Internal problem ID [9244]
Internal file name [OUTPUT/8180_Monday_June_06_2022_02_03_47_AM_3157412/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 911.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "exactByInspection"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)*y+H(x)]`]]
\[ \boxed {y^{\prime }+\left (-\frac {\ln \left (y\right )}{x}+\frac {\cos \left (x \right ) \ln \left (y\right )}{\sin \left (x \right )}-f_{1} \left (x \right )\right ) y=0} \]
Entering Exact first order ODE solver. (Form one type)
To solve an ode of the form\begin {equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A} \end {equation} We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \] Hence\begin {equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B} \end {equation} Comparing (A,B) shows that\begin {align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end {align*}
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\] If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is \[ M(x,y) \mathop {\mathrm {d}x}+ N(x,y) \mathop {\mathrm {d}y}=0 \tag {1A} \] Therefore \begin {align*} \left (x \sin \left (x \right )\right )\mathop {\mathrm {d}y} &= \left (y \left (\ln \left (y \right ) \sin \left (x \right )-\cos \left (x \right ) \ln \left (y \right ) x +f_{1} \left (x \right ) x \sin \left (x \right )\right )\right )\mathop {\mathrm {d}x}\\ \left (-y \left (\ln \left (y \right ) \sin \left (x \right )-\cos \left (x \right ) \ln \left (y \right ) x +f_{1} \left (x \right ) x \sin \left (x \right )\right )\right )\mathop {\mathrm {d}x} + \left (x \sin \left (x \right )\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}
Comparing (1A) and (2A) shows that \begin {align*} M(x,y) &= -y \left (\ln \left (y \right ) \sin \left (x \right )-\cos \left (x \right ) \ln \left (y \right ) x +f_{1} \left (x \right ) x \sin \left (x \right )\right )\\ N(x,y) &= x \sin \left (x \right ) \end {align*}
The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied \[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x} \] Using result found above gives \begin {align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (-y \left (\ln \left (y \right ) \sin \left (x \right )-\cos \left (x \right ) \ln \left (y \right ) x +f_{1} \left (x \right ) x \sin \left (x \right )\right )\right )\\ &= \left (-f_{1} \left (x \right ) x -\ln \left (y \right )-1\right ) \sin \left (x \right )+x \left (\ln \left (y \right )+1\right ) \cos \left (x \right ) \end {align*}
And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (x \sin \left (x \right )\right )\\ &= \sin \left (x \right )+\cos \left (x \right ) x \end {align*}
Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\), then the ODE is not exact. By inspection \(\frac {1}{x^{2} y}\) is an integrating factor. Therefore by multiplying \(M=-\left (\ln \left (y\right ) \sin \left (x \right )-\cos \left (x \right ) \ln \left (y\right ) x +f_{1} \left (x \right ) x \sin \left (x \right )\right ) y\) and \(N=x \sin \left (x \right )\) by this integrating factor the ode becomes exact. The new \(M,N\) are \begin{align*} M&=-\frac {\ln \left (y\right ) \sin \left (x \right )-\cos \left (x \right ) \ln \left (y\right ) x +f_{1} \left (x \right ) x \sin \left (x \right )}{x^{2}} \\ N&=\frac {\sin \left (x \right )}{x y} \\ \end{align*}
To solve an ode of the form\begin {equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A} \end {equation} We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \] Hence\begin {equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B} \end {equation} Comparing (A,B) shows that\begin {align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end {align*}
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\] If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is \[ M(x,y) \mathop {\mathrm {d}x}+ N(x,y) \mathop {\mathrm {d}y}=0 \tag {1A} \] Therefore \begin {align*} \left (\frac {\sin \left (x \right )}{y x}\right )\mathop {\mathrm {d}y} &= \left (\frac {\ln \left (y \right ) \sin \left (x \right )-\cos \left (x \right ) \ln \left (y \right ) x +f_{1} \left (x \right ) x \sin \left (x \right )}{x^{2}}\right )\mathop {\mathrm {d}x}\\ \left (-\frac {\ln \left (y \right ) \sin \left (x \right )-\cos \left (x \right ) \ln \left (y \right ) x +f_{1} \left (x \right ) x \sin \left (x \right )}{x^{2}}\right )\mathop {\mathrm {d}x} + \left (\frac {\sin \left (x \right )}{y x}\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}
Comparing (1A) and (2A) shows that \begin {align*} M(x,y) &= -\frac {\ln \left (y \right ) \sin \left (x \right )-\cos \left (x \right ) \ln \left (y \right ) x +f_{1} \left (x \right ) x \sin \left (x \right )}{x^{2}}\\ N(x,y) &= \frac {\sin \left (x \right )}{y x} \end {align*}
The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied \[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x} \] Using result found above gives \begin {align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (-\frac {\ln \left (y \right ) \sin \left (x \right )-\cos \left (x \right ) \ln \left (y \right ) x +f_{1} \left (x \right ) x \sin \left (x \right )}{x^{2}}\right )\\ &= \frac {-\sin \left (x \right )+\cos \left (x \right ) x}{x^{2} y} \end {align*}
And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (\frac {\sin \left (x \right )}{y x}\right )\\ &= \frac {-\sin \left (x \right )+\cos \left (x \right ) x}{x^{2} y} \end {align*}
Since \(\frac {\partial M}{\partial y}= \frac {\partial N}{\partial x}\), then the ODE is exact The following equations are now set up to solve for the function \(\phi \left (x,y\right )\) \begin {align*} \frac {\partial \phi }{\partial x } &= M\tag {1} \\ \frac {\partial \phi }{\partial y } &= N\tag {2} \end {align*}
Integrating (1) w.r.t. \(x\) gives \begin{align*} \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int M\mathop {\mathrm {d}x} \\ \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int -\frac {\ln \left (y \right ) \sin \left (x \right )-\cos \left (x \right ) \ln \left (y \right ) x +f_{1} \left (x \right ) x \sin \left (x \right )}{x^{2}}\mathop {\mathrm {d}x} \\ \tag{3} \phi &= \int _{}^{x}-\frac {\ln \left (y \right ) \sin \left (\textit {\_a} \right )-\cos \left (\textit {\_a} \right ) \ln \left (y \right ) \textit {\_a} +f_{1} \left (\textit {\_a} \right ) \textit {\_a} \sin \left (\textit {\_a} \right )}{\textit {\_a}^{2}}d \textit {\_a}+ f(y) \\ \end{align*} Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(x\) and \(y\). Taking derivative of equation (3) w.r.t \(y\) gives \begin{align*} \tag{4} \frac {\partial \phi }{\partial y} &= -\frac {-\frac {\sin \left (x \right )}{x}+\operatorname {Ci}\left (x \right )}{y}+\frac {\operatorname {Ci}\left (x \right )}{y}+f'(y) \\ &=\frac {\sin \left (x \right )}{y x}+f'(y) \\ \end{align*} But equation (2) says that \(\frac {\partial \phi }{\partial y} = \frac {\sin \left (x \right )}{y x}\). Therefore equation (4) becomes \begin{equation} \tag{5} \frac {\sin \left (x \right )}{y x} = \frac {\sin \left (x \right )}{y x}+f'(y) \end{equation} Solving equation (5) for \( f'(y)\) gives \[ f'(y) = 0 \] Therefore \[ f(y) = c_{1} \] Where \(c_{1}\) is constant of integration. Substituting this result for \(f(y)\) into equation (3) gives \(\phi \) \[ \phi = \int _{}^{x}-\frac {\ln \left (y \right ) \sin \left (\textit {\_a} \right )-\cos \left (\textit {\_a} \right ) \ln \left (y \right ) \textit {\_a} +f_{1} \left (\textit {\_a} \right ) \textit {\_a} \sin \left (\textit {\_a} \right )}{\textit {\_a}^{2}}d \textit {\_a}+ c_{1} \] But since \(\phi \) itself is a constant function, then let \(\phi =c_{2}\) where \(c_{2}\) is new constant and combining \(c_{1}\) and \(c_{2}\) constants into new constant \(c_{1}\) gives the solution as \[ c_{1} = \int _{}^{x}-\frac {\ln \left (y \right ) \sin \left (\textit {\_a} \right )-\cos \left (\textit {\_a} \right ) \ln \left (y \right ) \textit {\_a} +f_{1} \left (\textit {\_a} \right ) \textit {\_a} \sin \left (\textit {\_a} \right )}{\textit {\_a}^{2}}d \textit {\_a} \]
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{x}-\frac {\ln \left (y\right ) \sin \left (\textit {\_a} \right )-\cos \left (\textit {\_a} \right ) \ln \left (y\right ) \textit {\_a} +f_{1} \left (\textit {\_a} \right ) \textit {\_a} \sin \left (\textit {\_a} \right )}{\textit {\_a}^{2}}d \textit {\_a} &= c_{1} \\ \end{align*}
Verification of solutions
\[ \int _{}^{x}-\frac {\ln \left (y\right ) \sin \left (\textit {\_a} \right )-\cos \left (\textit {\_a} \right ) \ln \left (y\right ) \textit {\_a} +f_{1} \left (\textit {\_a} \right ) \textit {\_a} \sin \left (\textit {\_a} \right )}{\textit {\_a}^{2}}d \textit {\_a} = c_{1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+\left (-\frac {\ln \left (y\right )}{x}+\frac {\cos \left (x \right ) \ln \left (y\right )}{\sin \left (x \right )}-f_{1} \left (x \right )\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\left (-\frac {\ln \left (y\right )}{x}+\frac {\cos \left (x \right ) \ln \left (y\right )}{\sin \left (x \right )}-f_{1} \left (x \right )\right ) y \end {array} \]
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 Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying inverse_Riccati trying an equivalence to an Abel ODE differential order: 1; trying a linearization to 2nd order --- trying a change of variables {x -> y(x), y(x) -> x} differential order: 1; trying a linearization to 2nd order trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 3 `, `-> Computing symmetries using: way = 4 `, `-> Computing symmetries using: way = 5 trying symmetry patterns for 1st order ODEs -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] <- symmetry pattern of the form [0, F(x)*G(y)] successful`
✓ Solution by Maple
Time used: 0.218 (sec). Leaf size: 21
dsolve(diff(y(x),x) = -(-1/x*ln(y(x))+1/sin(x)*cos(x)*ln(y(x))-_F1(x))*y(x),y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{\csc \left (x \right ) x \left (c_{1} +\int \frac {f_{1} \left (x \right ) \sin \left (x \right )}{x}d x \right )} \]
✓ Solution by Mathematica
Time used: 0.838 (sec). Leaf size: 105
DSolve[y'[x] == (F1[x] + Log[y[x]]/x - Cot[x]*Log[y[x]])*y[x],y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^x\left (\frac {2 \log (y(x)) \sin (K[1])}{K[1]^2}+\frac {2 (\text {F1}(K[1]) \sin (K[1])-\cos (K[1]) \log (y(x)))}{K[1]}\right )dK[1]+\int _1^{y(x)}\left (-\frac {2 \sin (x)}{x K[2]}-\int _1^x\left (\frac {2 \sin (K[1])}{K[1]^2 K[2]}-\frac {2 \cos (K[1])}{K[1] K[2]}\right )dK[1]\right )dK[2]=c_1,y(x)\right ] \]