Solution of first order non-linear ODE

Question

I'm trying to prove that the ode $$ x'(t)=5\cos^2(tx)-x^2-5, \ \ \ x(t_0)=x_0 $$ has strictly decreasing solutions for $x_0\neq 0$. The problem is that i only find the solution $x(t)=0$. It also says that for $t_0=1,x_0=0$ the Cauchy problem has unique solution wich doesn't make sense since i can't find any other solution than 0. Any ideas?

Answer 1

The ODE is of the form $$ x'(t)=-x(t)^2-5\sin^2(tx(t))=-x(t)^2g(t,x(t)) $$ with a smooth function $g$. This implies that the zero solution separates the solution space. No other solution can take the value zero. For all other solutions the right side is strictly negative.