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differential equation solved by Laplace transform

We solve the differential equation


with initial condition




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


We have from convolution theorem


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

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


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

so we can write

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


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

Now, if

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

we have a countable set of solutions

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