74.15.53 problem 53 (f)
Internal
problem
ID
[16492]
Book
:
INTRODUCTORY
DIFFERENTIAL
EQUATIONS.
Martha
L.
Abell,
James
P.
Braselton.
Fourth
edition
2014.
ElScAe.
2014
Section
:
Chapter
4.
Higher
Order
Equations.
Exercises
4.7,
page
195
Problem
number
:
53
(f)
Date
solved
:
Tuesday, January 28, 2025 at 09:09:25 AM
CAS
classification
:
[[_2nd_order, _linear, _nonhomogeneous]]
\begin{align*} \left (x^{2}+1\right )^{2} y^{\prime \prime }+2 x \left (x^{2}+1\right ) y^{\prime }+4 y&=\arctan \left (x \right ) \end{align*}
With initial conditions
\begin{align*} y \left (0\right )&=0\\ y^{\prime }\left (0\right )&=1 \end{align*}
✓ Solution by Maple
Time used: 0.021 (sec). Leaf size: 26
dsolve([(1+x^2)^2*diff(y(x),x$2)+2*x*(1+x^2)*diff(y(x),x)+4*y(x)=arctan(x),y(0) = 0, D(y)(0) = 1],y(x), singsol=all)
\[
y = \frac {x^{2} \arctan \left (x \right )+\arctan \left (x \right )+3 x}{4 x^{2}+4}
\]
✓ Solution by Mathematica
Time used: 0.764 (sec). Leaf size: 420
DSolve[{(1+x^2)^2*D[y[x],{x,2}]+2*x*(1+x^2)*D[y[x],x]+4*y[x]==ArcTan[x],{y[0]==0,Derivative[1][y][0] ==1}},y[x],x,IncludeSingularSolutions -> True]
\[
y(x)\to \frac {\exp \left (\int _1^x\frac {K[1]+2 i}{K[1]^2+1}dK[1]\right ) \left (\int _1^x\exp \left (-2 \int _1^{K[2]}\frac {K[1]+2 i}{K[1]^2+1}dK[1]\right )dK[2] \left (\int _1^x\frac {\exp \left (\int _1^{K[4]}\frac {K[1]+2 i}{K[1]^2+1}dK[1]\right ) \arctan (K[4])}{\left (K[4]^2+1\right )^{3/2}}dK[4]-\int _1^0\frac {\exp \left (\int _1^{K[4]}\frac {K[1]+2 i}{K[1]^2+1}dK[1]\right ) \arctan (K[4])}{\left (K[4]^2+1\right )^{3/2}}dK[4]+\exp \left (\int _1^0\frac {K[1]+2 i}{K[1]^2+1}dK[1]\right )\right )+\int _1^x-\frac {\exp \left (\int _1^{K[3]}\frac {K[1]+2 i}{K[1]^2+1}dK[1]\right ) \arctan (K[3]) \int _1^{K[3]}\exp \left (-2 \int _1^{K[2]}\frac {K[1]+2 i}{K[1]^2+1}dK[1]\right )dK[2]}{\left (K[3]^2+1\right )^{3/2}}dK[3]-\int _1^0-\frac {\exp \left (\int _1^{K[3]}\frac {K[1]+2 i}{K[1]^2+1}dK[1]\right ) \arctan (K[3]) \int _1^{K[3]}\exp \left (-2 \int _1^{K[2]}\frac {K[1]+2 i}{K[1]^2+1}dK[1]\right )dK[2]}{\left (K[3]^2+1\right )^{3/2}}dK[3]-\exp \left (\int _1^0\frac {K[1]+2 i}{K[1]^2+1}dK[1]\right ) \int _1^0\exp \left (-2 \int _1^{K[2]}\frac {K[1]+2 i}{K[1]^2+1}dK[1]\right )dK[2]\right )}{\sqrt {x^2+1}}
\]