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timofeeve [1]
3 years ago
10

Find an implicit and an explicit solution of the given initial-value problem. (use x for x(t).) dx dt = 2(x2 + 1), x(π/4) = 1

Mathematics
1 answer:
Keith_Richards [23]3 years ago
6 0
\displaystyle
\dfrac{dx}{dt}=2(x^2+1)\\\\
\int_{t_0}^t dt=\int_{x_0}^x\dfrac{dx}{2(x^2+1)}\\\\
t-t_0=\dfrac{1}{2}\int_{x_0}^x\dfrac{dx}{(x^2+1)}\\\\
t-t_0=\dfrac{1}{2}\left[\arctan(x)\right]_{x_0}^x\\\\
t-t_0=\dfrac{1}{2}\left[\arctan(x)-\arctan(x_0)\right]

We'll use x(\pi/4)=1, considering that x_0=1, t_0=\dfrac{\pi}{4}:

t-t_0=\dfrac{1}{2}\left[\arctan(x)-\arctan(x_0)\right]\\\\
t-\dfrac{\pi}{4}=\dfrac{1}{2}\left[\arctan(x)-\arctan(1)\right]\\\\
t-\dfrac{\pi}{4}=\dfrac{1}{2}\left[\arctan(x)-\dfrac{\pi}{4}\right]\\\\
t-\dfrac{\pi}{4}=\dfrac{1}{2}\arctan(x)-\dfrac{\pi}{8}\\\\
t-\dfrac{\pi}{8}=\dfrac{1}{2}\arctan(x)\\\\
\boxed{\arctan(x)=2t-\dfrac{\pi}{4}}

Applying tan in the both sides:

\arctan(x)=2t-\dfrac{\pi}{4}\\\\
\tan(\arctan(x))=\tan\left(2t-\dfrac{\pi}{4}\right)\\\\
\boxed{x(t)=\tan\left(2t-\dfrac{\pi}{4}\right)}
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