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6.49 Solving differential equations with Laplace transforms

Example.

Solve the initial value problem

\frac{dy}{dx}+y=\frac{5}{2}\sin 2x,\;\;\;y(0)=0.

We apply the Laplace transform to both sides, to obtain:

{\mathcal{L}}\left(\frac{dy}{dx}+y\right)={\frac{5}{2}\frac{2}{s^{2}+4}=}\,{% \frac{5}{s^{2}+4}.}

Now the left-hand side is

{\mathcal{L}}\left(\frac{dy}{dx}\right)+{\mathcal{L}}(y)=(-y(0)+s{\mathcal{L}}% (y))+{\mathcal{L}}(y)=(s+1){\mathcal{L}}(y).

Now we divide through by $(s+1)$ to obtain ${\mathcal{L}}(y)={\frac{5}{(s+1)(s^{2}+4)}.}$ Now we apply ${\mathcal{L}}^{-1}$ to get $y=e^{-x}+\frac{1}{2}\sin 2x-\cos 2x$ .