{-as}F(s),$$

where $$F(s) = (\lap f)(s)$$.

I will leave the last bit to you.