Internal problem ID [5787]
Internal file name [OUTPUT/5035_Sunday_June_05_2022_03_18_21_PM_30532492/index.tex]
Book: Ordinary differential equations and calculus of variations. Makarets and Reshetnyak. Wold
Scientific. Singapore. 1995
Section: Chapter 1. First order differential equations. Section 1.2 Homogeneous equations
problems. page 12
Problem number: 37.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[[_homogeneous, `class C`], _rational, [_Abel, `2nd type`, `class A`]]
\[ \boxed {\left (4 y+x \right ) y^{\prime }-3 y=2 x -5} \]
This is ODE of type polynomial. Where the RHS of the ode is ratio of equations of two lines. Writing the ODE in the form \[ y^{\prime }= \frac {a_1 x + b_1 y + c_1}{ a_2 x + b_2 y + c_3 } \] Where \(a_1=2, b_1=3, c_1 =-5, a_2=1, b_2=4, c_2=0\). There are now two possible solution methods. The first case is when the two lines \(a_1 x + b_1 y + c_1\),\( a_2 x + b_2 y + c_3\) are not parallel and the second case is if they are parallel. If they are not parallel, then the transformation \(X=x-x_0\), \(Y=y-y_0\) converts the ODE to a homogeneous ODE. The values \( x_0,y_0\) have to be determined. If they are parallel then a transformation \(U(x)=a_1 x + b_1 y\) converts the given ODE in \(y\) to a separable ODE in \(U(x)\). The first case is when \(\frac {a_1}{b_1} \neq \frac {a_2}{b_2}\) and the second case when \(\frac {a_1}{b_1} = \frac {a_2}{b_2}\). From the above we see that \(\frac {a_1}{b_1}\neq \frac {a_2}{b_2}\). Hence this is case one where lines are not parallel. Using the transformation \begin {align*} X &=x-x_0 \\ Y &=y-y_0 \end {align*}
Where the constants \(x_0,y_0\) are obtained by solving the following two linear algebraic equations \begin {align*} a_1 x_0 + b_1 y_0 + c_1 &= 0\\ a_2 x_0 + b_2 y_0 + c_2 &= 0 \end {align*}
Substituting the values for \(a_1,b_1,c_1,a_2,b_2,c_2\) gives \begin {align*} 2 x_{0} +3 y_{0} -5 &= 0 \\ x_{0} +4 y_{0} &= 0 \\ \end {align*}
Solving for \(x_0,y_0\) from the above gives \begin {align*} x_0 &= 4 \\ y_0 &= -1 \end {align*}
Therefore the transformation becomes \begin {align*} X &=x-4 \\ Y &=y+1 \end {align*}
Using this transformation in \(\left (4 y+x \right ) y^{\prime }-3 y = 2 x -5\) result in \begin {align*} \frac {dY}{dX} &= \frac {2 X +3 Y}{4 Y +X} \end {align*}
This is now a homogeneous ODE which will now be solved for \(Y(X)\). In canonical form, the ODE is \begin {align*} Y' &= F(X,Y)\\ &= \frac {2 X +3 Y}{4 Y +X}\tag {1} \end {align*}
An ode of the form \(Y' = \frac {M(X,Y)}{N(X,Y)}\) is called homogeneous if the functions \(M(X,Y)\) and \(N(X,Y)\) are both homogeneous functions and of the same order. Recall that a function \(f(X,Y)\) is homogeneous of order \(n\) if \[ f(t^n X, t^n Y)= t^n f(X,Y) \] In this case, it can be seen that both \(M=2 X +3 Y\) and \(N=4 Y +X\) are both homogeneous and of the same order \(n=1\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {Y}{X}\), or \(Y=uX\). Hence \[ \frac { \mathop {\mathrm {d}Y}}{\mathop {\mathrm {d}X}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}}X + u \] Applying the transformation \(Y=uX\) to the above ODE in (1) gives \begin {align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}}X + u &= \frac {3 u +2}{4 u +1}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {\frac {3 u \left (X \right )+2}{4 u \left (X \right )+1}-u \left (X \right )}{X} \end {align*}
Or \[ \frac {d}{d X}u \left (X \right )-\frac {\frac {3 u \left (X \right )+2}{4 u \left (X \right )+1}-u \left (X \right )}{X} = 0 \] Or \[ 4 \left (\frac {d}{d X}u \left (X \right )\right ) X u \left (X \right )+\left (\frac {d}{d X}u \left (X \right )\right ) X +4 u \left (X \right )^{2}-2 u \left (X \right )-2 = 0 \] Or \[ -2+X \left (4 u \left (X \right )+1\right ) \left (\frac {d}{d X}u \left (X \right )\right )+4 u \left (X \right )^{2}-2 u \left (X \right ) = 0 \] Which is now solved as separable in \(u \left (X \right )\). Which is now solved in \(u \left (X \right )\). In canonical form the ODE is \begin {align*} u' &= F(X,u)\\ &= f( X) g(u)\\ &= -\frac {2 \left (2 u^{2}-u -1\right )}{X \left (4 u +1\right )} \end {align*}
Where \(f(X)=-\frac {2}{X}\) and \(g(u)=\frac {2 u^{2}-u -1}{4 u +1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {2 u^{2}-u -1}{4 u +1}} \,du &= -\frac {2}{X} \,d X \\ \int { \frac {1}{\frac {2 u^{2}-u -1}{4 u +1}} \,du} &= \int {-\frac {2}{X} \,d X} \\ \frac {5 \ln \left (u -1\right )}{3}+\frac {\ln \left (u +\frac {1}{2}\right )}{3}&=-2 \ln \left (X \right )+c_{3} \\ \end{align*} The above can be written as \begin {align*} \frac {5 \ln \left (u -1\right )+\ln \left (u +\frac {1}{2}\right )}{3} &= -2 \ln \left (X \right )+c_{3}\\ 5 \ln \left (u -1\right )+\ln \left (u +\frac {1}{2}\right ) &= \left (3\right ) \left (-2 \ln \left (X \right )+c_{3}\right ) \\ &= -6 \ln \left (X \right )+3 c_{3} \end {align*}
Raising both side to exponential gives \begin {align*} {\mathrm e}^{5 \ln \left (u -1\right )+\ln \left (u +\frac {1}{2}\right )} &= {\mathrm e}^{-6 \ln \left (X \right )+3 c_{3}} \end {align*}
Which simplifies to \begin {align*} \frac {\left (u -1\right )^{5} \left (2 u +1\right )}{2} &= \frac {3 c_{3}}{X^{6}}\\ &= \frac {c_{4}}{X^{6}} \end {align*}
Which simplifies to \[ u \left (X \right ) = \operatorname {RootOf}\left (2 \textit {\_Z}^{6}-9 \textit {\_Z}^{5}+15 \textit {\_Z}^{4}-10 \textit {\_Z}^{3}-\frac {2 c_{4} {\mathrm e}^{3 c_{3}}}{X^{6}}+3 \textit {\_Z} -1\right ) \] Now \(u\) in the above solution is replaced back by \(Y\) using \(u=\frac {Y}{X}\) which results in the solution \[ Y \left (X \right ) = X \operatorname {RootOf}\left (2 \textit {\_Z}^{6} X^{6}-9 \textit {\_Z}^{5} X^{6}+15 \textit {\_Z}^{4} X^{6}-10 \textit {\_Z}^{3} X^{6}+3 \textit {\_Z} \,X^{6}-X^{6}-2 c_{4} {\mathrm e}^{3 c_{3}}\right ) \] The solution is \[ Y \left (X \right ) = X \operatorname {RootOf}\left (2 \textit {\_Z}^{6} X^{6}-9 \textit {\_Z}^{5} X^{6}+15 \textit {\_Z}^{4} X^{6}-10 \textit {\_Z}^{3} X^{6}+3 \textit {\_Z} \,X^{6}-X^{6}-2 c_{4} {\mathrm e}^{3 c_{3}}\right ) \] Replacing \(Y=y-y_0, X=x-x_0\) gives \[ 1+y = \left (x -4\right ) \operatorname {RootOf}\left (2 \textit {\_Z}^{6} \left (x -4\right )^{6}-9 \textit {\_Z}^{5} \left (x -4\right )^{6}+15 \textit {\_Z}^{4} \left (x -4\right )^{6}-10 \textit {\_Z}^{3} \left (x -4\right )^{6}+3 \textit {\_Z} \left (x -4\right )^{6}-\left (x -4\right )^{6}-2 c_{4} {\mathrm e}^{3 c_{3}}\right ) \] Or \[ y = \left (x -4\right ) \operatorname {RootOf}\left (2 \textit {\_Z}^{6} \left (x -4\right )^{6}-9 \textit {\_Z}^{5} \left (x -4\right )^{6}+15 \textit {\_Z}^{4} \left (x -4\right )^{6}-10 \textit {\_Z}^{3} \left (x -4\right )^{6}+3 \textit {\_Z} \left (x -4\right )^{6}-\left (x -4\right )^{6}-2 c_{4} {\mathrm e}^{3 c_{3}}\right )-1 \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (x -4\right ) \operatorname {RootOf}\left (\left (2 x^{6}-48 x^{5}+480 x^{4}-2560 x^{3}+7680 x^{2}-12288 x +8192\right ) \textit {\_Z}^{6}+\left (-9 x^{6}+216 x^{5}-2160 x^{4}+11520 x^{3}-34560 x^{2}+55296 x -36864\right ) \textit {\_Z}^{5}+\left (15 x^{6}-360 x^{5}+3600 x^{4}-19200 x^{3}+57600 x^{2}-92160 x +61440\right ) \textit {\_Z}^{4}+\left (-10 x^{6}+240 x^{5}-2400 x^{4}+12800 x^{3}-38400 x^{2}+61440 x -40960\right ) \textit {\_Z}^{3}+\left (3 x^{6}-72 x^{5}+720 x^{4}-3840 x^{3}+11520 x^{2}-18432 x +12288\right ) \textit {\_Z} -x^{6}+24 x^{5}-240 x^{4}+1280 x^{3}-2 c_{4} {\mathrm e}^{3 c_{3}}-3840 x^{2}+6144 x -4096\right )-1 \\ \end{align*}
Verification of solutions
\[ y = \left (x -4\right ) \operatorname {RootOf}\left (\left (2 x^{6}-48 x^{5}+480 x^{4}-2560 x^{3}+7680 x^{2}-12288 x +8192\right ) \textit {\_Z}^{6}+\left (-9 x^{6}+216 x^{5}-2160 x^{4}+11520 x^{3}-34560 x^{2}+55296 x -36864\right ) \textit {\_Z}^{5}+\left (15 x^{6}-360 x^{5}+3600 x^{4}-19200 x^{3}+57600 x^{2}-92160 x +61440\right ) \textit {\_Z}^{4}+\left (-10 x^{6}+240 x^{5}-2400 x^{4}+12800 x^{3}-38400 x^{2}+61440 x -40960\right ) \textit {\_Z}^{3}+\left (3 x^{6}-72 x^{5}+720 x^{4}-3840 x^{3}+11520 x^{2}-18432 x +12288\right ) \textit {\_Z} -x^{6}+24 x^{5}-240 x^{4}+1280 x^{3}-2 c_{4} {\mathrm e}^{3 c_{3}}-3840 x^{2}+6144 x -4096\right )-1 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (4 y+x \right ) y^{\prime }-3 y=2 x -5 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 x +3 y-5}{4 y+x} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying homogeneous C trying homogeneous types: trying homogeneous D <- homogeneous successful <- homogeneous successful`
✓ Solution by Maple
Time used: 0.437 (sec). Leaf size: 186
dsolve((x+4*y(x))*diff(y(x),x)=2*x+3*y(x)-5,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (x -5\right ) \operatorname {RootOf}\left (\textit {\_Z}^{36}+\left (3 c_{1} x^{6}-72 c_{1} x^{5}+720 c_{1} x^{4}-3840 c_{1} x^{3}+11520 c_{1} x^{2}-18432 c_{1} x +12288 c_{1} \right ) \textit {\_Z}^{6}-2 c_{1} x^{6}+48 c_{1} x^{5}-480 c_{1} x^{4}+2560 c_{1} x^{3}-7680 c_{1} x^{2}+12288 c_{1} x -8192 c_{1} \right )^{6}-x +4}{\operatorname {RootOf}\left (\textit {\_Z}^{36}+\left (3 c_{1} x^{6}-72 c_{1} x^{5}+720 c_{1} x^{4}-3840 c_{1} x^{3}+11520 c_{1} x^{2}-18432 c_{1} x +12288 c_{1} \right ) \textit {\_Z}^{6}-2 c_{1} x^{6}+48 c_{1} x^{5}-480 c_{1} x^{4}+2560 c_{1} x^{3}-7680 c_{1} x^{2}+12288 c_{1} x -8192 c_{1} \right )^{6}} \]
✓ Solution by Mathematica
Time used: 60.076 (sec). Leaf size: 805
DSolve[(x+4*y[x])*y'[x]==2*x+3*y[x]-5,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {x}{4}+\frac {1}{4 \text {Root}\left [\text {$\#$1}^6 \left (-3125 x^6+75000 x^5-750000 x^4+4000000 x^3-12000000 x^2+19200000 x-12800000+3125 e^{\frac {15 c_1}{8}}\right )+\text {$\#$1}^4 \left (1875 x^4-30000 x^3+180000 x^2-480000 x+480000\right )+\text {$\#$1}^3 \left (-1000 x^3+12000 x^2-48000 x+64000\right )+\text {$\#$1}^2 \left (225 x^2-1800 x+3600\right )+\text {$\#$1} (96-24 x)+1\&,1\right ]} \\ y(x)\to -\frac {x}{4}+\frac {1}{4 \text {Root}\left [\text {$\#$1}^6 \left (-3125 x^6+75000 x^5-750000 x^4+4000000 x^3-12000000 x^2+19200000 x-12800000+3125 e^{\frac {15 c_1}{8}}\right )+\text {$\#$1}^4 \left (1875 x^4-30000 x^3+180000 x^2-480000 x+480000\right )+\text {$\#$1}^3 \left (-1000 x^3+12000 x^2-48000 x+64000\right )+\text {$\#$1}^2 \left (225 x^2-1800 x+3600\right )+\text {$\#$1} (96-24 x)+1\&,2\right ]} \\ y(x)\to -\frac {x}{4}+\frac {1}{4 \text {Root}\left [\text {$\#$1}^6 \left (-3125 x^6+75000 x^5-750000 x^4+4000000 x^3-12000000 x^2+19200000 x-12800000+3125 e^{\frac {15 c_1}{8}}\right )+\text {$\#$1}^4 \left (1875 x^4-30000 x^3+180000 x^2-480000 x+480000\right )+\text {$\#$1}^3 \left (-1000 x^3+12000 x^2-48000 x+64000\right )+\text {$\#$1}^2 \left (225 x^2-1800 x+3600\right )+\text {$\#$1} (96-24 x)+1\&,3\right ]} \\ y(x)\to -\frac {x}{4}+\frac {1}{4 \text {Root}\left [\text {$\#$1}^6 \left (-3125 x^6+75000 x^5-750000 x^4+4000000 x^3-12000000 x^2+19200000 x-12800000+3125 e^{\frac {15 c_1}{8}}\right )+\text {$\#$1}^4 \left (1875 x^4-30000 x^3+180000 x^2-480000 x+480000\right )+\text {$\#$1}^3 \left (-1000 x^3+12000 x^2-48000 x+64000\right )+\text {$\#$1}^2 \left (225 x^2-1800 x+3600\right )+\text {$\#$1} (96-24 x)+1\&,4\right ]} \\ y(x)\to -\frac {x}{4}+\frac {1}{4 \text {Root}\left [\text {$\#$1}^6 \left (-3125 x^6+75000 x^5-750000 x^4+4000000 x^3-12000000 x^2+19200000 x-12800000+3125 e^{\frac {15 c_1}{8}}\right )+\text {$\#$1}^4 \left (1875 x^4-30000 x^3+180000 x^2-480000 x+480000\right )+\text {$\#$1}^3 \left (-1000 x^3+12000 x^2-48000 x+64000\right )+\text {$\#$1}^2 \left (225 x^2-1800 x+3600\right )+\text {$\#$1} (96-24 x)+1\&,5\right ]} \\ y(x)\to -\frac {x}{4}+\frac {1}{4 \text {Root}\left [\text {$\#$1}^6 \left (-3125 x^6+75000 x^5-750000 x^4+4000000 x^3-12000000 x^2+19200000 x-12800000+3125 e^{\frac {15 c_1}{8}}\right )+\text {$\#$1}^4 \left (1875 x^4-30000 x^3+180000 x^2-480000 x+480000\right )+\text {$\#$1}^3 \left (-1000 x^3+12000 x^2-48000 x+64000\right )+\text {$\#$1}^2 \left (225 x^2-1800 x+3600\right )+\text {$\#$1} (96-24 x)+1\&,6\right ]} \\ \end{align*}