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*}
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.
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*}
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.
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*}
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*}
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.
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.
Post a Comment