Internal problem ID [1664]
Internal file name [OUTPUT/1665_Sunday_June_05_2022_02_26_21_AM_72125485/index.tex]
Book: Differential equations and their applications, 3rd ed., M. Braun
Section: Section 1.2. Page 9
Problem number: 20.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[_linear]
\[ \boxed {y^{\prime }+\frac {y}{t}=\frac {1}{t^{2}}} \]
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(t)y &= q(t) \end {align*}
Where here \begin {align*} p(t) &=\frac {1}{t}\\ q(t) &=\frac {1}{t^{2}} \end {align*}
Hence the ode is \begin {align*} y^{\prime }+\frac {y}{t} = \frac {1}{t^{2}} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {1}{t}d t} \\ &= t \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {1}{t^{2}}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left (t y\right ) &= \left (t\right ) \left (\frac {1}{t^{2}}\right )\\ \mathrm {d} \left (t y\right ) &= \frac {1}{t}\, \mathrm {d} t \end {align*}
Integrating gives \begin {align*} t y &= \int {\frac {1}{t}\,\mathrm {d} t}\\ t y &= \ln \left (t \right ) + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu =t\) results in \begin {align*} y &= \frac {\ln \left (t \right )}{t}+\frac {c_{1}}{t} \end {align*}
which simplifies to \begin {align*} y &= \frac {\ln \left (t \right )+c_{1}}{t} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\ln \left (t \right )+c_{1}}{t} \\ \end{align*}
Verification of solutions
\[ y = \frac {\ln \left (t \right )+c_{1}}{t} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+\frac {y}{t}=\frac {1}{t^{2}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y}{t}+\frac {1}{t^{2}} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime }+\frac {y}{t}=\frac {1}{t^{2}} \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (t \right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }+\frac {y}{t}\right )=\frac {\mu \left (t \right )}{t^{2}} \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d t}\left (y \mu \left (t \right )\right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }+\frac {y}{t}\right )=y^{\prime } \mu \left (t \right )+y \mu ^{\prime }\left (t \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (t \right ) \\ {} & {} & \mu ^{\prime }\left (t \right )=\frac {\mu \left (t \right )}{t} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )=t \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \left (\frac {d}{d t}\left (y \mu \left (t \right )\right )\right )d t =\int \frac {\mu \left (t \right )}{t^{2}}d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (t \right )=\int \frac {\mu \left (t \right )}{t^{2}}d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \frac {\mu \left (t \right )}{t^{2}}d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )=t \\ {} & {} & y=\frac {\int \frac {1}{t}d t +c_{1}}{t} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\ln \left (t \right )+c_{1}}{t} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 12
dsolve(diff(y(t),t)+1/t*y(t)=1/t^2,y(t), singsol=all)
\[ y \left (t \right ) = \frac {\ln \left (t \right )+c_{1}}{t} \]
✓ Solution by Mathematica
Time used: 0.024 (sec). Leaf size: 14
DSolve[y'[t]+1/t*y[t]==1/t^2,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {\log (t)+c_1}{t} \]