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differential equation solved by Laplace transform

We solve the differential equation

$$y"(t)+y(t)=f(t)$$

with initial condition

$$y(0)=0,\:y'(0)=0$$

where

$$f(t)=u(t-2\pi)-u(t-\pi)$$
$$\mathcal{L}(y(t)(s)=Y(s)$$
$$\mathcal{L}(y'(t)(s)=sY(s)$$
$$\mathcal{L}(y"\;(t)(s)=s^2Y(s)$$

We have

$$\mathcal{L}(y"(t)+y(t))(s)=s^2Y(s)+Y(s)=F(s)=\frac{1}{s}(e^{-2\pi s}-e^{-\pi s})$$

We have

$$Y(s)=\frac{1}{s^2+1}F(s)$$

We have from convolution theorem

$$y(t)=\sin*f(t)=\int_{0}^{t}f(\tau)\sin(t-\tau)d\tau$$

proof without words

Linear trigonometric equation

We solve the equation

$$a\sin x+b\cos x+c=0$$

where $ a,b,c \in R $ and $ a^2+b^2>0 $ .
Then we can write

$$\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\sin x+\frac{b}{\sqrt{a^2+b^2}}\cos x+\frac{c}{\sqrt{a^2+b^2}})=0$$

There is a unique $ t\in[0,2\pi) $ such as

$$\cos t=\frac{a}{\sqrt{a^2+b^2}}$$

and

$$\sin t=\frac{b}{\sqrt{a^2+b^2}}$$

.
so we can write

$$\sqrt{a^2+b^2}(\sin x\cos t+\cos x\sin t+\frac{c}{\sqrt{a^2+b^2}})=0$$

or

$$\sqrt{a^2+b^2}(\sin (x+t)+\frac{c}{\sqrt{a^2+b^2}})=0$$

Now, if

$$|\frac{c}{\sqrt{a^2+b^2}}|\leq 1$$

we have a countable set of solutions

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