Proving single solution of initial value problem is increasingGiven the initial value problemy′(x)=y(x)−sin⁡{y(x)},y(0)=1

Question

Ben Castillo · Accepted Answer

Step 1We use the simple fact that v(y)=y−sin⁡y is increasing and positive on y&amp;gt;0The fact that y0(x)=0 is a solution implies that y(x)&amp;gt;0 for all x  (by uniqueness). Theny′(x)=v(y(x))&amp;gt;0.Thus limx→−∞y(x) exists and is nonnegative. If it&#039;s not zero, then y(x)≥y1&amp;gt;0 for some fixed y1 and soy′(x)=v(y(x))≥v(y1)&amp;gt;0.That would imply for any x&amp;lt;0 (by the mean value theorem and that y(0)=1)y(x)=y(x)−y(0)+y(0)=y′(x0)x+1≤v(y1)x+1this is impossible as y(x)&amp;gt;0 for all x.

undodaonePvopxl24 · Accepted Answer

Step 1Let us introduce the new variable z defined by z=−x. Theny′(x)=dydx=dydzdzdx=−dydzTherefore, we have the differential equation with the reversed variable directiondydz=−y+sin⁡yThis is in fact the dynamics of a simple damped pendulum. Let V(y)=y22dVdz=ydydz=−y2+ysin⁡yand dVdz&amp;lt;0 for all y∈R{0} and dVdz=0 at y=0. Hence, R is positively invariant with respect to the ODE, meaning that every solution starting at z=z0 is defined for all zgivenz0. Let v be the solution of the ODE with y(0)=1. Then, we have that limz→∞v(z)=0. Finally, we conclude that limx→−∞u(x)=0.We can also show that v(z) cannot change its sign and monotonically decreases whenever v(z)&amp;gt;0.

Proving single solution of initial value problem is

Answered question

Answer & Explanation

New Questions in Calculus 2