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 Post subject: Differential equations
PostPosted: Thu Jul 14, 2016 10:34 am 
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Solve the differential equation:

$$f''(x)+5f'(x)+6f(x)=0$$

where \(f'(0)=3, \;\; f(0)=2\)

and the differential equation:

$$f''(x)+f(x)=\sin 2x$$

where \(f'(0)=1, \;\; f(0)=2\)

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PostPosted: Thu Jul 14, 2016 10:35 am 
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Hello Tolis.

The characterstic polynomial of the first equation is the \(\displaystyle{f(y)=y^2+5\,y+6\,y\in\mathbb{C}_[y]}\) with

\(\displaystyle{f(y)=0\iff y^2+5\,y+6=0\iff \left(y+2\right)\,\left(y+3\right)=0\iff y=-3\,\,\lor\,\,y=-2}\) .

So, the general solution of tthe equation is \(\displaystyle{f(x)=c_1\,e^{-3\,x}+c_2\,e^{-2\,x}\,,x\in\mathbb{R}}\) .

Since \(\displaystyle{f(0)=2}\), we get : \(\displaystyle{c_1+c_2=2}\) . Also,

\(\displaystyle{f^\prime(x)=-3\,c_1\,e^{-3\,x}-2\,c_2\,e^{-2\,x}\,,x\in\mathbb{R}}\) and then :

\(\displaystyle{f^\prime(0)=3\iff -3\,c_1-2\,c_2=3\iff -3\,(2-c_2)-2\,c_2=3\iff c_2=9\implies c_1=-7}\), so :

\(\displaystyle{f:\mathbb{R}\longrightarrow \mathbb{R}\,,f(x)=-7\,e^{-3\,x}+9\,e^{-2\,x}}\) .

Now, for the second equation :

Consider the equation \(\displaystyle{f^{\prime \prime}(x)+f(x)=0}\) with characteristic polynomial

\(\displaystyle{g(y)=y^2+1}\) and \(\displaystyle{g(y)=0\iff y=\pm i}\), so :

the general solution of the second equation is

\(\displaystyle{f(x)=h(x)+k_1\,\cos\,x+k_2\,\sin\,x\,,x\in\mathbb{R}}\) , where

\(\displaystyle{h}\) is a partial solution of \(\displaystyle{f^{\prime \prime}(x)+f(x)=\sin\,(2\,x)}\) given by :

\(\displaystyle{h(x)=\int_{0}^{x}\dfrac{\cos\,t\,\sin\,x-\sin\,t\,\cos\,x}{\cos^2\,t+\sin^2\,t}\cdot \sin\,(2\,t)\,\mathrm{d}t=...=\dfrac{1}{3}\,\sin\,x-\dfrac{1}{3}\,\sin\,(2\,x)\,,x\in\mathbb{R}}\) .


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