2.1.14 problem 14

Internal problem ID [8152]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 14
Date solved : Sunday, November 10, 2024 at 02:58:39 AM
CAS classification : [_separable]

Solve

\begin{align*} y^{\prime }&=\frac {\ln \left (y^{2}+1\right )}{\ln \left (x^{2}+1\right )} \end{align*}

Solved as first order separable ode

Time used: 0.271 (sec)

The ode \(y^{\prime } = \frac {\ln \left (y^{2}+1\right )}{\ln \left (x^{2}+1\right )}\) is separable as it can be written as

\begin{align*} y^{\prime }&= \frac {\ln \left (y^{2}+1\right )}{\ln \left (x^{2}+1\right )}\\ &= f(x) g(y) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{\ln \left (x^{2}+1\right )}\\ g(y) &= \ln \left (y^{2}+1\right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx}\\ \int { \frac {1}{\ln \left (y^{2}+1\right )}\,dy} &= \int { \frac {1}{\ln \left (x^{2}+1\right )} \,dx}\\ \int _{}^{y}\frac {1}{\ln \left (\tau ^{2}+1\right )}d \tau = \int \frac {1}{\ln \left (x^{2}+1\right )}d x +2 c_1 \end{align*}

We now need to find the singular solutions, these are found by finding for what values \(g(y)\) is zero, since we had to divide by this above. Solving \(g(y)=0\) or \(\ln \left (y^{2}+1\right )=0\) for \(y\) gives

\begin{align*} y&=0 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \int _{}^{y}\frac {1}{\ln \left (\tau ^{2}+1\right )}d \tau = \int \frac {1}{\ln \left (x^{2}+1\right )}d x +2 c_1\\ y = 0 \end{align*}

Figure 2.36: Slope field plot
\(y^{\prime } = \frac {\ln \left (y^{2}+1\right )}{\ln \left (x^{2}+1\right )}\)

Summary of solutions found

\begin{align*} \int _{}^{y}\frac {1}{\ln \left (\tau ^{2}+1\right )}d \tau &= \int \frac {1}{\ln \left (x^{2}+1\right )}d x +2 c_1 \\ y &= 0 \\ \end{align*}

Solved as first order Exact ode

Time used: 0.190 (sec)

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*} \mathop {\mathrm {d}y} &= \left (\frac {\ln \left (y^{2}+1\right )}{\ln \left (x^{2}+1\right )}\right )\mathop {\mathrm {d}x}\\ \left (-\frac {\ln \left (y^{2}+1\right )}{\ln \left (x^{2}+1\right )}\right ) \mathop {\mathrm {d}x} + \mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}

Comparing (1A) and (2A) shows that

\begin{align*} M(x,y) &= -\frac {\ln \left (y^{2}+1\right )}{\ln \left (x^{2}+1\right )}\\ N(x,y) &= 1 \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^{2}+1\right )}{\ln \left (x^{2}+1\right )}\right )\\ &= -\frac {2 y}{\left (y^{2}+1\right ) \ln \left (x^{2}+1\right )} \end{align*}

And

\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (1\right )\\ &= 0 \end{align*}

Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let

\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial y} - \frac {\partial N}{\partial x} \right ) \\ &=1\left ( \left ( -\frac {2 y}{\left (y^{2}+1\right ) \ln \left (x^{2}+1\right )}\right ) - \left (0 \right ) \right ) \\ &=-\frac {2 y}{\left (y^{2}+1\right ) \ln \left (x^{2}+1\right )} \end{align*}

Since \(A\) depends on \(y\), it can not be used to obtain an integrating factor. We will now try a second method to find an integrating factor. Let

\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} \right ) \\ &=-\frac {\ln \left (x^{2}+1\right )}{\ln \left (y^{2}+1\right )}\left ( \left ( 0\right ) - \left (-\frac {2 y}{\left (y^{2}+1\right ) \ln \left (x^{2}+1\right )} \right ) \right ) \\ &=-\frac {2 y}{\ln \left (y^{2}+1\right ) \left (y^{2}+1\right )} \end{align*}

Since \(B\) does not depend on \(x\), it can be used to obtain an integrating factor. Let the integrating factor be \(\mu \). Then

\begin{align*} \mu &= e^{\int B \mathop {\mathrm {d}y}} \\ &= e^{\int -\frac {2 y}{\ln \left (y^{2}+1\right ) \left (y^{2}+1\right )}\mathop {\mathrm {d}y} } \end{align*}

The result of integrating gives

\begin{align*} \mu &= e^{-\ln \left (\ln \left (y^{2}+1\right )\right ) } \\ &= \frac {1}{\ln \left (y^{2}+1\right )} \end{align*}

\(M\) and \(N\) are now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\).

\begin{align*} \overline {M} &=\mu M \\ &= \frac {1}{\ln \left (y^{2}+1\right )}\left (-\frac {\ln \left (y^{2}+1\right )}{\ln \left (x^{2}+1\right )}\right ) \\ &= -\frac {1}{\ln \left (x^{2}+1\right )} \end{align*}

And

\begin{align*} \overline {N} &=\mu N \\ &= \frac {1}{\ln \left (y^{2}+1\right )}\left (1\right ) \\ &= \frac {1}{\ln \left (y^{2}+1\right )} \end{align*}

So now a modified ODE is obtained from the original ODE which will be exact and can be solved using the standard method. The modified ODE is

\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \\ \left (-\frac {1}{\ln \left (x^{2}+1\right )}\right ) + \left (\frac {1}{\ln \left (y^{2}+1\right )}\right ) \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \end{align*}

The following equations are now set up to solve for the function \(\phi \left (x,y\right )\)

\begin{align*} \frac {\partial \phi }{\partial x } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial y } &= \overline {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 \overline {M}\mathop {\mathrm {d}x} \\ \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int -\frac {1}{\ln \left (x^{2}+1\right )}\mathop {\mathrm {d}x} \\ \tag{3} \phi &= \int -\frac {1}{\ln \left (x^{2}+1\right )}d x+ 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{equation} \tag{4} \frac {\partial \phi }{\partial y} = 0+f'(y) \end{equation}

But equation (2) says that \(\frac {\partial \phi }{\partial y} = \frac {1}{\ln \left (y^{2}+1\right )}\). Therefore equation (4) becomes

\begin{equation} \tag{5} \frac {1}{\ln \left (y^{2}+1\right )} = 0+f'(y) \end{equation}

Solving equation (5) for \( f'(y)\) gives

\[ f'(y) = \frac {1}{\ln \left (y^{2}+1\right )} \]

Integrating the above w.r.t \(y\) gives

\begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( \frac {1}{\ln \left (y^{2}+1\right )}\right ) \mathop {\mathrm {d}y} \\ f(y) &= \int _{}^{y}\frac {1}{\ln \left (\tau ^{2}+1\right )}d \tau + c_1 \\ \end{align*}

Where \(c_1\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \)

\[ \phi = \int -\frac {1}{\ln \left (x^{2}+1\right )}d x +\int _{}^{y}\frac {1}{\ln \left (\tau ^{2}+1\right )}d \tau + c_1 \]

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 the constant \(c_1\) gives the solution as

\[ c_1 = \int -\frac {1}{\ln \left (x^{2}+1\right )}d x +\int _{}^{y}\frac {1}{\ln \left (\tau ^{2}+1\right )}d \tau \]

Figure 2.37: Slope field plot
\(y^{\prime } = \frac {\ln \left (y^{2}+1\right )}{\ln \left (x^{2}+1\right )}\)

Summary of solutions found

\begin{align*} \int -\frac {1}{\ln \left (x^{2}+1\right )}d x +\int _{}^{y}\frac {1}{\ln \left (\tau ^{2}+1\right )}d \tau &= c_1 \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {\ln \left (1+y \left (x \right )^{2}\right )}{\ln \left (x^{2}+1\right )} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {\ln \left (1+y \left (x \right )^{2}\right )}{\ln \left (x^{2}+1\right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{\ln \left (1+y \left (x \right )^{2}\right )}=\frac {1}{\ln \left (x^{2}+1\right )} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{\ln \left (1+y \left (x \right )^{2}\right )}d x =\int \frac {1}{\ln \left (x^{2}+1\right )}d x +\mathit {C1} \\ \bullet & {} & \textrm {Cannot compute integral}\hspace {3pt} \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{\ln \left (1+y \left (x \right )^{2}\right )}d x =\int \frac {1}{\ln \left (x^{2}+1\right )}d x +\mathit {C1} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
<- separable successful`
 

Maple dsolve solution

Solving time : 0.009 (sec)
Leaf size : 30

dsolve(diff(y(x),x) = ln(1+y(x)^2)/ln(x^2+1), 
       y(x),singsol=all)
 

\[ \int \frac {1}{\ln \left (x^{2}+1\right )}d x -\left (\int _{}^{y}\frac {1}{\ln \left (\textit {\_a}^{2}+1\right )}d \textit {\_a} \right )+c_{1} = 0 \]

Mathematica DSolve solution

Solving time : 0.669 (sec)
Leaf size : 48

DSolve[{D[y[x],x] == Log[1+y[x]^2]/Log[1+x^2],{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{\log \left (K[1]^2+1\right )}dK[1]\&\right ]\left [\int _1^x\frac {1}{\log \left (K[2]^2+1\right )}dK[2]+c_1\right ] \\ y(x)\to 0 \\ \end{align*}