Internal problem ID [14891]
Internal file name [OUTPUT/14571_Monday_April_08_2024_06_27_32_AM_27579930/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Chapter 4 review exercises, page 219
Problem number: 73.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_integrable_as_is"
Maple gives the following as the ode type
[[_2nd_order, _exact, _nonlinear], [_2nd_order, _with_linear_symmetries], [_2nd_order, _reducible, _mu_y_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {t \left (y^{\prime \prime } y+{y^{\prime }}^{2}\right )+y^{\prime } y=1} \] With initial conditions \begin {align*} [y \left (1\right ) = 1, y^{\prime }\left (1\right ) = 1] \end {align*}
Integrating both sides of the ODE w.r.t \(t\) gives \begin {align*} \int \left (y^{\prime \prime } y t +\left (y^{\prime } t +y\right ) y^{\prime }\right )d t &= \int 1d t\\ t y y^{\prime } = t + c_{1} \end {align*}
Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(t,y)\\ &= f( t) g(y)\\ &= \frac {t +c_{1}}{y t} \end {align*}
Where \(f(t)=\frac {t +c_{1}}{t}\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= \frac {t +c_{1}}{t} \,d t \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {\frac {t +c_{1}}{t} \,d t} \\ \frac {y^{2}}{2}&=t +c_{1} \ln \left (t \right )+c_{2} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}-t -c_{1} \ln \left (t \right )-c_{2} = 0 \] Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} \frac {y^{2}}{2}-t -c_{1} \ln \left (t \right )-c_{2} = 0 \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 1\) and \(t = 1\) in the above gives \begin {align*} -\frac {1}{2}-c_{2} = 0\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {\frac {2 c_{1}}{t}+2}{2 \sqrt {2 c_{1} \ln \left (t \right )+2 c_{2} +2 t}} \end {align*}
substituting \(y^{\prime } = 1\) and \(t = 1\) in the above gives \begin {align*} 1 = \frac {c_{1} +1}{\sqrt {2 c_{2} +2}}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=0\\ c_{2}&=-{\frac {1}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} \frac {y^{2}}{2}-t +\frac {1}{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}-t +\frac {1}{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2}-t +\frac {1}{2} = 0 \] Verified OK.
Writing the ode as \[ y^{\prime \prime } y t +\left (y^{\prime } t +y\right ) y^{\prime } = 1 \] Integrating both sides of the ODE w.r.t \(t\) gives \begin {align*} \int \left (y^{\prime \prime } y t +\left (y^{\prime } t +y\right ) y^{\prime }\right )d t &= \int 1d t\\ t y y^{\prime } = t +c_{1} \end {align*}
Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(t,y)\\ &= f( t) g(y)\\ &= \frac {t +c_{1}}{y t} \end {align*}
Where \(f(t)=\frac {t +c_{1}}{t}\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= \frac {t +c_{1}}{t} \,d t \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {\frac {t +c_{1}}{t} \,d t} \\ \frac {y^{2}}{2}&=t +c_{1} \ln \left (t \right )+c_{2} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}-t -c_{1} \ln \left (t \right )-c_{2} = 0 \] Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} \frac {y^{2}}{2}-t -c_{1} \ln \left (t \right )-c_{2} = 0 \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 1\) and \(t = 1\) in the above gives \begin {align*} -\frac {1}{2}-c_{2} = 0\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {\frac {2 c_{1}}{t}+2}{2 \sqrt {2 c_{1} \ln \left (t \right )+2 c_{2} +2 t}} \end {align*}
substituting \(y^{\prime } = 1\) and \(t = 1\) in the above gives \begin {align*} 1 = \frac {c_{1} +1}{\sqrt {2 c_{2} +2}}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=0\\ c_{2}&=-{\frac {1}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} \frac {y^{2}}{2}-t +\frac {1}{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}-t +\frac {1}{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2}-t +\frac {1}{2} = 0 \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN differential order: 2; trying a linearization to 3rd order trying 2nd order ODE linearizable_by_differentiation trying 2nd order, 2 integrating factors of the form mu(x,y) trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type <- LODE of Euler type successful <- 2nd order, 2 integrating factors of the form mu(x,y) successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 11
dsolve([t*(diff(y(t),t$2)*y(t)+diff(y(t),t)^2)+diff(y(t),t)*y(t)=1,y(1) = 1, D(y)(1) = 1],y(t), singsol=all)
\[ y \left (t \right ) = \sqrt {2 t -1} \]
✓ Solution by Mathematica
Time used: 0.371 (sec). Leaf size: 14
DSolve[{t*(y''[t]*y[t]+y'[t]^2)+y'[t]*y[t]==1,{y[1]==1,y'[1]==1}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \sqrt {2 t-1} \]