Lecture 4

     First Order and first-degree differential equation

    A differential equation of the form  $M + N~\frac{dy}{dx}= 0$   or  $M~dx + N~dy = 0$   is called first-order and first-degree differential equation where both  $M$ and  $N$ are functions of x and y or constants.

    Types of differential equation of first order and first degree:

            (1) Separation of variables

            (2) Homogeneous equation 

            (3) Equation reducible  to homogeneous

            (4) Exact equation

            (5) Linear equation

            (6) Reducible to linear equation.

    Variable Separable:

If the differential equation    $M~dx + N~dy = 0$ can be expressed in the form  $f(x)~dx + \phi(y)~dy$ - then it is called variables are separable.

    

    Rule 1: If the differential equation is of the form  $f(x)~dx + f(y)~dy = 0$ ie function of $x$ with $dx$ and function of $y$ with $dy$, then only integrate we will find the solution

    ie  $$\begin{align*} & f(x)~dx + f(y) ~dy = 0 \\ & \int f(x)~dx + \int f(y)~dy = 0 \\ & F(x) + F(y) = c\\ & \textrm{when,} \int f(x)~dx = F(x) \end{align*}$$  

   

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    Question 1: Solve,  $$\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$$  

    Solution: Given equation is  $$\begin{align*} & \frac{dy}{dx} + \frac{y^2 + y + 1 }{x^2 + x + 1} = 0 \\ \implies & \frac{dy}{dx} = - \frac{y^2 + y + 1}{x^2 + x + 1} \\ \implies & \frac{dy}{y^2 + y + 1 } = - \frac{dx}{x^2 + x + 1} \\ \implies & \frac{dy }{y^2 + y + 1 } + \frac{dx}{x^2 + x + 1}=0 \\ \end{align*}$$  

Integrating both sides.  $$\begin{align*} & \int \frac{dy }{y^2 + y + 1 }+ \int \frac{dx}{x^2 + x + 1} = 0 \\ \implies & \int \frac{dy}{\left(y + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} + \int \frac{dx}{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = 0 \\ \implies & \frac{1}{\frac{\sqrt 3}{2}} \tan^{-1} \left(\frac{y + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + \frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} \left(\frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = c \\ \implies & \frac{2}{\sqrt{3}} \tan^{- 1} \left(\frac{2y + 1}{\sqrt{3}}\right) + \frac{2}{\sqrt{3}} \tan^{- 1} \left(\frac{2x + 1}{\sqrt{3}}\right) = c \end{align*}$$  

    Rule 2: If any differential equation of the form  $f(x) ~ \phi (y)~dx + f(y)~\phi (y)~dy = 0$   then divide the equation by  $\phi (x)~\phi (y)$ . the equation of the form  $\frac{f(x)}{\phi (x)}~dx + \frac{f(y)}{\phi (y)}~ dy = 0$   ie function of $x$ with $dx$ and function of $y$ with $dy$. Then only integrate we can find the solution.

   

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    Question 2: solve,  $$x \sqrt{1 - y^2}~dx + y \sqrt{1 - x^2 }~ dy = 0$$  

    Solution: Given equation is  $$x \sqrt{1 - y^2}~dx + y \sqrt{1 - x^2 }~ dy = 0$$  

Dividing both sides by  $\sqrt{1 - x^2}~ \sqrt{1 - y^2}$  

  $$\begin{align*} & \frac{X ~DX }{\sqrt{1 - x^2}}~dx + \frac{y~dy}{\sqrt{1 - y^2}} = 0 \\ \implies & \frac{- 1}{2} \int \frac{- 2x ~dx }{\sqrt{1 - x^2}} - \frac{1}{2}\int \frac{- 2y~dy}{\sqrt{1 - y^2}} = 0 \\ \implies & - \frac{1}{2} ~2 \sqrt{1 - x^2} - \frac{1}{2}~ 2\sqrt{1 - y^2} = - c ~~~~~\left[\textrm{Since} ~~ \frac{f'(x)~dx}{\sqrt{f(x)}} = 2\sqrt{f(x)}\right]\\ \implies & \sqrt{1 - x^2} + \sqrt{1 - y^2} = c \end{align*}$$  

   

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    Question 3: Solve,  $\sec^2 x~ \tan y ~dx + \sec^2 y ~ \tan x ~dy = 0$  

    Solution: Given equation is,  $$\sec^2 x~ \tan y ~dx + \sec^2 y ~ \tan x ~dy = 0$$  

Dividing both sides by  $\tan x~.\tan y$   we get,

  $$\frac{\sec^2 x~dx }{\tan x} + \frac{\sec^2 y~ dy}{\tan y} = 0$$  

Now integrating both sides,  $$\begin{align*} & \int \frac{\sec^2 x~dx }{\tan x} + \int \frac{\sec^2 y~ dy}{\tan y} = 0\\ \implies & \ln (\tan x) + \ln (\tan y) = \ln c ~~~~~\left[\textrm{Since}~~ \int \frac{f'(x)}{f(x)}~dx = \ln(f(x))\right]\\ \implies & \ln (\tan x~. \tan y) = \ln c \\ \implies & \tan x~.\tan y = c \end{align*}$$  

    Rule 3: If the differential equation is of the form  $\frac{dy}{dx} = f(ax + by + c)$   then put  $ax + by + c = z$   and simplify. Finally integrating we will get the solution.

   

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    Question 4: Solve,  $$\sin^{- 1} \left(\frac{dy}{dx}\right) = x + y$$      

    Solution: Given equation is  $$\begin{align*} & \sin^{- 1} \left(\frac{dy}{dx}\right) = x + y \\ \implies & \frac{dy}{dx} = \sin (x + y) \tag{4.1}\label{eq:4.1} \end{align*}$$  

We put  $x + y = z$  

Then  $$\begin{align*} & 1 + \frac{dy}{dx} = \frac{dz}{dx} \\ \implies & \frac{dy}{dx} = \frac{dz}{dx} - 1 \tag{4.2}\label{eq:4.2}\\ \end{align*}$$  

From (4.1) and (4.2) we get,  $$\begin{align*} & \frac{dz}{dx} - 1 = \sin z \\ \implies & \frac{dz}{dx} = 1 + \sin z \\ \implies & \frac{dz}{1 + \sin z} = dx \\ \end{align*}$$  

Integrating both sides,  $$\begin{align*} & \int \frac{dz}{1 + \sin z} = \int dx \\ \implies & \int \frac{1 - \sin z}{1 - \sin^2 z}~dz = \int dx \\ \implies & \int \frac{1 - \sin z}{\cos^2 z}~dz = \int dx \\ \implies & \int \sec^2 z ~dz ~ \int \sec z ~.\tan z ~dz = \int dx \\ \implies & \tan z - \sec z = x + c \\ \implies & \tan (x + y ) - \sec (x + y ) = x + c \end{align*}$$  

   

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    Question 5: Solve,  $$\frac{dy}{dx}= (4x + y + 1)^2$$  

    Solution: Given equation is  $$\frac{dy}{dx} = (4x + y + 1)^2 \tag{5.1}\label{eq:5.1}$$  

we put  $4x + y + 1 = z$  

then  $$\begin{align*} & 4 + \frac{dy}{dx} = \frac{dz}{dx} \\ \implies & \frac{dy}{dx}= \frac{dz}{dx} - 4 \tag{5.2}\label{eq:5.2} \end{align*}$$  

From (5.1) and (5.2) we get  $$\begin{align*} & \frac{dz}{dx}- 4 = z^2 \\ \implies & \frac{dz}{dx} = 4 + z^2 \\ \implies & \frac{dz}{4 + z^2}= dx \end{align*}$$  

Integrating both sides we get,  $$\begin{align*} & \frac{1}{2} \tan^{- 1} \frac{z}{2} = x + \frac{c}{2}\\ \implies & \tan^{- 1} \frac{z}{2} = 2x + c \\ \implies & ~~~~~~~~\frac{z}{2}= \tan (2x + c) \\ \implies & 4x + y + 1 = 2 \tan (2x + c) \end{align*}$$  

   

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    Question 6: Solve,  $$(e^x + 1) y ~dy = (y + 1 ) e^x ~dx$$  

    Solution: The given equation is  $$\begin{align*} & (e^x + 1)y ~dy = (y + 1 ) e^x ~dx \\ \implies & \frac{y~dy}{y + 1} = \frac{e^x ~ dx}{e^x + 1} \\ \implies & \frac{(y + 1 - 1)~dy}{y + 1} = \frac{e^x ~dx}{e^x + 1}\\ \implies & \left(1 - \frac{1}{y + 1}\right)~dy = \frac{d(e^x + 1)}{e^x + 1}\\ \therefore & dy - \frac{dy}{y + 1} = \frac{d*e^x + 1}{e^x + 1} \end{align*}$$  

Where the variables have been separated, so the general solution will be obtained by integrating on both sides and adding a constant.

    $$\begin{align*} & \int dy - \int \frac{dy }{y + 1} = \int \frac{d(e^x + 1)}{e^x + 1}\\ & y - \ln (y + 1 ) = \ln (e^x + 1) + \ln c ~~~~~\textrm{[where} \ln c ~\textrm{is the integrating constant]}\\ & y = \ln (y + 1) + \ln (e^x + 1) + \ln c\\ & y = \ln (y + 1) (e^x + 1).c \\ & e^y = (y + 1)(e^x + 1)c \end{align*}$$  

Which is the required solution.

   

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     Question 7: Solve,  $$y (1 + xy ) ~dx + x (1 - xy ) ~dy = 0$$  

    Solution: The given equation is,  $$\begin{align*} & y ( 1 + xy )~dx + x (1 - xy)~dy = 0 \\ \implies & y~dx + x y^2~dx + x ~dy - x^2 y ~dy = 0 \\ \implies & x ~dy + y ~dx + xy (y~dx - x~dy) = 0 \\ \implies & d(xy) + xy ( y~dx - x~dy) = 0 \\ \implies & \frac{d(xy)}{(xy)^2} + \frac{dx}{x} - \frac{dy }{y}= 0 ~~~ \textrm{[Dividing}~ \left(xy\right)^2 ~\textrm{on both sides]} \end{align*}$$  

Here the variables have been separated, so the general solution will be obtained by integrating on both sides and adding a constant 

  $$\begin{align*} & \int \frac{d(xy)}{(xy)^2} + \int \frac{dx}{x} - \int \frac{dy}{y}= 0 \\ \implies & - \frac{1}{xy} + \ln x - \ln y = c \\ \implies & \ln x - \ln y - \frac{1}{xy} = c \\ \implies & \ln \frac{x}{y} - \frac{1}{xy} = c \\ \end{align*}$$  

Which is the required solution.