How to find the solution to following IVP?

Question

If the IVP is

$$\frac{dx}{dt}=x\sin(t)+\sin(x^2) $$ subjected to $$x(a)=b. $$

I want to know in which interval this IVP has unique global solution? Further what is that solution?

Answer 1

Only a comment:

$$x'(t)=x(t) \sin (t)+\sin \left(x^2\right) => t'(x)=\frac{1}{x \sin (t(x))+\sin \left(x^2\right)}$$

$$ t'(x)=\frac{1}{x \sin (t(x))+\sin \left(x^2\right)}=\\t'(x)=\frac{1}{x \sin (t(x))+\sum _{k=0}^{\infty } \frac{(-1)^k x^{4 k+2}}{(2 k+1)!}}$$

With CAS help:

$$C+\int_{} ^{t(x)} \left(\sum _{k=0}^{\infty } \frac{(-1)^k e^{-(1+4 k) \cos (s)}}{(1+2 k)!}\right) \, ds+\sum _{k=0}^{\infty } \frac{e^{-(1+4 k) \cos (t(x))} x^{-1-4 k}}{1+4 k}=0$$

$$C+\int_{}^{t(x)} \left(\sum _{j=0}^{\infty } \left(\sum _{k=0}^{\infty } \frac{(-1)^k (-1-4 k)^j \cos ^j(s)}{j! (1+2 k)!}\right)\right) \, ds+\frac{e^{-\cos (t(x))} \, _2F_1\left(\frac{1}{4},1;\frac{5}{4};\frac{e^{-4 \cos (t(x))}}{x^4}\right)}{x}=0$$ $C+\sum _{j=0}^{\infty } \left(\sum _{k=0}^{\infty } \frac{(-1)^{1+k} (-1-4 k)^j \cos ^{1+j}(t(x)) \, _2F_1\left(\frac{1}{2},\frac{1+j}{2};\frac{3+j}{2};\cos ^2(t(x))\right) \sin (t(x))}{(1+j) j! (1+2 k)! \sqrt{\sin ^2(t(x))}}\right)+\frac{e^{-\cos (t(x))} \, _2F_1\left(\frac{1}{4},1;\frac{5}{4};\frac{e^{-4 \cos (t(x))}}{x^4}\right)}{x}=0$

subjected to $$t(b)=a$$ then: $\color{Blue}{\sum _{j=0}^{\infty } \left(\sum _{k=0}^{\infty } \frac{(-1)^{1+k} (-1-4 k)^j \cos ^{1+j}(t(x)) \, _2F_1\left(\frac{1}{2},\frac{1+j}{2};\frac{3+j}{2};\cos ^2(t(x))\right) \sin (t(x))}{(1+j) j! (1+2 k)! \sqrt{\sin ^2(t(x))}}\right)+\frac{e^{-\cos (t(x))} \, _2F_1\left(\frac{1}{4},1;\frac{5}{4};\frac{e^{-4 \cos (t(x))}}{x^4}\right)}{x}-\sum _{j=0}^{\infty } \left(\sum _{k=0}^{\infty } \frac{(-1)^{1+k} (-1-4 k)^j \cos ^{1+j}(a) \, _2F_1\left(\frac{1}{2},\frac{1+j}{2};\frac{3+j}{2};\cos ^2(a)\right) \sin (a)}{(1+j) j! (1+2 k)! \sqrt{\sin ^2(a)}}\right)-\frac{e^{-\cos (a)} \, _2F_1\left(\frac{1}{4},1;\frac{5}{4};\frac{e^{-4 \cos (a)}}{b^4}\right)}{b}=0}$

probably can't be solvelable.

Answer 2

You have $$|x'|\le |x|+1.$$ Such linear bounds on the derivative are sufficient to show the existence of a solution over all of $\Bbb R$.

Indeed consider $\phi(t)=\ln(1+|x(t)|)$, then $$ |\phi'(t)|=\frac{|x'(t)|}{1+|x(t)|}\le 1 $$ so that $$\phi(t)\le |t-a|+\phi(a)$$ or $$ |x(t)|\le (1+|x(a)|)e^{|t-a|}-1 $$ so that $x$ is bounded at all times, no divergence to infinity in finite time is possible.