0
$\begingroup$

$xy'-\cos \left(y\right)=x^2y'\cos \left(y\right)\left(1-\sin \left(y\right)\right)$

First I rewrite the equation:

$$xy'-x^2y'\cos \left(y\right)\left(1-\sin \left(y\right)\right)=\cos \left(y\right)$$

$$y'=\frac{\cos \left(y\right)}{x-x^2\cos \left(y\right)\left(1-\sin \left(y\right)\right)}$$

$$x'=\frac{x-x^2\cos \left(y\right)\left(1-\sin \left(y\right)\right)}{\cos \left(y\right)}$$

$$x'=\frac{x}{\cos \:\:\left(y\right)}-x^2\left(1-\sin \:\left(y\right)\right)$$

This is a Riccati equation, however I lack a given particular solution to substitute, any help?

  • 0
    can you find a particular solution to this problem? – 2017-01-26
  • 0
    x = 0? seems to be one – 2017-01-26

2 Answers 2

2

Writing the equation as Riccati as you have done $\displaystyle\frac{dx}{dy}=\frac{x}{\cos y}+x^2(\sin y-1).$

Next we use the substitution $v=x(\sin y-1)$. Differentiating wrt $y$ we have, $\frac{dv}{dy}=x\cos y+\frac{dx}{dy}(\sin y-1)$ and substituting back $\frac{dx}{dy}$ and $x=\frac{v}{\sin y-1}$ we have

\begin{array} $\displaystyle\frac{dv}{dy}&=&x\cos y+(\sin y-1)[\frac{x}{\cos y}+x^2(\sin y-1)]\\ &=&x\cos y+\frac{v}{\cos y}+v^2&\\ \end{array} Hence $\displaystyle\frac{dv}{dy}-v\bigg(\frac{\cos y}{\sin y-1}+\sec y\bigg)=v^2$ which is a Bernoulli equation of degree 2.

Next we use the substitution $v=\frac{1}{u}$ which gives $\frac{dv}{dy}=-\frac{1}{u^2}\frac{du}{dy}$.

Hence we have $$ -\frac{1}{u^2}\frac{du}{dy}-\frac{1}u\bigg(\frac{\cos y}{\sin y-1}+\sec y\bigg)=\frac{1}{u^2}$$ or,

$$\frac{du}{dy}+u\bigg(\frac{\cos y}{\sin y-1}+\sec y\bigg)=-1 ...... (1)$$

Here our integrating factor is \begin{array} $\displaystyle I_f&=&e^{\int\big(\frac{\cos y}{\sin y-1}+\sec y\big)dy}\\ &=&e^{\ln|\sin y-1|+\ln|\sec y+\tan y|}\\ &=&e^{\ln|(\sin y-1)\frac{\sin y+1}{\cos y}|}\\ &=&e^{\ln|\cos y|}\\ &=&\cos y \end{array}

Multiply (1) by the integrating factor $I_f=\cos y$ we get

$\cos y\frac{du}{dy}+u\bigg(\frac{\cos^2 y}{\sin y-1}+1\bigg)=-\cos y$

$\cos y\frac{du}{dy}-u\sin y=-\cos y$, which implies

$$\frac{d(u\cos y)}{dy}=-\cos y.$$ Integrating wrt $y$ we have

$u\cos y=-\sin y +c$ or $u=-\tan y+c\sec y$

But $\displaystyle v=\frac{1}{u}=\frac{1}{-\tan y+c\sec y}.$

Finally $v=x(\sin y-1)$ from which we get

$\displaystyle x(\sin y-1)=\frac{1}{-\tan y+c\sec y}=\frac{\cos y}{c-\sin y}.$

Lets check the answer by taking derivative. Differentiating wrt $x$ of $$\displaystyle c-\sin y=\frac{\cos y}{x(\sin y -1)}$$ we have

\begin{array} $\displaystyle -\cos y\frac{dy}{dx}&=&\displaystyle-\frac{1}{x^2}\frac{\cos y}{\sin y-1}+\frac{1}{x}\bigg[\frac{-\sin y(\sin y -1)-\cos^2 y}{(\sin y-1)^2}\frac{dy}{dx}\bigg]\\ &=&\displaystyle-\frac{1}{x^2}\frac{\cos y}{\sin y-1}+\frac{1}{x}\big[\frac{1}{\sin y-1}\big]\frac{dy}{dx}\\ &=&\displaystyle\frac{x\frac{dy}{dx}-\cos y}{-x^2(1-\sin y)} \end{array} from which the result follows.

  • 0
    Unless I made some mistake, but your final answer doesn't satisfy main equation. – 2017-01-26
  • 0
    I am correct. I am going to prove to you now by taking derivatives. Please see edited answer. I am adding the derivative. – – 2017-01-27
  • 0
    I see. Good Job. – 2017-01-27
  • 0
    As we are all learning by solving problems, please refer to the link https://en.wikipedia.org/wiki/Riccati_equation, to see how I got the substitution and was able to solve problem. Thanks and I am grateful to help you. – 2017-01-27
0

HINT, we have that:

$$x\cdot\text{y}'\left(x\right)-\cos\left(\text{y}\left(x\right)\right)=x^2\cdot\text{y}'\left(x\right)\cdot\cos \left(\text{y}\left(x\right)\right)\cdot\left(1-\sin\left(\text{y}\left(x\right)\right)\right)\tag1$$

Write the differential equation in terms of $x$, since:

$$\frac{\text{d}\space\text{y}}{\text{d}x}\cdot\frac{\text{d}x}{\text{d}\space\text{y}}=1\tag2$$

So, we get:

$$\frac{1}{\frac{\text{d}x\left(\text{y}\right)}{\text{d}\space\text{y}}}=\frac{\cos\left(\text{y}\right)}{x\left(\text{y}\right)\cdot\left(1+x\left(\text{y}\right)\cdot\sin\left(\text{y}\right)\cdot\cos\left(\text{y}\right)-\cos\left(\text{y}\right)\cdot x\left(\text{y}\right)\right)}\tag3$$

From here follows that:

$$\frac{\sec\left(\text{y}\right)}{x\left(\text{y}\right)}-\frac{\frac{\text{d}x\left(\text{y}\right)}{\text{d}\space\text{y}}}{x\left(\text{y}\right)^2}=1-\sin\left(\text{y}\right)\tag4$$

Let $\mathcal{V}\left(\text{y}\right)=\frac{1}{x\left(\text{y}\right)}$:

$$\frac{\text{d}\space\mathcal{V}\left(\text{y}\right)}{\text{d}\space\text{y}}+\sec\left(\text{y}\right)\cdot\mathcal{V}\left(\text{y}\right)=1-\sin\left(\text{y}\right)\tag5$$

Now, use:

$$\rho\left(\text{y}\right)=\exp\left\{\int\sec\left(\text{y}\right)\space\text{d}\space\text{y}\right\}=\exp\left(\ln\left|\sec\left(\text{y}\right)+\tan\left(\text{y}\right)\right|+\text{C}\right)\tag6$$