2.4.7 Problem 7

Internal problem ID [18487]
Book : Elementary Differential Equations. By Thornton C. Fry. D Van Nostrand. NY. First Edition (1929)
Section : Chapter IV. Methods of solution: First order equations. section 31. Problems at page 85
Problem number : 7
Date solved : Monday, March 31, 2025 at 05:37:15 PM
CAS classification : [_Bernoulli]

Solved using first_order_ode_bernoulli

Time used: 0.246 (sec)

Solve

$$\begin{array}{rl}y-\cos \left(x\right)y^{\prime }& =y^2\cos \left(x\right)(1-\sin \left(x\right))\end{array}$$

In canonical form, the ODE is

$$\begin{array}{rl}{y}^{\prime }& =F(x,y)\\ & =\frac{y(y\cos \left(x\right)\sin \left(x\right)-y\cos \left(x\right)+1)}{\cos \left(x\right)}\end{array}$$

This is a Bernoulli ODE.

$$\begin{array}{}\text{(1)}& y^{\prime }=\left(\frac{1}{\cos \left(x\right)}\right)y+\left(\frac{\cos \left(x\right)\sin \left(x\right)-\cos \left(x\right)}{\cos \left(x\right)}\right)y^2\end{array}$$

The standard Bernoulli ODE has the form

$$\begin{array}{}\text{(2)}& y^{\prime }=f_0(x)y+f_1(x)y^n\end{array}$$

Comparing this to (1) shows that

$$\begin{array}{rl}{f}_0& =\frac{1}{\cos \left(x\right)}\\ f_1& =\frac{\cos \left(x\right)\sin \left(x\right)-\cos \left(x\right)}{\cos \left(x\right)}\end{array}$$

The first step is to divide the above equation by $y^n$ which gives

$$\begin{array}{}\text{(3)}& \frac{y^{\prime }}{y^n}=f_0(x)y^{1-n}+f_1(x)\end{array}$$

The next step is use the substitution $v=y^{1-n}$ in equation (3) which generates a new ODE in $v\left(x\right)$ which will be linear and can be easily solved using an integrating factor. Backsubstitution then gives the solution $y(x)$ which is what we want.

This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows that

$$\begin{array}{rl}{f}_0(x)& =\frac{1}{\cos \left(x\right)}\\ f_1(x)& =\frac{\cos \left(x\right)\sin \left(x\right)-\cos \left(x\right)}{\cos \left(x\right)}\\ n& =2\end{array}$$

Dividing both sides of ODE (1) by $y^n=y^2$ gives

$$\begin{array}{}\text{(4)}& y^{\prime }\frac{1}{y^2}& =\frac{1}{\cos \left(x\right)y}+\frac{\cos \left(x\right)\sin \left(x\right)-\cos \left(x\right)}{\cos \left(x\right)}\end{array}$$

Let

$$\begin{array}{rl}v& =y^{1-n}\\ \text{(5)}& & =\frac{1}{y}\end{array}$$

Taking derivative of equation (5) w.r.t $x$ gives

$$\begin{array}{}\text{(6)}& v^{\prime }& =-\frac{1}{y^2}{y}^{\prime }\end{array}$$

Substituting equations (5) and (6) into equation (4) gives

$$\begin{array}{rl}-v^{\prime }\left(x\right)& =\frac{v\left(x\right)}{\cos \left(x\right)}+\frac{\cos \left(x\right)\sin \left(x\right)-\cos \left(x\right)}{\cos \left(x\right)}\\ \text{(7)}& v^{\prime }& =-\frac{v}{\cos \left(x\right)}-\frac{\cos \left(x\right)\sin \left(x\right)-\cos \left(x\right)}{\cos \left(x\right)}\end{array}$$

The above now is a linear ODE in $v\left(x\right)$ which is now solved.

In canonical form a linear first order is

$$\begin{array}{rl}{v}^{\prime }\left(x\right)+q(x)v\left(x\right)& =p(x)\end{array}$$

Comparing the above to the given ode shows that

$$\begin{array}{rl}q(x)& =\sec \left(x\right)\\ p(x)& =1-\sin \left(x\right)\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdx}\\ & ={\mathrm{e}}^{\int \sec \left(x\right)dx}\\ & =\sec \left(x\right)+\tan \left(x\right)\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\left(\mu v\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(\mu v\right)& =\left(\mu \right)(1-\sin \left(x\right))\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(v(\sec \left(x\right)+\tan \left(x\right))\right)& =(\sec \left(x\right)+\tan \left(x\right))(1-\sin \left(x\right))\\ \mathrm{d}\left(v(\sec \left(x\right)+\tan \left(x\right))\right)& =\left((1-\sin \left(x\right))(\sec \left(x\right)+\tan \left(x\right))\right)\mathrm{d}x\end{array}$$

Integrating gives

$$\begin{array}{rl}v(\sec \left(x\right)+\tan \left(x\right))& =\int (1-\sin \left(x\right))(\sec \left(x\right)+\tan \left(x\right))dx\\ & =\sin \left(x\right)+c_1\end{array}$$

Dividing throughout by the integrating factor $\sec \left(x\right)+\tan \left(x\right)$ gives the final solution

$$v\left(x\right)=\frac{(\sin \left(x\right)+c_1)(\cos \left(x\right)-\sin \left(x\right)+1)}{\cos \left(x\right)+\sin \left(x\right)+1}$$

The substitution $v=y^{1-n}$ is now used to convert the above solution back to $y$ which results in

$$\frac{1}{y}=\frac{(\sin \left(x\right)+c_1)(\cos \left(x\right)-\sin \left(x\right)+1)}{\cos \left(x\right)+\sin \left(x\right)+1}$$

Solving for $y$ gives

$$\begin{array}{rl}y& =\frac{\cos \left(x\right)+\sin \left(x\right)+1}{\cos \left(x\right)\sin \left(x\right)+\cos \left(x\right)c_1-\sin {\left(x\right)}^2-c_1\sin \left(x\right)+\sin \left(x\right)+c_1}\end{array}$$

Which simplifies to

$$\begin{array}{rl}y& =\frac{\cos \left(x\right)+\sin \left(x\right)+1}{(\sin \left(x\right)+c_1)(\cos \left(x\right)-\sin \left(x\right)+1)}\end{array}$$

Figure 2.51: Slope field $y-\cos \left(x\right)y^{\prime }=y^2\cos \left(x\right)(1-\sin \left(x\right))$

Summary of solutions found

$$\begin{array}{rl}y& =\frac{\cos \left(x\right)+\sin \left(x\right)+1}{(\sin \left(x\right)+c_1)(\cos \left(x\right)-\sin \left(x\right)+1)}\end{array}$$

✓ Maple. Time used: 0.003 (sec). Leaf size: 27

ode:=y(x)-cos(x)*diff(y(x),x) = y(x)^2*cos(x)*(1-sin(x)); 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{\cos \left(x\right)+\sin \left(x\right)+1}{(\sin \left(x\right)+c_1)(-\sin \left(x\right)+\cos \left(x\right)+1)}$$

Maple trace

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

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & y\left(x\right)-\cos \left(x\right)\left(\frac{d}{dx}y\left(x\right)\right)=y{\left(x\right)}^2\cos \left(x\right)(1-\sin \left(x\right))\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}y\left(x\right)=-\frac{-y\left(x\right)+y{\left(x\right)}^2\cos \left(x\right)(1-\sin \left(x\right))}{\cos \left(x\right)}\end{array}$$

✓ Mathematica. Time used: 0.419 (sec). Leaf size: 41

ode=y[x]-Cos[x]*D[y[x],x]==y[x]^2*Cos[x]*(1-Sin[x]); 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$\begin{array}{r}y(x)\to \frac{e^{2\text{arctanh}(\tan \left(\frac{x}{2}\right))}}{\cos (x)e^{2\text{arctanh}(\tan \left(\frac{x}{2}\right))}+c_1}\\ y(x)\to 0\end{array}$$

✓ Sympy. Time used: 2.096 (sec). Leaf size: 39

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-(1 - sin(x))*y(x)**2*cos(x) + y(x) - cos(x)*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

$$y\left(x\right)=\frac{\sqrt{\sin \left(x\right)+1}}{(C_1-\int \sqrt{\sin \left(x\right)-1}\sqrt{\sin \left(x\right)+1}dx)\sqrt{\sin \left(x\right)-1}}$$