Internal problem ID [3211]
Internal file name [OUTPUT/2703_Sunday_June_05_2022_08_39_00_AM_74786523/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: 68.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[_linear]
\[ \boxed {\left (1+x \right ) y^{\prime }-y=x \left (1+x \right )^{2}} \]
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}{1+x}\\ q(x) &=\frac {x^{3}+2 x^{2}+x}{1+x} \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {y}{1+x} = \frac {x^{3}+2 x^{2}+x}{1+x} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {1}{1+x}d x} \\ &= \frac {1}{1+x} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {x^{3}+2 x^{2}+x}{1+x}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{1+x}\right ) &= \left (\frac {1}{1+x}\right ) \left (\frac {x^{3}+2 x^{2}+x}{1+x}\right )\\ \mathrm {d} \left (\frac {y}{1+x}\right ) &= x\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {y}{1+x} &= \int {x\,\mathrm {d} x}\\ \frac {y}{1+x} &= \frac {x^{2}}{2} + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {1}{1+x}\) results in \begin {align*} y &= \frac {\left (1+x \right ) x^{2}}{2}+c_{1} \left (1+x \right ) \end {align*}
which simplifies to \begin {align*} y &= \frac {\left (1+x \right ) \left (x^{2}+2 c_{1} \right )}{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (1+x \right ) \left (x^{2}+2 c_{1} \right )}{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (1+x \right ) \left (x^{2}+2 c_{1} \right )}{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (1+x \right ) y^{\prime }-y=x \left (1+x \right )^{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+x \left (1+x \right )^{2}}{1+x} \\ \bullet & {} & \textrm {Collect w.r.t.}\hspace {3pt} y\hspace {3pt}\textrm {and simplify}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y}{1+x}+\left (1+x \right ) x \\ \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}{1+x}=\left (1+x \right ) x \\ \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}{1+x}\right )=\mu \left (x \right ) \left (1+x \right ) x \\ \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}{1+x}\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 )}{1+x} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )=\frac {1}{1+x} \\ \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 \mu \left (x \right ) \left (1+x \right ) x d x +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (x \right )=\int \mu \left (x \right ) \left (1+x \right ) x d x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (x \right ) \left (1+x \right ) x d x +c_{1}}{\mu \left (x \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )=\frac {1}{1+x} \\ {} & {} & y=\left (1+x \right ) \left (\int x d x +c_{1} \right ) \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\left (1+x \right ) \left (\frac {x^{2}}{2}+c_{1} \right ) \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left (1+x \right ) \left (x^{2}+2 c_{1} \right )}{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: 16
dsolve((1+x)*diff(y(x),x)-y(x)=x*(1+x)^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (x^{2}+2 c_{1} \right ) \left (x +1\right )}{2} \]
✓ Solution by Mathematica
Time used: 0.029 (sec). Leaf size: 20
DSolve[(1+x)*y'[x]-y[x]==x*(1+x)^2,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2} (x+1) \left (x^2+2 c_1\right ) \]