Answer to Question #215854 in Differential Equations for Zerin

Question

Solve the problem :
(1/(1+x^2) + cos t -2xt) dx/dt = x(x+sin t), x(0)=1

Answer 1

$(\frac{1}{1+x^2} + \cos t -2xt) \frac{dx}{dt}- x(x+\sin t)=0$

$(\frac{1}{1+x^2} + \cos t -2xt) \frac{dx}{dt}- x(x+\sin t)=\frac{d}{dt}(\arctan x+x\cos t-x^2t)=0$

$\arctan x+x\cos t-x^2t=C$

if t=0 and x(0)=1 then

$C=\arctan x+x\cos t-x^2t=\arctan 1+1\cdot \cos 0-1^2\cdot0=\pi/4+1$

Answer: $\arctan x+x\cos t-x^2t=\pi/4+1$