Solution of system of ODE.

Question

Let $x$ and $y$ be continuously differentiable functions on $[0,\infty)$ satisfying the following differential equations

$$\frac{dx}{dt}+(\sin(t)-1)x=\log(1+t),\;x(0)=1$$

$$\frac{dy}{dt}+(\sin(t)-1)y=t,\;y(0)=2$$

Define $z(t)=y(t)-x(t),t\geq 0$. Which of the following statements are true ?

$1.$ $z(t)\leq 1$.

$2.$ $z(2)>z(1)$.

$3.$ $z(1)>1$.

$4.$ $z(2)\leq z(1).$

Firstly I thought to apply the Sturm comparison theorem but it’s for homogeneous differential equations. Secondly I thought that $z$ is solution of the differential equation $$\frac{dz}{dt}+(\sin(t)-1)z=\log(1+t)-t,\;z(0)=1.$$

Answer 1

Well, your DE for $z$ is correct. The resulting DE is first-order linear: \begin{align*} \dot{z}+\underbrace{(\sin(t)-1)}_{p(t)}z&=\ln(1+t)-t\\ e^{\int p(t)\,dt}&=e^{-\cos(t)-t}=\text{integrating factor}\\ e^{-\cos(t)-t}\dot{z}+e^{-\cos(t)-t}(\sin(t)-1)z&=e^{-\cos(t)-t}[\ln(1+t)-t]\\ \frac{d}{dt}\left[e^{-\cos(t)-t}\,z\right]&=e^{-\cos(t)-t}[\ln(1+t)-t]\\ e^{-\cos(t)-t}\,z(t)-e^{-1}&=\int_0^t e^{-\cos(\tau)-\tau}[\ln(1+\tau)-\tau]\,d\tau\\ e^{-\cos(t)-t}\,z(t)&=e^{-1}+\int_0^t e^{-\cos(\tau)-\tau}[\ln(1+\tau)-\tau]\,d\tau\\ z(t)&=e^{\cos(t)+t-1}+e^{\cos(t)+t}\int_0^t e^{-\cos(\tau)-\tau}[\ln(1+\tau)-\tau]\,d\tau\\ &=e^{\cos(t)+t-1}\left[1+e\int_0^t e^{-\cos(\tau)-\tau}[\ln(1+\tau)-\tau]\,d\tau\right] \end{align*} The integral on the RHS is intractable, but there are some things we can say about $z(t)$ qualitatively from this expression. First, the integrand is non-positive, which implies that the integral is non-positive. If $t=0,$ the integral is zero, of course, making $z(0)=1,$ as required. As $t$ increases, the integral will become more negative, eventually overcoming the $1.$ The option 1 is false: $z$ definitely gets bigger than $1.$ Option 2 is also false. You can approximate $z(2)\approx 1.06518,$ whereas $z(1)\approx 1.59294.$ By the same token, Option 3 is true, and if Option 2 is false, which it is, then Option 4, being its negation, must be true. Hence, we conclude that Options 3 and 4 are true.

[EDIT] Here is some Mathematica code to plot this function:

g[t_]:=Exp[Cos[t]+t-1](1+Exp[1]NIntegrate[Exp[-Cos[s]-s](Log[1+s]-s),{s,0,t}])
Plot[g[t],{t,0,5}]

with the results:

So we can see that the function starts increasing, but then decreases. If we go further out:

Plot[g[t],{t,0,10}]

we get

So it appears that eventually, the function is always decreasing.