77.1.160 problem 187 (page 297)
Internal
problem
ID
[18050]
Book
:
V.V.
Stepanov,
A
course
of
differential
equations
(in
Russian),
GIFML.
Moscow
(1958)
Section
:
All
content
Problem
number
:
187
(page
297)
Date
solved
:
Tuesday, January 28, 2025 at 08:28:31 PM
CAS
classification
:
system_of_ODEs
\begin{align*} y^{\prime \prime }+z^{\prime }\left (x \right )-2 z \left (x \right )&={\mathrm e}^{2 x}\\ z^{\prime }\left (x \right )+2 y^{\prime }-3 y&=0 \end{align*}
✓ Solution by Maple
Time used: 0.088 (sec). Leaf size: 117
dsolve([diff(y(x),x$2)+diff(z(x),x)-2*z(x)=exp(2*x),diff(z(x),x)+2*diff(y(x),x)-3*y(x)=0],singsol=all)
\begin{align*}
y &= \frac {{\mathrm e}^{2 x}}{4}+{\mathrm e}^{x} c_{1} +c_{2} {\mathrm e}^{\frac {x}{2}} \cos \left (\frac {\sqrt {23}\, x}{2}\right )+c_{3} {\mathrm e}^{\frac {x}{2}} \sin \left (\frac {\sqrt {23}\, x}{2}\right ) \\
z \left (x \right ) &= -\frac {{\mathrm e}^{2 x}}{8}+{\mathrm e}^{x} c_{1} -\frac {7 c_{2} {\mathrm e}^{\frac {x}{2}} \cos \left (\frac {\sqrt {23}\, x}{2}\right )}{4}+\frac {c_{2} {\mathrm e}^{\frac {x}{2}} \sqrt {23}\, \sin \left (\frac {\sqrt {23}\, x}{2}\right )}{4}-\frac {7 c_{3} {\mathrm e}^{\frac {x}{2}} \sin \left (\frac {\sqrt {23}\, x}{2}\right )}{4}-\frac {c_{3} {\mathrm e}^{\frac {x}{2}} \sqrt {23}\, \cos \left (\frac {\sqrt {23}\, x}{2}\right )}{4} \\
\end{align*}
✓ Solution by Mathematica
Time used: 2.152 (sec). Leaf size: 199
DSolve[{D[y[x],{x,2}]+D[z[x],x]-2*z[x]==Exp[2*x],D[z[x],x]+2*D[y[x],x]-3*y[x]==0},{y[x],z[x]},x,IncludeSingularSolutions -> True]
\begin{align*}
y(x)\to \frac {1}{276} e^{x/2} \left (23 e^{x/2} \left (3 e^x+6 c_1+2 c_2+4 c_3\right )+46 (3 c_1-c_2-2 c_3) \cos \left (\frac {\sqrt {23} x}{2}\right )-2 \sqrt {23} (9 c_1-11 c_2+2 c_3) \sin \left (\frac {\sqrt {23} x}{2}\right )\right ) \\
z(x)\to -\frac {1}{552} e^{x/2} \left (23 e^{x/2} \left (3 e^x-4 (3 c_1+c_2+2 c_3)\right )+92 (3 c_1+c_2-4 c_3) \cos \left (\frac {\sqrt {23} x}{2}\right )-4 \sqrt {23} (33 c_1-25 c_2-8 c_3) \sin \left (\frac {\sqrt {23} x}{2}\right )\right ) \\
\end{align*}