Internal problem ID [3206]
Internal file name [OUTPUT/2698_Sunday_June_05_2022_08_38_57_AM_38294634/index.tex]
Book: Differential equations for engineers by Wei-Chau XIE, Cambridge Press 2010
Section: Chapter 2. First-Order and Simple Higher-Order Differential Equations. Page
78
Problem number: 62.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[_linear]
\[ \boxed {\left (2 x +3\right ) y^{\prime }-y=\sqrt {2 x +3}} \]
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {1}{2 x +3}\\ q(x) &=\frac {1}{\sqrt {2 x +3}} \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {y}{2 x +3} = \frac {1}{\sqrt {2 x +3}} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {1}{2 x +3}d x} \\ &= \frac {1}{\sqrt {2 x +3}} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {1}{\sqrt {2 x +3}}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{\sqrt {2 x +3}}\right ) &= \left (\frac {1}{\sqrt {2 x +3}}\right ) \left (\frac {1}{\sqrt {2 x +3}}\right )\\ \mathrm {d} \left (\frac {y}{\sqrt {2 x +3}}\right ) &= \frac {1}{2 x +3}\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {y}{\sqrt {2 x +3}} &= \int {\frac {1}{2 x +3}\,\mathrm {d} x}\\ \frac {y}{\sqrt {2 x +3}} &= \frac {\ln \left (2 x +3\right )}{2} + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {1}{\sqrt {2 x +3}}\) results in \begin {align*} y &= \frac {\sqrt {2 x +3}\, \ln \left (2 x +3\right )}{2}+c_{1} \sqrt {2 x +3} \end {align*}
which simplifies to \begin {align*} y &= \left (\frac {\ln \left (2 x +3\right )}{2}+c_{1} \right ) \sqrt {2 x +3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (\frac {\ln \left (2 x +3\right )}{2}+c_{1} \right ) \sqrt {2 x +3} \\ \end{align*}
Verification of solutions
\[ y = \left (\frac {\ln \left (2 x +3\right )}{2}+c_{1} \right ) \sqrt {2 x +3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (2 x +3\right ) y^{\prime }-y=\sqrt {2 x +3} \\ \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+\sqrt {2 x +3}}{2 x +3} \\ \bullet & {} & \textrm {Collect w.r.t.}\hspace {3pt} y\hspace {3pt}\textrm {and simplify}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y}{2 x +3}+\frac {1}{\sqrt {2 x +3}} \\ \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}{2 x +3}=\frac {1}{\sqrt {2 x +3}} \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (x \right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }-\frac {y}{2 x +3}\right )=\frac {\mu \left (x \right )}{\sqrt {2 x +3}} \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d x}\left (y \mu \left (x \right )\right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }-\frac {y}{2 x +3}\right )=y^{\prime } \mu \left (x \right )+y \mu ^{\prime }\left (x \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (x \right ) \\ {} & {} & \mu ^{\prime }\left (x \right )=-\frac {\mu \left (x \right )}{2 x +3} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )=\frac {1}{\sqrt {2 x +3}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}\left (y \mu \left (x \right )\right )\right )d x =\int \frac {\mu \left (x \right )}{\sqrt {2 x +3}}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (x \right )=\int \frac {\mu \left (x \right )}{\sqrt {2 x +3}}d x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \frac {\mu \left (x \right )}{\sqrt {2 x +3}}d x +c_{1}}{\mu \left (x \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )=\frac {1}{\sqrt {2 x +3}} \\ {} & {} & y=\sqrt {2 x +3}\, \left (\int \frac {1}{2 x +3}d x +c_{1} \right ) \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\left (\frac {\ln \left (2 x +3\right )}{2}+c_{1} \right ) \sqrt {2 x +3} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left (\ln \left (2 x +3\right )+2 c_{1} \right ) \sqrt {2 x +3}}{2} \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: 23
dsolve((2*x+3)*diff(y(x),x)=y(x)+sqrt(2*x+3),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\ln \left (3+2 x \right )+2 c_{1} \right ) \sqrt {3+2 x}}{2} \]
✓ Solution by Mathematica
Time used: 0.049 (sec). Leaf size: 29
DSolve[(2*x+3)*y'[x]==y[x]+Sqrt[2*x+3],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2} \sqrt {2 x+3} (\log (2 x+3)+2 c_1) \]