## 1 Introduction
### The Strategy and Tactics of the Submarine Fleet in Modern Conditions
So, when considering only the friction of the hull against the medium with the submarine's motion, the differential equation of motion has the form:  
\[M\ddot{U}(t)+\int\limits_{0}^{t}[m_{0}\delta_{1}(t-\tau)+\alpha]\ \ddot{U}(\tau)d\tau=\] \[=M_{0}[\delta_{1}(t)-\delta_{1}(t-1)]+\int\limits_{0}^{t}[m_{0} \delta_{1}(t-\tau)+\alpha][\delta_{1}(\tau)-\delta_{1}(\tau-1)]d\tau\.\]  
Let us transform it:  
\[M\ddot{U}(t)+m_{0}\int\limits_{0}^{t}\delta_{1}(t-\tau)\ddot{U}(\tau)d\tau+ \alpha\int\limits_{0}^{t}\ddot{U}(\tau)d\tau\ =\] \[=M_{0}[\delta_{1}(t)-\delta_{1}(t-\tau)]d\tau+m_{0}\int\limits_{0}^ {t}\delta_{1}(t-\tau)\delta_{1}(\tau)d\tau-m_{0}\int\limits_{0}^{t}\delta_{1}( t-\tau)\delta_{1}(t-1)d\tau+\] \[+\alpha\int\limits_{0}^{t}[\delta_{1}(\tau)-\delta_{1}(\tau-1)]d\tau\.\]  
Or use it in this form:  
\[(M+m_{0})\ddot{U}(t)+\alpha\dot{U}(t)\ =(M_{0}+m_{0})[\delta_{1}(t)-\delta_{1}(t-1) ]+\alpha[\delta_{0}(t)-\delta_{0}(t-1)]\.\]  
By applying the Laplace transformation, upon the zero initial conditions, we obtain:  
\[(M+m_{0})\rho^{2}U+\alpha\rho U\ =(M_{0}+m_{0})[1-e^{-\rho}]+\alpha\left( \frac{1}{\rho}-\frac{1}{\rho}e^{-\rho}\right).\]  
Or in another way:  
\[\rho[(M+m_{0})\rho+\alpha]\ U=(1-e^{-\rho})\left(M_{0}+m_{0}+\alpha\frac{1}{ \rho}\right).\]  
Hence, we have:  
\[U=\frac{(1-e^{-\rho})\left(M_{0}+m_{0}+\frac{\alpha}{\rho}\right)}{\rho[(M+m_ {0})\rho+\alpha]}=\frac{(1-e^{-\rho})[(M_{0}+m_{0})\rho+\alpha]}{\rho^{2}[(M+ m_{0})\rho+\alpha]}=\]  
Let us represent the first fraction in the form of the following:  
\[\frac{(M_{0}+m_{0})\rho+\alpha}{\rho^{2}[(M+m_{0})\rho+\alpha]}=\frac{A}{\rho ^{2}}+\frac{B}{\rho}+\frac{C}{(M+m_{0})\rho+\alpha}\.\]  
Then:  
\[=M_{0}[\delta_{1}(t)-\delta_{1}(t-1)]+\int\limits_{0}^{t}[m_{0} \delta_{1}(t-\tau)+\alpha][\delta_{1}(\tau)-\delta_{1}(\tau-1)]d\tau\.\]0  
This expression allows for determining the coefficients A, B, C:  
\[ \begin{array}{ll}\rho=0&A=1;\\ \rho^{2}&0=B(M+m_{0})+C;\\ \rho&M_{0}+m_{0}=M+m_{0}+\alpha B;\ B=\frac{M_{0}-M}{\alpha};\ C=\frac{(M-M_{0 })(M+m_{0})}{\alpha}\.\end{array} \]Taking into account the found solutions, we have:  
\[U(\rho)=\frac{1}{\rho^{2}}-\frac{M-M_{0}}{\alpha}\cdot\frac{1}{\rho}+\frac{M-M_{0 }}{\alpha}\cdot\frac{1}{\rho+\frac{\alpha}{M+m_{0}}}-e^{-\rho}\left(\frac{1}{ \rho^{2}}-\frac{M-M_{0}}{\alpha}\cdot\frac{1}{\rho}\cdot\frac{1}{\rho+\frac{ \alpha}{M+m_{0}}}\right).\]  
Specify:  
\[\frac{\alpha}{M+m_{0}}=\
u\]  
and transfer to the original:  
\[ \begin{array}{l}U(t)=\left[t-\frac{M-M_{0}}{\alpha}\big{(}1-e^{-\
u t}\big{)} \right]\delta_{0}(t)-\left[t-1-\frac{M-M_{0}}{\alpha}\big{(}1-e^{-\
u(t-1)} \big{)}\right]\delta_{0}(t-1)=\\ =\left(t-\frac{M-M_{0}}{\alpha}+\frac{M-M_{0}}{\alpha}e^{-\
u t}\big{)}\delta_ {0}(t)-\left(t-1-\frac{M-M_{0}}{\alpha}+\frac{M-M_{0}}{\alpha}e^{-\
u(t-1)} \right)\delta_{0}(t-1)\.\end{array} \]  
If the multipliers \(\left(1-e^{-\
u t}\right)\) and \(\left(1-e^{-\
u(t-1)}\right)\) expand into a series \(\left(t\geq 1\right)\) then:  
\[ \begin{array}{l}\left(1-e^{-\
u t}\right)=1-\left(1-\frac{\
u}{1!}t+\
u^{2 }\frac{t^{2}}{2!}-\
u^{3}\frac{t^{3}}{3!}+\cdots\right)=\
u\frac{1}{1!}-\
u^{ 2}\frac{t^{2}}{2!}+\
u^{3}\frac{t^{2}}{3!}-\cdots\ ;\\ \left(1-e^{-\
u(t-1)}\right)=1-\left(1-\
u\frac{t-1}{1!}+\
u^{2}\frac{(t-1)^{ 2}}{2!}-\
u^{3}\frac{(t-1)^{3}}{3!}+\cdots\right)=\\ =\
u\frac{t-1}{1!}-\
u^{2}\frac{(t-1)^{2}}{2!}+\
u^{3}\frac{(t-1)^{3}}{3!}- \cdots\ ;\end{array} \]  
In this case, the first and second square brackets can be written in another way:  
\[