Functions P(x),Q(x),R(x) are differentiable on some open interval around 0 and satisfy the below equations as well as the initial conditions.P′(x)=2P2(x)Q(x)R(x)+1Q(x)R(x), P(0)=1Q′(x)=P(x)Q2(x)R(x)+4P(x)R(x), Q(0)=1 R′(x)=3P(x)Q(x)R2(x)+1P(x)Q(x), R(0)=1.Then P(x)Q(x)R(x)=tan(nx+π4). The value of n is

Functions P(x),Q(x),R(x) are differentiable on some open interval arou

1.	Functions P(x),Q(x),R(x) are differentiable on some open interval around 0 and satisfy the below equations as well as the initial conditions.P′(x)=2P2(x)Q(x)R(x)+1Q(x)R(x), P(0)=1Q′(x)=P(x)Q2(x)R(x)+4P(x)R(x), Q(0)=1 R′(x)=3P(x)Q(x)R2(x)+1P(x)Q(x), R(0)=1.Then P(x)Q(x)R(x)=tan(nx+π4). The value of n is
Answer» Functions $P (x), Q (x), R (x)$ are differentiable on some open interval around $0$ and satisfy the below equations as well as the initial conditions. $P^{'} (x) = 2 P^{2} (x) Q (x) R (x) + \frac{1}{Q (x) R (x)},$ $P (0) = 1$ $Q^{'} (x) = P (x) Q^{2} (x) R (x) + \frac{4}{P (x) R (x)},$ $Q (0) = 1$ $R^{'} (x) = 3 P (x) Q (x) R^{2} (x) + \frac{1}{P (x) Q (x)},$ $R (0) = 1.$ Then $P (x) Q (x) R (x) = tan (n x + \frac{π}{4}) .$ The value of $n$ is